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