Question

If $(ta{n}^{-1}x{)}^2+(co{t}^{-1}x{)}^2=\frac{5{\pi }^2}{8}$, then x is equal to

• A. 0
• B. 2
• C. 1
• D. -1

Answer 1

The correct option is D

-1

Given,
$(ta{n}^{-1}x{)}^2+(co{t}^{-1}x{)}^2=\frac{5{\pi }^2}{8}$
$\therefore (ta{n}^{-1}x+co{t}^{-1}x{)}^2-2ta{n}^{-1}x(\frac{\pi }{2}-ta{n}^{-1}x)\frac{5{\pi }^2}{8}$
$\Rightarrow \frac{{\pi }^2}{4}-2\times \frac{\pi }{2}ta{n}^{-1}x-2(ta{n}^{-1}x{)}^2=\frac{5{\pi }^2}{8}$
$\Rightarrow 2(ta{n}^{-1}x{)}^2-\pi ta{n}^{-1}x-\frac{3{\pi }^2}{8}=0$
$\Rightarrow ta{n}^{-1}x=-\frac{\pi }{4},\frac{3\pi }{4}$
$\Rightarrow x=-1$