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Page No 474:

Question 1:

In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.

(i) sin A=23

(ii) cos A=45

(iii) tan θ = 11

(iv) cot θ=125

Answer:

(i) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = 2 and

Hypotenuse = 3

Therefore, by Pythagoras theorem,

Now we substitute the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)

Therefore,

Hence, Base =

Now,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(ii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 4 and

Hypotenuse = 5

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

Hence, Perpendicular side = 3

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(iii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 1 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)

Hence, Hypotenuse =

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(iv) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 12 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)

Hence, Hypotenuse = 13

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(x) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 5 and

Hypotenuse = 13

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)


Hence, Perpendicular side = 12

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(xi) Given:

…… (1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = 1 and

Hypotenuse =

Therefore,

By Pythagoras theorem,

Now we substitute the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)

Hence, Base side = 3

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

 

(xii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 12 and

Hypotenuse = 15

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

Hence, Perpendicular side = 9

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Page No 474:

Question 2:

In a ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine

(i) sin A, cos A
(ii) sin C, cos C

Answer:

(i) The given triangle is below:-

Given: In ,

To Find:

In this problem, Hypotenuse side is unknown

Hence we first find Hypotenuse side by Pythagoras theorem

By Pythagoras theorem,

We get,

By definition,

By definition,

Answer:

(ii) The given triangle is below:

Given: In ΔABC,

To Find:

In this problem, Hypotenuse side is unknown

Hence we first find Hypotenuse side by Pythagoras theorem

By Pythagoras theorem,

We get,

By definition,

By definition,

Answer:

Page No 474:

Question 3:

Given 15 cot A = 8, find sin A and sec A.

Answer:

Given: 15 = 8

To Find:

Since

By taking 15 on R.H.S

We get,

By definition,

Hence,

Comparing equation (1) and (2)

We get,

= 8

= 15

can be drawn as shown below using above information

Hypotenuse side AC is unknown.

Therefore, we find side AC of by Pythagoras theorem.

So, by applying Pythagoras theorem to

We get,

Therefore, Hypotenuse = 17

Now by definition,

Substituting values of sides from the above figure

By definition,

Hence,

Answer: and

Page No 474:

Question 4:

If cot θ=78, evaluate :

(i) 1+ sin θ 1- sin θ1+cos θ 1-cos θ
(ii) cot2 θ

Answer:

(i) Given:

To evaluate:

…… (1)

We know the following formula

By applying the above formula in the numerator of equation (1) ,

We get,

… (Where a = 1 and b = )

…… (2)

Similarly,

By applying formula in the denominator of equation (1) ,

We get,

… (Where a = 1 and b = )

…… (3)

Substituting the value of numerator and denominator of equation (1) ,from equation (2) and(3)

Therefore,

…… (4)

Since,

Therefore,

Putting the value of and in Equation (4)

We get,

We know that,

Since,

Therefore,

Answer:

(ii) Given:

To evaluate:

Squaring on both sides,

We get,

Answer:

Page No 474:

Question 5:

If 3 cot A = 4, check whether 1-tan2 A1+tan2 A=cos2 A-sin2 A or not.

Answer:

Given:

To check whether or not

Dividing by 3 on both sides,

We get,

…… (1)

By definition,

Therefore,

…… (2)

Comparing Equation (1) and (2)

We get,

= 4

= 3

Hence, is as shown in figure below

In, Hypotenuse is unknown

Hence, It can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem in

We get

Hence, Hypotenuse = 5

To check whether or not

We get the values of

By definition,

Substituting the value from Equation (1)

We get,

….… (3)

Now by definition,

…… (4)

Now by definition,

…… (5)

Now we first take L.H.S of Equation

Substituting value of from equation (3)

We get,

Taking L.C.M on both numerator and denominator

We get,

…… (6)

Now we take R.H.S of Equation

Substituting value of and from equation (4) and (5)

We get,

…… (7)

Comparing Equation (6) and (7)

We get,

Answer: Yes

Page No 474:

Question 6:

If tan θ=ab, find the value of cos θ+sin θ cos θ-sin θ.

Answer:

Given:

…… (1)

Now, we know that

Therefore equation (1) becomes as follows

Now, by applying invertendo

We get,

Now, by applying Compenendo-dividendo

We get,

Therefore,

Page No 474:

Question 7:

If 3 cot θ = 2, find the value of 4 sin θ-3 cos θ2 sin θ+6 cos θ.

Answer:

Given:

Therefore,

…… (1)

Now, we know that

Therefore equation (1) becomes

…… (2)

Now, by applying Invertendo to equation (2)

We get,

…… (3)

Now, multiplying by on both sides

We get,

Therefore, 3 cancels out on R.H.S and

We get,

Now by applying dividendo in above equation

We get,

…… (4)

Now, multiplying by on both sides of equation (3)

We get,

Therefore, 2 cancels out on R.H.S and

We get,

Now by applying componendo in above equation

We get,

…… (5)

Now, by dividing equation (4) by equation (5)

We get,

Therefore,

Therefore, on L.H.S cancels out and we get,

Now, by taking 2 in the numerator of L.H.S on the R.H.S

We get,

Therefore, 2 cancels out on R.H.S. and

We get,

Hence,

Page No 474:

Question 8:

If tan θ=ab, prove that a sin θ+b cos θa sin θ+b cos θ=a2-b2a2+b2.

Answer:

Given:

…… (1)

Now, we know that

Therefore equation (1) becomes

…… (2)

Now, multiplying by on both sides of equation (2)

We get,

Therefore,

…... (3)

Now by applying dividendo in above equation (3)

We get,

…… (4)

Now by applying componendo in equation (3)

We get,

…… (5)

Now, by dividing equation (4) by equation (5)

We get,

Therefore,

Therefore, and cancels on L.H.S and R.H.S respectively and we get,

Hence, it is proved that

Page No 474:

Question 9:

If sec θ=135, show that 2 sin θ-3 cos θ4 sin θ- 9 cos θ=3.

