if tanA=2. evaluate secAsinA + tan²A-cosecA

its answer is 5....

 it is 6 -(root 5)/2

secA sinA =sinA / cos A =tan A =2

tan2A=22 =4

cosec A is 1 / sinA and when we draw the triangle it comes root5 / 2

so according to the question my answer is right...

hope it helps!!

cheers!!!

hope it helps!!

cheers!!

ay!

it is 

tan²A = 2² =4

you plz correct that...

 thumbs up plzzz...

 your answer is wrong

 secA.sina =  2    

tan²A = 4    

1+ sec²A = 4

1/cos²A = 3 

sin²A =  4/3 = 2/sqrt3

Ans is            6sqrt3 - 2/ sqrt3

 1+tan²A=sec²A

no i am right.

SecAsinA +  tan²A - CosecA

tanA + 4 -CosecA

2 + 4 - cosecA

6 - cosecA ------ 1

1 + Sec²A = tan²A

Sec²A = 4-1

1/cos²A = 3 

cos²A = 1/3

Sin²A/cos²A = 4

Sin²A X 1/cos²A = 4

3Sin²A = 4

Sin²A = 4/3

Sin A =  2/sqrt3 ---------- 2

From 1 & 2

you'll get the answer

hey!!  Pradeep17Ch is right!!

thnx pradeep

 Can you make out any error in my calculation?

 i am right....!

• 12- ROOT 5/2