Question

Let $f:A\to B$ be a function defined by $y=f\left(x\right)$ where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping $g:B\to A$ such that $f\left(x\right)=y$ if and only if $g\left(y\right)=x\forall xϵA,yϵB$ Then function g is said to be inverse of f and vice versa so we write $g=f^{-1}:B\to A[\left\{f\right(x),x\}:\{x,f(x\left)\right\}ϵf^{-1}]$when branch of an inverse function is not given (define) then we consider its principal value branch.

If $x<\frac{-1}{\sqrt{3}}$, then the value of $3{\tan }^{-1}x-{\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$ equals?

• A. $-\pi$
• B. $\pi$
• C. $\frac{-\pi }{2}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is A $-\pi$

Let $x=tany$
$x<-\frac{1}{\sqrt{3}}\Rightarrow tany<-\frac{1}{\sqrt{3}}\Rightarrow secy(siny+\frac{cosy}{\sqrt{3}})<0\Rightarrow \frac{sin2y}{cos2y+1}<0\Rightarrow -\frac{\pi }{2}<y<-\frac{\pi }{6}\Rightarrow -\frac{3\pi }{2}<3y<-\frac{\pi }{2}\Rightarrow -\frac{\pi }{2}<\pi +3y<\frac{\pi }{2}$
Therefore,
$tan3y=\frac{3tany-{tan}^{3}y}{1-3{tan}^{2}y}=\frac{3x-x^3}{1-3x^2}\Rightarrow tan(\pi +3y)=\frac{3x-x^3}{1-3x^2}\Rightarrow \pi +3y={tan}^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)\Rightarrow 3{tan}^{-1}x-{tan}^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)=-\pi$