Question

The number of real values of $x$ satisfying the equation $\frac{{\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3}{{\left({\tan }^{-1}x+{\cot }^{-1}x\right)}^3}=7$, is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

As, ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}\forall x\in R$
So, the given equation can be written as
${\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3=7\left(\frac{{\pi }^3}{8}\right)$
$\Rightarrow {\left({\sin }^{-1}x+{\cos }^{-1}x\right)}^3-3\left({\sin }^{-1}x\right)\left({\cos }^{-1}x\right)\left({\sin }^{-1}x+{\cos }^{-1}x\right)=7{\left(\frac{\pi }{2}\right)}^3$
$\Rightarrow {\left(\frac{\pi }{2}\right)}^3-3\left({\sin }^{-1}x\right)\left({\cos }^{-1}x\right)\frac{\pi }{2}=7{\left(\frac{\pi }{2}\right)}^3$
$\Rightarrow \left({\sin }^{-1}x\right)\left({\cos }^{-1}x\right)=\frac{-{\pi }^2}{2}$
Now, as maximum value of ${\cos }^{-1}x$ is $\pi \left({\cos }^{-1}x\ge 0\right)$ and minimum value of ${\sin }^{-1}x$ is $\frac{-\pi }{2}$ and this happens at the same $x$ i.e., at $x=-1$.
Hence, $x=-1$ is the only solution.