Answer:

Given:

To show that

Now, we know that

Therefore,

Therefore,

…… (1)

Now, we know that

…… (2)

Now, by comparing equation (1) and (2)

We get,

= 5

And

Hypotenuse = 13

Therefore from above figure

Base side

Hypotenuse

Side AB is unknown, It can be determined by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

Therefore by substituting the values of known sides

We get,

Therefore,

Therefore,

…… (3)

Now, we know that

Now from figure (a)

We get,

Therefore,

…… (4)

Now L.H.S. of the equation to be proved is as follows

Substituting the value of and from equation (1) and (4) respectively

We get,

Therefore,

Hence proved that,

Page No 474:

Question 10:

If cos θ=1213, show that sin θ (1 − tan θ)=35156.

Answer:

Given: ……(1)

To show that

Now, we know that …… (2)

Therefore, by comparing equation (1) and (2)

We get,

= 12

And

Hypotenuse = 13

Therefore from above figure

Base side

Hypotenuse

Side AB is unknown and it can be determined by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

Therefore by substituting the values of known sides

We get,

Therefore,

Therefore,

…… (3)

Now, we know that

Now from figure (a)

We get,

Therefore,

…… (4)

Now, we know that

Now from figure (a)

We get,

Therefore,

…… (5)

Now L.H.S. of the equation to be proved is as follows

…… (6)

Substituting the value of and from equation (4) and (5) respectively

We get,

Taking L.C.M inside the bracket

We get,

Therefore,

Now, by opening the bracket and simplifying

We get,

…… (7)

From equation (6) and (7) , it can be shown that

Page No 474:

Question 11:

If tan θ=17, show that cosec2 θ-sec2 θcosec2 θ+sec2 θ=34

Answer:

Given: ……(1)

To show that

Now, we know that

Since ……(2)

Therefore,

Comparing Equation (1) and (2)

We get,

Therefore, Triangle representing angle is as shown below

Hypotenuse AC is unknown and it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

Therefore by substituting the values of known sides

We get,

Therefore,

Therefore,

…… (3)

Now, we know that

Now from figure (a)

We get,

Therefore from figure (a) and equation (3) ,

…… (4)

Now, we know that

Therefore, from equation (4)

We get,

Therefore,

…… (5)

Now, we know that

Now from figure (a)

We get,

Therefore from figure (a) and equation (3) ,

…… (6)

Now, we know that

Therefore, from equation (6)

We get,

Therefore,

…… (7)

Now, L.H.S of the equation to be proved is as follows

Substituting the value of and from equation (6) and (7)

We get,

Now by taking L.C.M. in numerator as well as denominator

We get,

Therefore,

Therefore,

Therefore,

Hence proved that

Page No 474:

Question 12:

If sec θ=54, find the value of sin θ-2 cos θtan θ-cot θ.

Answer:

Given: ……(1)

To find the value of

Now we know that

Therefore,

Therefore from equation (1)

…… (2)

Also, we know that cos2θ+sin2θ=1

Therefore,

Substituting the value of from equation (2)

We get,

        

Therefore

…… (3)

Also, we know that sec2θ=1+tan2θ.

Therefore,

tan2θ=sec2θ-1

Therefore

tan2θ=542-1       =2516-1       =916

Therefore,

tanθ=916      =34

Therefore,

tanθ=34 …… (4)

Also cotθ=1tanθ

Therefore, from equation (4)

We get,

cotθ=134

 cotθ=43…… (5)

Substituting the value of,,and from equation (2) (3) (4) and (5) respectively in the expression below

We get,

sinθ-2cosθtanθ-cotθ=35-24534-43                    =35-853×3-4×44×3                    =35-853×3-4×44×3                    =3-859-164×3                    =-55-712                    =127

 

Therefore,

Page No 474:

Question 13:

If cos θ=35, find the value of sin θ-1tan θ2 tan θ.

Answer:

Given: ……(1)

To find the value of

Now, we know the following trigonometric identity

Therefore, by substituting the value of from equation (1) ,

We get,

Therefore,

Therefore by taking square root on both sides

We get,

Therefore,

…… (2)

Now, we know that

Therefore by substituting the value of and from equation (2) and (1) respectively

We get,

…… (4)

Now, by substituting the value of and from equation (2) and (4) respectively in the expression below

We get,

Therefore,

Therefore,

Page No 474:

Question 14:

In the given figure, find tan P and cot R. Is tan P = cot R?
 

Answer:

The given figure is below:

To Find:

In the given right angled ΔPQR, length of side QR is unknown.

Therefore, to find length of side QR we use Pythagoras Theorem

Hence, by applying Pythagoras theorem in ΔPQR,

We get,

Now, we substitute the length of given side PR and PQ in the above equation

By definition, we know that

… (1)

Also, by definition, we know that

… (2)

Comparing equation (1) and (2) , we come to know that R.H.S of both the equation are equal

Therefore, L.H.S of both the equation are also equal

Answer:



Page No 475:

Question 15:

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Answer:

Given:

…… (1)

To show:

is as shown in figure below

Now since ……from (1)

Therefore

Now observe that denominator of above equality is same that is AB

Hence only when

Therefore …… (2)

We know that when two sides of a triangle are equal, then angle opposite to the sides are also equal.

Therefore from equation (2)

We can say that

Angle opposite to side AC = Angle opposite to side BC

Therefore,

Hence,

Page No 475:

Question 16:

In a ∆ABC, right angled at A, if tan C=3, find the value of sin B cos C + cos B sin C.

Answer:

Given:

To find:

The givenis as shown in figure below

Side BC is unknown and can be found using Pythagoras theorem

Therefore,

Now by substituting the value of known sides from figure (a)

We get,

Now by taking square root on both sides

We get,

Therefore Hypotenuse side BC = 2 …… (1)

Now

Therefore,

Now by substituting the values from equation (1) and figure (a)

We get,

…… (2)

Now

Therefore,

Now by substituting the values from equation (1) and figure (a)

We get,

…… (3)

Now

Therefore,

Now by substituting the values from equation (1) and figure (a)

We get,

…… (4)

Now by definition,

Therefore,

Now by substituting the value of and from equation (4) and given data respectively

We get,

Now gets cancelled as it is present in both numerator and denominator

Therefore,

…… (5)

Now by substituting the value of and from equation (2) , (3) , (4) and (5) respectively in

We get,

Page No 475:

Question 17:

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.
(ii) sec A=125 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ=43for some angle θ.

