Question

Prove that `cos^(-1)  12/13 + sin^(-1)  3/5 = sin^(-1)  56/65`.

Answer 1

Let `sin^-1  3/5 = x`.

Then, `sin x = 3/5`

⇒ `cos x = sqrt(1 - (3/5)^2`

= `sqrt(16/25)`

= `4/5`

∴ `tan x = 3/4` ⇒ `x = tan^-1  3/4`

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

Now, let `cos^-1  12/13 = y`.

Then, `cos y = 12/13` ⇒ `sin y = 5/13`.

∴ `tan y = 5/12` ⇒ `y = tan^-1  5/12`

∴ `cos^-1  12/13 = tan^-1  5/12`   ...(2)

Let `sin^-1  56/65 = z`.

Then, `sin z = 56/65` ⇒ `cos z = 33/65`.

∴ `tan z = 56/33` ⇒ `z = tan^-1  56/33`

∴ `sin^-1  56/65 = tan^-1  56/33`   ...(3)

Now, we have:

L.H.S. = `cos^-1  12/13 + sin^-1  3/5`

= `tan^-1  5/12 + tan^-1  3/4`   ...[Using (1) and (2)]

= `tan^-1  (5/12 + 3/4)/(1 - 5/12 * 3/4)`   ...`[tan^-1x + tan^-1y = tan^-1  (x + y)/(1 - xy)]`

= `tan^-1  (20 + 36)/(48 - 15)`

= `tan^-1  56/33`

= `sin^-1  56/65` = R.H.S.   ...[Using (3)]