Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $-1$

We have, $ta{n}^{-1}x+co{t}^{-1}x=\frac{\pi }{2}$

The given equation can be written as :
$(ta{n}^{-1}x+co{t}^{-1}x{)}^2-2ta{n}^{-1}x(\frac{\pi }{2}-ta{n}^{-1}x)=\frac{5{\pi }^2}{8}$

$2(ta{n}^{-1}x{)}^2-2\left(\frac{\pi }{2}\right)ta{n}^{-1}(x)-\frac{3{\pi }^2}{8}=0$

$ta{n}^{-1}x=-\frac{\pi }{4}or\frac{3\pi }{4}$

Hence, $ta{n}^{-1}x=1$
$x=-1$