Question

`tan^-1  1/4+tan^-1  2/9=1/2cos^-1  3/2=1/2sin^-1(4/5)`

Answer 1

LHS = `tan^-1  1/4+tan^-1  2/9`

`=tan^-1((1/4+2/9)/(1-1/4xx2/9))`     `[becausetan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))]`

`=tan^-1((17/36)/(34/36))`

`=tan^-1  1/2`

`=1/2cos^-1((1-1/4)/(1+1/4))`         `[becausetan^-1x=1/2cos^-1((1-x^2)/(1+x^2))]`

`=1/2cos^-1((3/4)/(5/4))`

`=1/2cos^-1(3/5)`

Now,

`tan^-1  1/2=1/2sin^-1((2/2)/(1+1/4))`     `[becausetan^-1x=1/2sin^-1((2x)/(1+x^2))]`

`=1/2sin^-1 (1/(5/4))`

`=1/2sin^-1(4/5)`