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प्रश्न
Prove that `tan^(-1) 63/16 = sin^(-1) 5/13 + cos^(-1) 3/5`.
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उत्तर
Let `sin^(-1) 5/13 = x`.
Then, `sin x = 5/13` ⇒ `cos x = 12/13`.
∴ `tan x = 5/12` ⇒ `x = tan^-1 5/12`
∴ `sin^-1 5/13 = tan^-1 5/12` ...(1)
Let `cos^-1 3/5 = y`.
Then, `cos y = 3/5` ⇒ `sin y = 4/5`.
∴ `tan y = 4/3` ⇒ `y = tan^-1 4/3`
∴ `cos^-1 3/5 = tan^-1 4/3` ...(2)
Using (1) and (2), we have
R.H.S. = `sin^-1 5/13 + cos^-1 3/5`
= `tan^-1 5/12 + tan^-1 4/3`
= `tan^-1 ((5/12 + 4/3)/(1 - 5/12 xx 4/3))` ...`[tan^-1x + tan^-1y = tan^-1 (x + y)/(1 - xy)]`
= `tan^-1 ((15 + 48)/(36 - 20))`
= `tan^-1 63/16`
= L.H.S.
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