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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`
= `cos^(-1) 33/65` .....[by (3)]
= R.H.S.
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