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 $x<0$ then ${\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}$equals?

• A. $\frac{-\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{-\pi }{2}$
• D. $\pi$

Answer 1

The correct option is C $\frac{-\pi }{2}$

Consider, ${\tan }^{-1}x+{\tan }^{-1}\left(\frac{1}{x}\right)$
$={\tan }^{-1}x+(-\pi )+{\cot }^{-1}x$ ($\because {\tan }^{-1}\left(\frac{1}{x}\right)=\left\{\begin{array}{ccc}co{t}^{-1}x,& if& x>0\\ -\pi +{\cot }^{-1}x& if& x<0\end{array}\right\}$)
$={\tan }^{-1}x+{\cot }^{-1}x-\pi$
$=\frac{\pi }{2}-\pi$
$=\frac{-\pi }{2}$