# prove that tan -1(1) + tan-1(2) + tan-1(3) = pie.

=tan-11+(tan-12+tan-13)

=^/4  +(^/2-cot-12 ) + (^/2-cot-13)

=^/4+^ - tan-11/2+tan-11/3

=^/4 - ^ - tan-1{ (1/2+1/3)/(1-1/2*1/3) }

=^+  ^/4 - tan-11

=^

• -11

tan-1(1) + tan-1(2) + tan-1(3)

tan-1(1+2/1-1*2) + tan(3)     (tan(A+B) = tanA + tanB/1-tanAtanB)

tan-1(-3) + tan-1(3)

tan-1(-3+3/1+3*3)

tan-1(0/1+9)

tan-1(0)

pie

hence proved

• 8

one best way to look is

sides 1 , 2 and 3 resembles a triangle with 0 area.

thus, sum of all the three angles must be 180o.

• -27

nah my approach to this question as mentioned is wrong... !!

• -11

srry... i don't listen tribal songs... :P

• -18
Pink panther
• -21
its very simple refer pg 166
• -24
tan-11+tan-12 +tan-13=pi
we know that tan-1x +tan-1y +tan-1z=pi then  x +y +z =xyz
​so 1+2+3=1*2*3
6=6
hence proved

• -14
All the above anwers are wrong
• -15
Refer rd pg 4.72..
• -17
Solution :

Tan-11+ Tan-12+ Tan-13=?
Solution:
Tan-11=π/4
Tan-12+ Tan-13= Tan-12+ Tan-13/ 1-Tan-12. Tan-13= Tan-1 -1=nπ-π/4=π-π/4

• -15
Tan^-1 (1)+tan^-1 (1/2)+tan^-1 (1/3)
• -15
by this we can solve it

• 44
tan^-1(2)=π/2-tan^-1(1/2)
• -4
Easily solved I hope it helped to all😊

• 7

• 2