Question

Prove that `tan^(-1)  63/16 = sin^(-1)  5/13 + cos^(-1)  3/5`.

Answer 1

Let `sin^(-1)  5/13 = x`.

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

∴ `tan x = 5/12` ⇒ `x = tan^-1  5/12`

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

Let `cos^-1  3/5 = y`.

Then, `cos y = 3/5` ⇒ `sin y = 4/5`.

∴ `tan y = 4/3` ⇒ `y = tan^-1  4/3`

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

Using (1) and (2), we have

R.H.S. = `sin^-1  5/13 + cos^-1  3/5`

= `tan^-1  5/12 + tan^-1  4/3`

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

= `tan^-1  ((15 + 48)/(36 - 20))`

= `tan^-1  63/16`

= L.H.S.