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.

Which of the following is not correct?

• A. $\cos \left({\tan }^{-1}\tan 4\right)=-\cos 4$
• B. ${\tan }^{-1}\left(\tan \right(-6\left)\right)=2\pi -6$
• C. ${\cos }^{-1}\left(\cos 10\right)=4\pi -10$
• D. ${\sin }^{-1}\left(\sin 12\right)+{\cos }^{-1}\left(\cos 12\right)=2\pi$

Answer 1

Answer: (D)

${\sin }^{-1}\left(\sin 12\right)+{\cos }^{-1}\left(\cos 12\right)=2\pi$

A: ${tan}^{-1}tanx=\{-\pi -xxϵ[-\frac{3\pi }{2},-\frac{\pi }{2}]xxϵ[-\frac{\pi }{2},\frac{\pi }{2}]x-\pi xϵ[\frac{\pi }{2},\frac{3\pi }{2}]\backslash 4={229}^0\Rightarrow \frac{\pi }{2}\le 4<\frac{3\pi }{2}cos\left({tan}^{-1}tan4\right)=cos(\pi -4)=-cos4$
B: ${tan}^{-1}tan(-6)={tan}^{-1}tan(2\pi -6)=(2\pi -6)$ as $(2\pi -6)ϵ[-\frac{\pi }{2},\frac{\pi }{2}]$
C: ${cos}^{-1}cos\left(10\right)={cos}^{-1}cos(4\pi -10)=(4\pi -10)$ as $(4\pi -10)ϵ(0,\pi )$
D: ${sin}^{-1}sin\left(12\right)+{cos}^{-1}cos\left(12\right)={sin}^{-1}sin(12-4\pi )+{cos}^{-1}cos(4\pi -12)=12-4\pi +4\pi -12=0$
as $(12-4\pi )ϵ[-\frac{\pi }{2},\frac{\pi }{2}]$ and $(4\pi -12)ϵ(0,\pi )$
Hence, option 'D' is correct.