Question

Prove that `2 sin^-1  3/5 = tan^-1  24/7`.

Answer 1

Let `sin^-1  3/5 = x`.

Then, `sin x = 3/5`.

⇒ `cos x = sqrt(1 - (3/5)^2) = 4/5`

∴ `tan x = 3/4`

∴ `x = tan^-1  3/4`

⇒ `sin^-1  3/5 = tan^-1  3/4`

Now, we have:

L.H.S. = `2 sin^-1  3/5`

= `2 tan^-1  3/4`

= `tan^-1 ((2 xx 3/4)/(1 - (3/4)^2))`   ...`[∵ 2 tan^-1 x = tan^-1  ((2x)/(1 - x^2))]`

= `tan^-1 ((3/2)/(1 - 9/16))`

= `tan^-1  ((3/2)/(7/16))`

= `tan^-1  (3/2 xx 16/7)`

= `tan^-1  24/7` = R.H.S.