Question

Prove that `2sin^-1  3/5 - tan^-1  17/31 = pi/4`

Answer 1

Let `sin^-1  3/5` = θ

Then sin θ = `3/5`

Where θ ∈ `[(-pi)/2, pi/2]`

Thus tan θ = `3/4`

Which gives θ = `tan^-1  3/4`.

Therefore, `2sin^-1  3/5 - tan^-1  17/31`

= `2theta - tan^-1  17/31`

= `2tan^-1  3/4 - tan^-1  17/31`

= `tan^-1 ((2 * 3/4)/(1 - 9/16)) - tan^-1  17/31`

= `tan^-1  24/7 - tan^-1  17/31`

= `tan^-1 ((24/7 - 17/31)/(1 + 24/7 * 17/31))`

= `pi/4`