Question

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

• A. 0
• B. -1
• C. -2
• D. -3

Answer 1

Answer: (B)

-1

As we know that

${\text{tan}}^{-1}x+{\text{cot}}^{-1}x=\frac{\pi }{2}⟹{\text{cot}}^{-1}x=\frac{\pi }{2}-{\text{tan}}^{-1}x$

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

Let ${\text{tan}}^{-1}x=y$

$⟹y^2+(\frac{\pi }{2}-y{)}^2=\frac{5{\pi }^2}{8}$

$⟹y^2+\frac{{\pi }^2}{4}+y^2-\pi y=\frac{5{\pi }^2}{8}$

$⟹2y^2-\pi y-\frac{3{\pi }^2}{8}=0$

$⟹y=\frac{-(-\pi )\pm \sqrt{(-\pi {)}^2+3{\pi }^2}}{4}=\frac{\pi }{4}\pm \frac{\pi }{2}=\frac{3\pi }{4},-\frac{\pi }{4}$

$⟹x=\tan \frac{3\pi }{4},\tan (-\frac{\pi }{4})=-1,-1$

So, $x=-1$