Question

If ${\sin }^{-1}\left(\frac{x}{13}\right)+{\csc }^{-1}\left(\frac{13}{12}\right)=\frac{\pi }{2}$, then the value of $x$ is

• A. $5$
• B. $4$
• C. $12$
• D. $11$

Answer 1

The correct option is A $5$

Given, ${\sin }^{-1}\left(\frac{x}{13}\right)+{\csc }^{-1}\left(\frac{13}{12}\right)=\frac{\pi }{2}$ .....(i)
Let ${\csc }^{-1}\frac{13}{12}=y$
Then, $\csc y=\frac{13}{12}\Rightarrow \sin y=\frac{12}{13}$
$\therefore \cos y=\sqrt{1-{\sin }^{2}y}$
$=\sqrt{1-{\left(\frac{12}{13}\right)}^2}=\sqrt{1-\frac{144}{169}}$
$=\sqrt{\frac{25}{169}}=\frac{5}{13}\Rightarrow y={\cos }^{-1}\frac{5}{13}$
Equation (i) becomes
${\sin }^{-1}\left(\frac{x}{13}\right)+{\cos }^{-1}\left(\frac{5}{13}\right)=\frac{\pi }{2}$ ........(i)
We know that
${\sin }^{-1}\theta +{\cos }^{-1}\theta =\frac{\pi }{2}$
$\therefore$ Both angles of equation (i) should be same.
$\therefore x=5$