Answer:

(i) In, is acute an angle

Therefore,

Minimum value of is 0° and

Maximum value of is 90°

We know that and

tan90° =

Therefore the statement that;

“The value of is always less than1” is false

(ii)

In and, is acute angle

Therefore,

Minimum value of is 0°and

Maximum value of is

We know that cos0° = 1 and

cos90° = 0

Now,

Therefore minimum value of is …… (1)

Now,

Therefore maximum value of is …… (2)

Now consider the given value

Here,

This value 2.4 lies in between 1 and

Now from equation (1) and (2) , we can say that the value lies in between minimum value of (that is 1) and maximum value of (that is)

Hence, , for some value of angle A is true

(iii) Cosecant of angle A is defined as

Also, is defined as

Therefore,

…… (1)

And

is defined as …… (2)

Therefore from equation (1) and (2) , it is clear that and (that is cosecant of angle A) are two different trigonometric angles

Hence, is the abbreviation used for cosecant of angle A is False

(iv) cot A is a trigonometric ratio which means cotangent of angle A

Hence, is the product of cot and A is False

(v)

The value

In, is acute an angle

Therefore,

Minimum value of is 0° and

Maximum value of is 90°

We know that and

sin90° = 1

Therefore the value of should lie between 0 and 1 and must not exceed 1

Hence the given value for (that is) is not possible

Therefore, , for some angle = False

Page No 475:

Question 18:

If sin θ=1213, find the value of sin2 θ-cos2 θ2 sin θ cos θ×1tan2 θ.

Answer:

Given: ……(1)

To Find: The value of expression

Now, we know that

…… (2)

Now when we compare equation (1) and (2)

We get,

= 12

And

Hypotenuse = 13

Therefore, Triangle representing angle is as shown below

Base side BC is unknown and it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

Therefore by substituting the values of known sides

We get,

Therefore,

Therefore,

…… (3)

Now, we know that

Now from figure (a)

We get,

Therefore from figure (a) and equation (3) ,

…… (4)

Now we know that,

Therefore, substituting the value of and from equation (1) and (4)

We get,

Therefore 13 gets cancelled and we get

…… (5)

Now we substitute the value of , and from equation (1) , (4) and (5) respectively in the expression below

Therefore,

We get,

Therefore by further simplifying we get,

 

Now 169 gets cancelled and gets reduced to

Therefore

Therefore the value of is

That is

Page No 475:

Question 19:

If sec A=54, verify that 3 sin A-4 sin3 A4 cos3 A-3 cos A=3 tan A-tan3 A1-3 tan2 A.

Answer:

Given:

…… (1)

To verify:

…… (2)

Now we know that

Therefore

Now, by substituting the value of from equation (1)

We get,

Therefore,

…… (3)

Now, we know the following trigonometric identity

Therefore,

Now by substituting the value of from equation (3)

We get,

Now by taking L.C.M

We get,

Now, by taking square root on both sides

We get,

Therefore,

…… (4)

Now, we know that

Now by substituting the value of and from equation (3) and (4) respectively

We get,

Therefore

…… (5)

Now from the expression of equation (2)

Now by substituting the value of and from equation (3) and (4)

We get,

Therefore,

Now by taking L.C.M of both numerator and denominator

We get,

…… (6)

Now from the expression of equation (2)

Now by substituting the value of from equation (5)

We get,

Now by taking L.C.M

We get,

Now,

Therefore,

Therefore,

…… (7)

Now by comparing equation (6) and (7)

We get,

Page No 475:

Question 20:

If cot θ=34, prove that sec θ-cosec θsec θ+cosec θ=17.

Answer:

Given:

…… (1)

To prove:

Now we know is defined as follows

…… (2)

Now by comparing equation (1) and (2)

We get,

= 3

= 4

Therefore triangle representing angle is as shown below

Side AC is unknown and can be found using Pythagoras theorem

Therefore,

Now by substituting the value of known sides from figure

We get,

Now by taking square root on both sides

We get,

Therefore Hypotenuse side AC = 5 …… (3)

Now we know, is defined as follows

Therefore from figure (a) and equation (3)

We get,

…… (4)

Now we know

Therefore by substituting the value of from equation (4)

We get,

Therefore,

…… (5)

Now we know, is defined as follows

Therefore from figure (a) and equation (3)

We get,

…… (6)

Now we know

Therefore by substituting the value of from equation (6)

We get,

Therefore,

…… (7)

Now, in expression, by substituting the value of andfrom equation (6) and (7) respectively, we get,

L.C.M of 3 and 4 is 12

Now by taking L.C.M in above expression

We get,

Now 12 gets cancelled and we get,

Now

Therefore,

Now 5 gets cancelled and we get,

Therefore, it is proved that

Page No 475:

Question 21:

If 3 cos θ − 4 sin θ = 2 cos θ + sin θ, find tan θ.

Answer:

Given:

To find:

Now consider the given expression

Now by dividing both sides of the above expression by

We get,

Now by separating the denominator for each terms

We get,

Now in the above expression present in both numerator and denominator gets cancelled

Therefore,

…… (1)

Now we know that,

Therefore by substituting in equation (1)

We get,

Now by taking on L.H.S

We get,

Therefore,

Hence



Page No 485:

Question 1:

Evaluate each of the following

sin 45° sin 30° + cos 45° cos 30°

Answer:

We have,

…… (1)

Now

So by substituting above values in equation (1)

We get,

Therefore,

Page No 485:

Question 2:

Evaluate each of the following

cos 60° cos 45° − sin 60° sin 45°

Answer:

We have to find the value of the following expression

…… (1)

Now,,

So by substituting above values in equation (1)

We get,

Therefore,

Page No 485:

Question 3:

Evaluate each of the following

sin2 30° cos2 45° + 4 tan2 30° + 12 sin2 90°-2 cos2 90°+124 cos2 0°

Answer:

We have,

…… (1)

Now,

,, ,,

So by substituting above values in equation (1)

We get,

LCM of 8, 3, 2 and 24 is 48

Therefore by taking LCM

We get,

In the above equation the first term gets reduced to

Therefore,

Page No 485:

Question 4:

Evaluate each of the following

4 (sin460° + cos4 30°) − 3 (tan2 60° − tan2 45°) + 5 cos2 45°

Answer:

