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Question: Answered & Verified by Expert
Prove that $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}$
MathematicsInverse Trigonometric Functions
Solution:
2657 Upvotes Verified Answer
Let $\sin ^{-1} \mathrm{x}=\alpha, \quad \therefore \sin \alpha=\mathrm{x}$,
$\cos \alpha=\sqrt{1-x^2}, \quad \sin ^{-1} y=\beta$
$\therefore \quad \sin \beta=\mathrm{y}, \quad \cos \beta=\sqrt{1-\mathrm{y}^2}$,
$\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$
$=x \sqrt{1-y^2}+\sqrt{1-x^2} y$
$\alpha+\beta=\sin ^{-1}\left[x \sqrt{1-y^2}+y \sqrt{1-x^2}\right]$
i.e. $\sin ^{-1} x+\sin ^{-1} y$
$=\sin ^{-1}\left[x \sqrt{1-y^2}+y \sqrt{1-x^2}\right]$
$\therefore \sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{77}{85}$.

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