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प्रश्न
Prove that `sin^(-1) 8/17 + sin^(-1) 3/5 = tan^(-1) 77/36`.
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उत्तर
Let `sin^-1 8/17 = x`.
Then, `sin x = 8/17`
⇒ `cos x = sqrt(1 - (8/17)^2`
= `sqrt((225)/(289)`
= `15/17`
∴ `tan x= 8/15` ⇒ `x = tan^-1 8/15`
∴ `sin^-1 8/17 = tan^-1 8/15` ...(1)
Now, let `sin^-1 3/5 = y`.
Then, `sin y = 3/5`
⇒ `cos y = sqrt(1 - (3/5)^2`
= `sqrt(16/25)`
= `4/5`
∴ `tan y = 3/5` ⇒ `y = tan^-1 3/4`
∴ `sin^-1 3/5 = tan^-1 3/4` ...(2)
Now, we ahve:
L.H.S. = `sin^-1 8/17 + sin^-1 3/5`
= `tan^-1 8/15 + tan^-1 3/4` ...[Using (1) and (2)]
= `tan^-1 (8/15 + 3/4)/(1 - 8/15 xx 3/4)`
= `tan^-1 ((32 + 45)/(60 - 24))` ...`[tan^-1x + tan^-1y = tan^-1 (x + y)/(1 - xy)]`
= `tan^-1 77/36` = R.H.S.
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