We have,

…… (1)

Now,

,, ,

So by substituting above values in equation (1)

We get,

Now, gets reduced to

Therefore,

Now, gets reduced to

Therefore,

Now by taking LCM

We get,

Therefore,



Page No 486:

Question 5:

Evaluate each of the following

(cosec2 45° sec2 30°) (sin2 30° + 4 cot2 45° − sec2 60°)

Answer:

We have,

…… (1)

Now,

, , , ,

So by substituting above values in equation (1)

We get,

Now, in above equation 4 cancel 8 and 2 remains

Hence,

Therefore,

Page No 486:

Question 6:

Evaluate each of the following

cosec3 30° cos 60° tan3 45° sin2 90° sec2 45° cot 30°

Answer:

We have,

…… (1)

Now,

, , , , ,

So by substituting above values in equation (1)

We get,

Now, 2 gets cancelled and we get,

Page No 486:

Question 7:

Evaluate each of the following

cot2 30°-2 cos2 60°-34sec2 45°-4 sec2 30°

Answer:

We have,

…… (1)

Now,

,, ,

So by substituting above values in equation (1)

We get,

Now, in the third term 4 gets cancelled by 2 and 2 remains

Therefore,

Now in the second term, 4 gets cancelled by 2 and 2 remains

Therefore,

Now, LCM of denominator in the above expression is 6

Therefore by taking LCM

We get,

Now in the above expression, gets reduced to

Therefore,

Page No 486:

Question 8:

Evaluate each of the following

tan2 60°+4 cos2 45°+3 sec2 30°+5 cos2 90°cosec 30°+sec 60°-cot2 30°

Answer:

We have,

…… (1)

Now,

, , , , ,

So by substituting above values in equation (1)

We get,

Now,

3 gets cancel in numerator and we get,

Now, in the numerator get reduced to 2and we get,

Therefore,

Page No 486:

Question 9:

Evaluate each of the following

tan 45°cosec 30°+sec 60°cot 45°-5 sin 90°2 cos 0°

Answer:

We have,

…… (1)

Now,

,,

So by substituting above values in equation (1)

We get,

Now by taking terms with denominator 2 together and solving

We get,

Now gets reduced to -2

Therefore,

Therefore,

Page No 486:

Question 10:

Find the value of x in each of the following :

(i) 2 sin x2=1

(ii) 3 sin x=cos x

(iii) 3 tan 2x=cos 60°+sin 45° cos 45°

Answer:

(i) We have,

Since,

Now,

(ii) We have,

Now by cross multiplying we get,

…… (1)

Now we know that

…… (2)

Therefore from equation (1) and (2)

We get,

…… (3)

Since,

…… (4)

By comparing equation (3) and (4) we get,

(iii) We have,

…… (1)

Now we know that

and

Now by substituting above values in equation (1), we get,

Therefore,

…… (2)

Since,

…… (3)

Therefore by comparing equation (2) and (3)

We get,

Page No 486:

Question 11:

If θ = 30°, verify that

(i) sin 2θ=2 tan θ1+tan2 θ

(ii) cos2 θ=1-tan2 θ1+tan2 θ

Answer:

(i) Given:

…… (1)

To verify:

…… (2)

Now consider right hand side

Hence it is verified that,

(ii) Given:

…… (1)

To verify:

…… (2)

Now consider left hand side of the equation (2)

Therefore,

Now consider right hand side of equation (2)

Therefore,

Hence it is verified that,

Page No 486:

Question 12:

If A = B = 60°, verify that

(i) cos (A − B) = cos A cos B + sin A sin B
(ii) sin (A − B) = sin A cos B − cos A sin B
(iii) tan A-B=tan A-tan B1+tan A tan B

Answer:

(i) Given:

…… (1)

To verify:

…… (2)

Now consider left hand side of the expression to be verified in equation (2)

Therefore,

Now consider right hand side of the expression to be verified in equation (2)

Therefore,

Hence it is verified that,

(ii) Given:

…… (1)

To verify:

…… (2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

Now by substituting the value of A and B from equation (1) in the above expression

We get,

Hence it is verified that,

(iii) Given:

…… (1)

To verify:

…… (2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

Now consider RHS of the expression to be verified in equation (2)

Therefore,

Now by substituting the value of A and B from equation (1) in the above expression

We get,

Hence it is verified that,

Page No 486:

Question 13:

Prove that 3 + 1  3 - cot 30° = tan360° - 2sin 60° .

Answer:

Consider the left hand side.
3+13-cot30°=3+13-3                 cot30°=3=3+1×3×3-1=32-1×3=23
Now, consider the right hand side.
tan360°-2sin60°=33-2×32              sin60°=32=3×32-1=23
So, LHS = RHS
Hence proved.

Page No 486:

Question 14:

(i) If tan A-B=13 and tan A+B=3, 0° < A+B90°, A>B find A and B.

(ii) If tan A+B=1 and tan  A-B=13, 0°<A+B<90°, A>B, then find the values of A and B.

Answer:

(i) 

Given:

 …… (1)

 …… (2)

We know that,

 …… (3)

 …… (4)

Now by comparing equation (1) and (3)

We get,

 …… (5)

Now by comparing equation (2) and (4)

We get,

 …… (6)

Now to get the values of A and B, let us solve equation (5) and (6) simultaneously

Therefore by adding equation (5) and (6)

We get,

Therefore,

Hence 

Now by subtracting equation (5) from equation (6)

We get,

Therefore,

Hence 

Therefore the values of A and B are as follows

 and 

(ii)
tanA+B=1tanA+B=tan45°A+B=45°                 .....1

Also,
tanA-B=13tanA-B=tan30°A-B=30°             .....2

Adding (1) and (2), we get

A+B+A-B=45°+30°2A=75°A=37.5°

Putting A=37.5° in (1), we get

37.5°+B=45°B=45°-37.5°B=7.5°

Thus, the values of A and B are 37.5º and 7.5º, respectively.

Page No 486:

Question 15:

If sin A-B=12 and cos A+B=12, 0° <A+B90°, A<B find A and B.

