Question

If ${\cos }^{-1}\frac{x}{a}+{\cot }^{-1}\frac{y}{b}=\frac{5\pi }{12}$ and ${\sin }^{-1}\frac{x}{a}-{\sin }^{-1}\frac{y}{b}=\frac{\pi }{12}$ then value of $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is

• A. $3/4$
• B. $5/4$
• C. $1$
• D. $1/2$

Answer 1

The correct option is B $5/4$

$\begin{array}{c}{\cos }^{-1}\frac{x}{a}+{\cos }^{-1}\frac{y}{b}=\frac{5\pi }{12}\to \left(i\right)\\ {\sin }^{-1}\frac{x}{a}-{\sin }^{-1}\frac{y}{b}=\frac{\pi }{12}\to \left(ii\right)\\ Addingequation\left(i\right)and\left(ii\right)\\ \Rightarrow \left({\sin }^{-1}\frac{x}{a}+{\cos }^{-1}\frac{x}{a}\right)+{\cos }^{-1}\frac{y}{b}.{\sin }^{-1}\frac{y}{b}=\frac{6\pi }{12}\\ \Rightarrow \frac{\pi }{2}+\frac{\pi }{2}-{\sin }^{-1}\frac{y}{b}-{\sin }^{-1}\frac{y}{b}=\frac{\pi }{2}\\ \Rightarrow \frac{\pi }{2}-2{\sin }^{-1}\frac{y}{b}=0\left[\because {\sin }^{-1}\theta +{\cos }^{-1}\theta =1\right]\\ \Rightarrow {\sin }^{-1}\frac{y}{b}=\frac{\pi }{4}\\ \therefore \frac{y}{b}=\frac{1}{\sqrt{2}}\\ Putting{\sin }^{-1}\frac{y}{b}=\frac{\pi }{4}inequation\left(ii\right)\\ \Rightarrow {\sin }^{-1}\frac{x}{a}-\frac{\pi }{4}=\frac{\pi }{12}\\ \Rightarrow {\sin }^{-1}\frac{x}{a}=\frac{\pi }{12}=\frac{3\pi }{12}\\ \Rightarrow {\sin }^{-1}\frac{x}{a}=\frac{\pi }{3}\\ \therefore \frac{x}{a}=\frac{\sqrt{3}}{2}\\ Hence,\frac{x^2}{a^2}+\frac{y^2}{12}=\frac{1}{2}+\frac{3}{4}=\frac{5}{4}Ans.\end{array}$