Question

$ sin^{-1} \frac{8}{17} + sin^{-1} \frac{3}{5} $ is equal to

• A. $ sin^{-1}(\frac{77}{85}) $
• B. $ sin^{-1}(\frac{77}{36}) $
• C. $ sin^{-1}(\frac{1}{36}) $
• D. Both (a) and (b)

Answer 1

Answer: (D)

We have, $\sin ^{-1}\left(\frac{8}{17}\right)+\sin ^{-1}\left(\frac{3}{5}\right)$
$=\sin ^{-1}\left\{\frac{8}{17} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \sqrt{1-\left(\frac{8}{17}\right)^{2}}\right\}$
$\left[\because \sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right\}\right]$
$=\sin ^{-1}\left\{\left(\frac{8}{17} \times \frac{4}{5}\right)+\left(\frac{3}{5} \times \frac{15}{17}\right)\right\} $
$=\sin ^{-1}\left\{\frac{32}{85}+\frac{45}{85}\right\}=\sin ^{-1}\left(\frac{77}{85}\right)$
$=\tan ^{-1} \frac{\frac{77}{85}}{\sqrt{1-\left(\frac{77}{85}\right)^{2}}} \left[\because \sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}}\right]$
$=\tan ^{-1}\left(\frac{77}{36}\right)$