Answer:

Given:

…… (1)

…… (2)

We know that,

…… (3)

…… (4)

Now by comparing equation (1) and (3)

We get,

…… (5)

Now by comparing equation (2) and (4)

We get,

…… (6)

Now to get the values of A and B, let us solve equation (5) and (6) simultaneously

Therefore by adding equation (5) and (6)

We get,

Therefore,

Hence

Now by subtracting equation (5) from equation (6)

We get,

Therefore,

Hence

Therefore the values of A and B are as follows

and

Page No 486:

Question 16:

In a ∆ABC right angled at B, ∠A = ∠C. Find the values of

(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B

Answer:

(i) We have drawn the following figure related to given information

To find:

…… (1)

Now we have,

,

,

Now by substituting the above values in equation (1)

We get,

Therefore,

…… (2)

Now in right angled

By applying Pythagoras theorem

We get,

Now, by substituting above value of AC2 in equation (2)

We get,

Now both numerator and denominator contains

Therefore it gets cancelled and 1 remains

Hence

(ii) We have drawn the following figure

e

To find:

…… (1)

Now we know that sum of all the angles of any triangle is 180°

Therefore,

Since and

Therefore,

It is given that

Therefore,

…… (2)

Now we have,

,

,

Now by substituting the above values in equation (1)

We get,

Since

Therefore

Page No 486:

Question 17:

Find acute angles A and B, if sin A+2B=32 and cos A+4B=0, A>B.

Answer:

Given:

…… (1)

…… (2)

We know that,

…… (3)

…… (4)

Now by comparing equation (1) and (3)

We get,

…… (5)

Now by comparing equation (2) and (4)

We get,

…… (6)

Now to get the values of A and B, let us solve equation (5) and (6) simultaneously

Therefore by subtracting equation (5) from (6)

We get,

Therefore,

Hence

Now by multiplying equation (5) by 2

We get,

…… (7)

Now by subtracting equation (6) from (7)

We get,

Therefore,

Hence

Therefore the values of A and B are as follows and

Page No 486:

Question 18:

In ∆PQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.

Answer:

We are given the following information in the form of triangle

To find: and

Now, in

…… (1)

Now we know that

…… (2)

Now by comparing equation (1) and (2)

We get,

…… (3)

Now we have

Now we know that

Therefore,

Now by cross multiplying

We get,

Therefore,

cm …… (4)

Now we know that

Now we know,

…… (6)

Now by comparing equation (5) and (6)

We get,

…… (7)

Hence from equation (3) and (7)

and

Page No 486:

Question 19:

If sin (AB) = sin A cos B − cos A sin B and cos (AB) = cos A cos B + sin A sin B, find the values of sin 15° and cos 15°.

Answer:

Given:

…… (1)

…… (2)

To find:

The values of and

In this problem we need to find and

Hence to get angle we need to choose the value of A and B such that

So If we choose and

Then we get,

Therefore by substituting and in equation (1)

We get,

Therefore,

…… (3)

Now we know that,

, ,

Now by substituting above values in equation (3)

We get,

Therefore,

…… (4)

Now by substituting and in equation (2)

We get,

Therefore,

…… (5)

Now we know that,

, ,

Now by substituting above values in equation (5)

We get,

Therefore,

…… (6)

Therefore from equation (4) and (6)

Page No 486:

Question 20:

In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15 units. Find the remaining angles and sides.

Answer:

We are given the following triangle with related information

It is required to find , and length of sides AC and BC

is right angled at C

Therefore,

Now we know that sum of all the angles of any triangle is

Therefore,

…… (1)

Now by substituting the values of known angles and in equation (1)

We get,

Therefore,

Therefore,

Now,

We know that,

Now we have,

AB=15 units and

Therefore by substituting above values in equation (2)

We get,

Now by cross multiplying we get,

Therefore,

…… (3)

Now,

We know that,

Now we have,

AB=15 units and

Therefore by substituting above values in equation (4)

We get,

Now by cross multiplying we get,

Therefore,

Hence,

Page No 486:

Question 21:

If ∆ABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units. Find ∠B, AB and AC.

Answer:

We are given the following information in the form of the triangle

It is required to find and length of sides AB and AC

In

Now we know that sum of all the angles of any triangle is

Therefore,

…… (1)

Now by substituting the values of known angles and in equation (1)

We get,

Therefore,

Therefore,

…… (2)

Now,

We know that,

Now we have,

BC = 7 units and

Therefore by substituting above values in equation (3)

We get,

Now by cross multiplying we get,

Therefore,

…… (4)

Now,

We know that,

……(5)

Now we have,

and

Therefore by substituting above values in equation (5)

We get,

Now by cross multiplying we get,

Therefore,

…… (6)

Therefore,

From equation (2), (4) and (6)

, ,

Page No 486:

Question 22:

If A and B are acute angles such that tan A=12, tan B=13 and tan A+B=tan A+tan B1-tan A tan B, find A + B.

Answer:

Given:

…… (1)

…… (2)

…… (3)

Now by substituting the value of and from equation (1) and (2) in equation (3)

We get,

Therefore,

…… (3)

Now we know that

…… (4)

Now by comparing equation (3) and (4)

We get,



Page No 492:

Question 1:

Evaluate the following :

(i) sin 20°cos 70°

(ii) cos 19°sin 71°

(iii) sin 21°cos 69°

Answer:

(i) Given that

Since

Therefore

(ii) Given that

Since

Therefore

(iii) Given that

Since

Page No 492:

Question 2:

Evaluate the following :

(i) sin 49°cos 41°2+cos 41°sin 49°2

(ii) cot 40°tan 50°-12 cos 35°sin 55°

(iii) tan 35°cot 55°+cot 78°tan 12°-1

(iv) sec 70°cosec 20°+sin 59°cos 31°

(v) tan 48° tan 23° tan 42° tan 67°

(vi) sec50° sin 40° + cos40° cosec 50°

Answer:

(i) We have to find:

Since and

So

So the value of is .

(ii) We have to find:

Since and

So the value of is .

(v) We have to find:

Since and

tan 35°cot 55°+cot 78°tan 12°-1=tan (90°-55°)cot 55°+cot 90°-78°tan 12°-1=cot 55°cot 55°+tan 12°tan 12°-1=1+1-1=2-1=1

So the value of is .

(iv) We have to find:

Sinceand

So

So the value of is .

(v) We have to find

Since.So

So the value of is .

