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 $\pi\left(\cos ^{-1} x \geq 0\right)$ 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.