Question

$({\tan }^{-1}x{)}^2+({\cot }^{-1}x{)}^2=\frac{5{\pi }^2}{8}\Rightarrow x=$

• A. -1
• B. 1
• C. 0
• D. $\pi \sqrt{\frac{5}{8}}$

Answer 1

Answer: (A)

-1

Given ,
${\left(ta{n}^{-1}x\right)}^2+{\left(co{t}^{-1}x\right)}^2=\frac{5{\pi }^2}{8}$
As we know that ,
$ta{n}^{-1}x+co{t}^{-1}x=\frac{\pi }{2}$
Squaring on both side , we get
${(ta{n}^{-1}x+co{t}^{-1}x)}^2=\frac{{\pi }^2}{4}$
${\left(ta{n}^{-1}x\right)}^2+{\left(co{t}^{-1}x\right)}^2+2ta{n}^{-1}xco{t}^{-1}x=\frac{{\pi }^2}{4}$
$\frac{5{\pi }^2}{8}+2ta{n}^{-1}xco{t}^{-1}x=\frac{{\pi }^2}{4}$
$2ta{n}^{-1}xco{t}^{-1}x=\frac{-3{\pi }^2}{8}$
$ta{n}^{-1}xco{t}^{-1}x=\frac{-3{\pi }^2}{16}$
$ta{n}^{-1}x[\frac{\pi }{2}-ta{n}^{-1}x]=\frac{-3{\pi }^2}{16}$
Now, Let
$y=ta{n}^{-1}x$
The Above Equation becomes ,
$16y^2-8\pi y-3{\pi }^2=0$
On solving this Quadratic Equation in y , we get
$y=\frac{-\pi }{4},y=\frac{3\pi }{4}$
Also
$y=ta{n}^{-1}x$
$ta{n}^{-1}x=\frac{-\pi }{4}$ $ta{n}^{-1}x=\frac{3\pi }{4}$
So , $x=-1$ $x=-1$