Question

Prove that `cos^(-1)  4/5 + cos^(-1)  12/13 = cos^(-1)  33/65`.

Answer 1

Let `cos^(-1)  4/5 = x`.

Then, `cos x = 4/5`

⇒ `sin x = sqrt (1 - (4/5)^2)`

⇒ `sin x = sqrt(1 - 16/25)`

⇒ `sin x = sqrt(9/25)`

⇒ `sin x = 3/5`

∴ `tan x = 3/4` ⇒ `x = tan^(-1)  3/4`

∴ `cos^(-1)  4/5 = 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 `cos^(-1)  33/65 = z`.

Then, `cos z = 33/65`

⇒ `sin z = 56/65`

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

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

Now, we will prove that:

L.H.S = `cos^(-1)  4/5 + cos^(-1)  12/13`

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

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

= `tan^(-1)  (36+20)/(48-15)`

= `tan^(-1)  56/33`

= `tan^(-1)  56/33`   ...[By (3)]

= R.H.S.