Question

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

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

Answer 1

Answer: (D)

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