(vi) We find to find

Since, and .So

So the value of is .

Page No 492:

Question 3:

Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°

(i) cos 78° + sec 78°

(ii) cosec 54° + sin 72°

(iii) cot 85° + cos 75°

(iv) sin 67° + cos 75°

Answer:


(i) We knowand

Thus the desired expression is .

(ii) We know and.So

Thus the desired expression is .

(iii) We know thatand.So

Thus the desired expression is.

(iv) We know thatand.So

Thus the desired expression is .

Page No 492:

Question 4:

Express cos 75° + cot 75° in terms of angles between 0° and 30°.

Answer:

Given that:

Hence the correct answer is

Page No 492:

Question 5:

If sin 3A = cos (A − 26°), where 3A is an acute angles, find the value of A.

Answer:

We are given 3A is an acute angle

We have:

Hence the correct answer is

Page No 492:

Question 6:

If A, B, C are the interior angles of a triangle ABC, prove that

(i) tan C+A2=cot B2
(ii) sin B+C2=cos A2

Answer:

(i) We have to prove:

Since we know that in triangle

Proved

 

(ii) We have to prove:

Since we know that in triangle

Proved

Page No 492:

Question 7:

Prove the following :

(i) sin θ sin (90° − θ) − cos θ cos (90° − θ) = 0
(ii) cos 90°-θ sec 90°-θ tan θcosec 90°-θ sin 90°-θ cot 90°-θ+tan 90°-θcot θ=2
(iii) tan 90°-A cot Acosec2 A-cos2 A=0
(iv) cos 90°-A sin 90°-Atan 90°-A=sin2 A
(v) sin (50° − θ) − cos (40° − θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1

Answer:

(i) We have to prove:

Left hand side

=Right hand side

Proved

(ii) We have to prove:

Left hand side

= right hand side

Proved

(iii) We have to prove:

Left hand side

= right hand side

Proved

(iv) We have to prove:

Left hand side

= Right hand side

Proved

(v) We have to prove:

Left hand side

Since .So

=Right hand side

Proved

Page No 492:

Question 8:

Evaluate :

(i) tan 7° tan 23° tan 60° tan 67° tan 83°

(ii) 2 sin 68°cos 22°-2 cot 15°5 tan 75°-3 tan 45° tan 20° tan 40° tan 50° tan 70°5

(iii) sin 18°cos 72°+3 tan 10° tan 30° tan 40° tan 50° tan 80°

(iv) 3 tan 41°cot 49°2-sin 35° sec 55°tan10° tan20° tan60° tan70° tan80°2
 

Answer:

We have to evaluate the following values-

(i) We will use the properties of complementary angles.

(ii) We will use the properties of complementary angles.

(iii) We will use the properties of complementary angles.

(xi)

3tan41°cot49°2-sin35°sec55°tan10°tan20°tan60°tan70°tan80°2=3tan90°-49°cot49°2-sin35°sec90°-35°tan10°tan20°tan60°tan90°-20°tan90°-10°2=3cot49°cot49°2-sin35°cosec35°tan10°tan20° ×3× cot20°cot10°2                      tan90°-θ=cotθ and sec90°-θ=cosecθ
=32-1tan10°cot10°×tan20°cot20°×32                  cosecθ=1sinθ=9-11×1×32                                                              cotθ=1tanθ=9-13=263



Page No 493:

Question 9:

If sin θ = cos (θ − 45°), where θ and θ − 45° are acute angles, find the degree measure of θ.

Answer:

Given that: where and are acute angles

We have to find

Therefore

Page No 493:

Question 10:

If A, B, C are the interior angles of a ∆ABC, show that :

(i) sin B+C2=cos A2
(ii) cos B+C2=sin A2

Answer:

(i) We have to prove:

Since we know that in triangle

Dividing by 2 on both sides, we get

Proved

(ii) We have to prove:

Since we know that in triangle

Dividing by 2 on both sides, we get

Proved

Page No 493:

Question 11:

If sin 3 θ = cos (θ − 6°), where 3 θ and θ − 6° are acute angles, find the value of θ.

Answer:

We have: where and are acute angles

We have to find

Now we proceed as to find

Therefore 

Page No 493:

Question 12:

If sec 4A = cosec (A − 20°), where 4A is an acute angles, find the value of A.

Answer:

Given: and is an acute angle

We have to find

Now

Hence the value of is

Page No 493:

Question 13:

If sec 2A = cosec (A − 42°), where 2A is an acute angles, find the value of A.

Answer:

Given: and is an acute angle

We have to find

So we proceed as follows to calculate

Hence the value of is

Page No 493:

Question 14:

If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Answer:

Given that,

tan2A = cot(A− 18°)

cot(90° − 2A) = cot(A − 18°)               (∵ tanθ = cot(90° − θ))

90° − 2A = − 18°

108° = 3A

A = 36°

Page No 493:

Question 15:

Prove that: sin (50° + θ) − cos (40° − θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1

Answer:

We have to prove: sin (50° + θ) − cos (40° − θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1

Left hand side

= 1 × 1 × 1    []

= Right hand side

Proved.

Page No 493:

Question 16:

Evaluate :

(i) 23cos430°-sin445°-3sin260°-sec245°+14cot230°

(ii) 3 cos 55°7 sin 35°-4cos 70° cosec 20°7tan 5° tan 25° tan 45° tan 65° tan 85°

(iii) cos 58°sin 32°+sin 22°cos 68°-cos 38° cosec 52°tan 18° tan 35° tan 60° tan 72° tan 55°

Answer:

We have to evaluate the following values-

(i) We will use the values of known angles of different trigonometric ratios.

(ii) We will use the properties of complementary angles.

(iii) We will use the properties of complementary angles.


 

Page No 493:

Question 17:

If 2θ + 45° and 30° − θ are acute angles, find the degree measure of θ satisfying sin (2θ + 45°) = cos (30° − θ).

Answer:

Given that: where and are acute angles

We have to find

So we have

Hence the value of is

Page No 493:

Question 1:

Write the maximum and minimum values of sin θ.

Answer:

The maximum value of is and the minimum value of is because value of lies between −1 and 1

 

Page No 493:

Question 2:

Write the maximum and minimum values of cos θ.

