Question

Prove that `sin^(-1)  8/17 + sin^(-1)  3/5 = tan^(-1)  77/36`.

Answer 1

Let `sin^-1  8/17 = x`.

Then, `sin x = 8/17`

 ⇒ `cos x = sqrt(1 - (8/17)^2`

= `sqrt((225)/(289)`

= `15/17`

∴ `tan x= 8/15` ⇒ `x = tan^-1  8/15`

∴ `sin^-1  8/17 = tan^-1  8/15`   ...(1)

Now, let `sin^-1  3/5 = y`.

Then, `sin y = 3/5`

⇒ `cos y = sqrt(1 - (3/5)^2`

= `sqrt(16/25)`

= `4/5`

∴ `tan y = 3/5` ⇒ `y = tan^-1  3/4`

∴ `sin^-1  3/5 = tan^-1  3/4`   ...(2)

Now, we ahve:

L.H.S. = `sin^-1  8/17 + sin^-1  3/5`

= `tan^-1  8/15 + tan^-1  3/4`   ...[Using (1) and (2)]

= `tan^-1  (8/15 + 3/4)/(1 - 8/15 xx 3/4)`

= `tan^-1 ((32 + 45)/(60 - 24))`   ...`[tan^-1x + tan^-1y = tan^-1  (x + y)/(1 - xy)]`

= `tan^-1  77/36` = R.H.S.