Question

The number of real solution(s) of the equation $({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3=7({\tan }^{-1}x+{\cot }^{-1}x{)}^3$ is

Answer 1

$({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3=7({\tan }^{-1}x+{\cot }^{-1}x{)}^3$
We know that,
${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}\forall x\in R{\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}\text{for}x\in [-1,1]$
Assuming ${\sin }^{-1}x=t$, we get
$t^3+{(\frac{\pi }{2}-t)}^3=7{\left(\frac{\pi }{2}\right)}^3\Rightarrow {\left(\frac{\pi }{2}\right)}^3-3\times \frac{\pi }{2}\times t(\frac{\pi }{2}-t)=7{\left(\frac{\pi }{2}\right)}^3\Rightarrow 2t(\frac{\pi }{2}-t)=-{\pi }^2\Rightarrow 2t^2-\pi t-{\pi }^2=0\Rightarrow (2t+\pi )(t-\pi )=0\Rightarrow t=-\frac{\pi }{2},\pi$
As ${\sin }^{-1}x\in [-\frac{\pi }{2},\frac{\pi }{2}]$, so $t=-\frac{\pi }{2}$
$\therefore x=-1$

Hence, there is only $1$ solution.