Answer:

The maximum value of is and the minimum value of is because value of lies between −1 and 1

 

Page No 493:

Question 3:

What is the maximum value of 1sec θ?

Answer:

The maximum value of is because the maximum value of is that is

 

Page No 493:

Question 4:

What is the maximum value of 1cosec θ?

Answer:

The maximum value of is because the maximum value of is that is

 

Page No 493:

Question 5:

If tan θ=45, find the value of cos θ-sin θcos θ+sin θ.

Answer:

 

It is given that .

We have to find .

 

 



Page No 494:

Question 6:

If cos θ=23, find the value of sec θ-1sec θ+1.

Answer:

Given in question:

We have to find

Hence the value of is

 

Page No 494:

Question 7:

If 3 cot θ = 4, find the value of 4 cos θ-sin θ2 cos θ+sin θ.

Answer:

We have:

Since we know that in right angle triangle

Now, we find

Hence the value of is

 

Page No 494:

Question 8:

Given tan θ=15, what is the value of cosec2 θ-sec2 θcosec2 θ+sec2 θ?

Answer:

Given: ,

We know that:

Now we find,

Hence the value of is

 

Page No 494:

Question 9:

If cot θ=13, write the value of 1-cos2 θ2-sin2 θ.

Answer:

Given:

Now we find,

Hence the value of is

 

Page No 494:

Question 10:

If tan A=34 and A+B=90°, then what is the value of cot B?

Answer:

Given that:

Hence the value of is

 

Page No 494:

Question 11:

If A + B = 90° and cos B=35, what is the value of sin A?

Answer:

We have:

Hence the value of is

 

Page No 494:

Question 12:

Write the acute angle θ satisfying 3 sin θ=cos θ.

Answer:

We have:

Hence the acute angle is

 

 

Page No 494:

Question 13:

Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.

Answer:

Given that:

Hence the value of is

 

Page No 494:

Question 14:

Write the value of tan 10° tan 15° tan 75° tan 80°?

Answer:

We have to find:

Hence the value of is

 

 

Page No 494:

Question 15:

If A + B = 90° and tan A=34, what is cot B?

Answer:

Given in question:

Hence the value of is

 

Page No 494:

Question 16:

Find A, if tan 2A = cot(A – 24°).

Answer:


tan2A=cotA-24°tan2A=tan90°-A-24°                tan90°-θ=cotθ2A=90°-A-24°2A=90°-A+24°
3A=114°A=114°3=38°

Thus, the value of A is 38º.

Page No 494:

Question 17:

Find the value of sin233° + sin257°.

Answer:


We have
sin57°=sin90°-33°=cos33°                    sin90°-θ=cosθ
sin233°+sin257°=sin233°+cos233° =1                                   sin2θ+cos2θ=1

Thus, the value of sin233° + sin257° is 1.

Page No 494:

Question 18:

Evaluate sin2 60° + 2tan 45° – cos2 30°.

Answer:


sin260°+2tan45°-cos230°=322+2×1-322=34+2-34=2

Page No 494:

Question 19:

If sin A=34, calculate sec A.

Answer:


We know
sin2A+cos2A=1342+cos2A=1cos2A=1-916=716cosA=74
Now,
secA=1cosAsecA=174secA=47=477

Page No 494:

Question 20:

If tan A=512, find the value of (sin A + cos A) sec A.

Answer:

Given:

We know that:

Now we find,

Hence the value of is

 

Page No 494:

Question 21:

In the given figure, PS = 3 cm, QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, PQ RQ and RQ = 9 cm. Evaluate tan θ.

Answer:


In right ∆PQS,

PQ2=PS2+SQ2        Pythagoras theoremPQ2=32+42PQ2=9+16=25PQ=25=5 cm

In right ∆PQR,

tanθ=PQRQtanθ=5 cm9 cm=59

Thus, the value of tanθ is 59.

Page No 494:

Question 1:

The value of (sin30° + cos30°) – (sin60° + cos60°) is ________.

Answer:


sin30°+cos30°-sin60°+cos60°=12+32-32+12=3+12-3+12=0

The value of (sin30° + cos30°) – (sin60° + cos60°) is ___0___.

Page No 494:

Question 2:

The value of tan 30°cot 60° is ______.

Answer:


tan30°cot60°=1313=1

The value of tan 30°cot 60° is ___1___.



Page No 495:

Question 3:

The value of (sin 45° + cos 45°)3 is _______.

Answer:


sin45°+cos45°3=12+123=223
=23=22

The value of (sin 45° + cos 45°)3 is     22   .

Page No 495:

Question 4:

The value of (sin 30° + cos 30°)2 – (sin 60° – cos 60°)2 is __________.

Answer:


sin30°+cos30°2-sin60°-cos60°2=12+322-32-122=3+12-3-124
=3+1+23-3+1-234=4+23-4+234=434=3

The value of (sin 30° + cos 30°)2 – (sin 60° – cos 60°)2 is       3      .

Page No 495:

Question 5:

The value of (cos2 23° – sin2 67°) is ____________.

Answer:


We know that,
cos23°=cos90°-67°=sin67°             cos90°-θ=sinθ
cos223°-sin267°=sin267°-sin267°=0

The value of (cos2 23° – sin2 67°) is ______0______.

Page No 495:

Question 6:

The value of the expression sin222°+sin268°cos222°+cos268°+sin263°+cos 63° sin 27° is ________.

Answer:


We know that,
sin68°=sin90°-22°=cos22°               sin90°-θ=cosθsin27°=sin90°-63°=cos63°cos68°=cos90°-22°=sin22°              cos90°-θ=sinθ
sin222°+sin268°cos222°+cos268°+sin263°+cos63°sin27°=sin222°+cos222°cos222°+sin222°+sin263°+cos63°×cos63°=11+sin263°+cos263°             sin2θ+cos2θ=1
=1+1=2

The value of the expression sin222°+sin268°cos222°+cos268°+sin263°+cos 63° sin 27° is ____2____.

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Question 7:

Given that sin α = 12 and cos β = 12, then α + β = _________.

Answer:


sinα=12sinα=sin30°α=30°
Also,
cosβ=12cosβ=cos60°β=60°
α+β=30°+60°=90°

Given that sin α = 12 and cos β = 12, then α + β = ____90°____.

