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`

= `cos^(-1)  33/65`   .....[by (3)]

= R.H.S.