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Question 8:

Given that sin (α – β ) = 12 and cos (α + β) = 12,  then α = _______ β = _________.

Answer:


sinα-β=12sinα-β=sin30°α-β=30°                 .....1
Also,
cosα+β=12cosα+β=cos60°α+β=60°              .....2
Adding (1) and (2), we get
α-β+α+β=30°+60°2α=90°α=45°
Putting α=45° in (2), we get
45°+β=60°β=60°-45°=15°
α=3β

Given that sin (α – β ) = 12 and cos (α + β) = 12,  then α = ___3___ β = ____45º___.

Page No 495:

Question 9:

If 4 tan θ = 3, then 4sinθ-cosθ4sinθ+cosθ is equal to _________.

Answer:


4sinθ-cosθ4sinθ+cosθ
=4sinθcosθ-14sinθcosθ+1          (Dividing numerator and denominator by cosθ)
=4tanθ-14tanθ+1
=3-13+1                4tanθ=3=24=12

If 4 tan θ = 3, then 4sinθ-cosθ4sinθ+cosθ is equal to      12      .

Page No 495:

Question 10:

If cos 5θ = sinθ and 5θ < 90°, then the value of tan 3θ  is ___________.

Answer:


cos5θ=sinθcos5θ=cos90°-θ5θ=90°-θ6θ=90°
3θ=90°2=45°tan3θ=tan45°=1

If cos 5θ = sinθ and 5θ < 90°, then the value of tan 3θ  is ____1____.

Page No 495:

Question 11:

The value of sec2 60° – tan2 60° is __________.

Answer:


sec260°-tan260°=22-32               sec60°=2 and tan60°=3=4-3=1

The value of sec2 60° – tan2 60° is _____1_____.

Page No 495:

Question 12:

If A + B = 90°, then the value of tan2A – cot2B is _________.

Answer:


A+B=90°A=90°-BtanA=tan90°-BtanA=cotB                         tan90°-θ=cotθ
tan2A-cot2B=cot2B-cot2B=0               tanA=cotB

If A + B = 90°, then the value of tan2A – cot2B is ____0____.

Page No 495:

Question 13:

The value of cos 1° cos 2° cos 3° ...... cos 120° is _______.

Answer:


cos1°cos2°cos3°...cos120°=cos1°×cos2°×cos3°×...×cos90°×...×cos120°=cos1°×cos2°×cos3°×...×0×...×cos120°                     cos90°=0=0

The value of cos 1° cos 2° cos 3° ...... cos 120° is ___0___.

Page No 495:

Question 14:

If tan θ + cot θ = 2 and, 0° < θ < 90° then tan10 θ + cot10 θ is equal to _________.

Answer:


tanθ+cotθ=2tanθ+1tanθ=2tan2θ+1tanθ=2
tan2θ+1=2tanθtan2θ-2tanθ+1=0tanθ-12=0
tanθ-1=0tanθ=1cotθ=1tanθ=1
So,
tan10θ+cot10θ=110+110=1+1=2

If tan θ + cot θ = 2 and, 0° < θ < 90° then tan10 θ + cot10 θ is equal to ____2____.

Page No 495:

Question 15:

If sin θ + cos θ = 1 and 0° ≤ θ ≤ 90°, then the possible values of θ are ________.

Answer:


The given equation is sinθ + cosθ = 1.

When θ = 0°,

LHS = sinθ + cosθ = sin0° + cos0° = 0 + 1 = 1 = RHS

When θ = 90°,

LHS = sinθ + cosθ = sin90° + cos90° = 1 + 0 = 1 = RHS

Thus, the possible values of θ (0° ≤ θ ≤ 90°) satisfying the given equation are 0° and 90°.

If sin θ + cos θ = 1 and 0° ≤ θ ≤ 90°, then the possible values of θ are _0° and 90°_.

Page No 495:

Question 16:

If sinA+B=cosA-B=32, then cot 2A = _________.

Answer:


sinA+B=32sinA+B=sin60°A+B=60°      .....1

Also,
cosA-B=32cosA-B=cos30°A-B=30°          .....2

Adding (1) and (2), we get

A+B+A-B=60°+30°2A=90°cot2A=cot90°=0

If sinA+B=cosA-B=32, then cot 2A = ____0____.

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Question 17:

If ΔABC is an isosceles right triangle right angled at B, then tan A+cot Ccot A+cot C=____________.

Answer:

It is given that ΔABC is an isosceles right triangle right angled at B.
B=90° and A=C      .....(1)
In ΔABC,
A+B+C=180°          (Angle sum property of a triangle)
A+90°+C=180°                  GivenA+C=180°-90°=90°A=90°-C                .....2
tanA+cotCcotA+cotC=tan90°-C+cotCcotA+cotC            Using 2=cotC+cotCcotA+cotC=2cotC2cotC                 Using 1
= 1

If ΔABC is an isosceles right triangle right angled at B, then tan A+cot Ccot A+cot C=       1       .

Page No 495:

Question 18:

If α + β = 90° and α=β2, then tan α tan β = ________.

Answer:


α+β=90°β2+β=90°               α=β23β2=90°β=90°×23=60°
α=60°2=30°
Now,
tanαtanβ=tan30°×tan60°=13×3=1

If α + β = 90° and α=β2, then tan α tan β = ____1____.

Page No 495:

Question 19:

If in a triangle ABC, angles A and B are complementary, then the value of cot C is ________.

Answer:


It is given that angles A and B are complementary.

A+B=90°     .....(1)

In ∆ABC,

A+B+C=180°         (Angle sum property of a triangle)

90°+C=180°                Using 1C=180°-90°=90°cotC=cot90°=0

If in a triangle ABC, angles A and B are complementary, then the value of cot C is ____0____.

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Question 20:

The value of tan 25° tan 10° tan 80° tan 65° is __________.

Answer:


tan25°tan10°tan80°tan65°=tan25°tan10°tan90°-10°tan90°-25°=tan25°tan10°cot10°cot25°                  tan90°-θ=cotθ
=tan25°cot25°×tan10°cot10°=1×1                                     cotθ=1tanθ=1

The value of tan 25° tan 10° tan 80° tan 65° is _____1_____.



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