Question

Let $f\left(x\right)={\sin }^{-1}\left(\tan x\right)+{\cos }^{-1}\left(\cot x\right)$ then

• A. $f\left(x\right)=\frac{\pi }{2}$ wherever defined
• B. domain of $f\left(x\right)$ is $x=n\pi \pm \frac{\pi }{4},n\in I$
• C. period of $f\left(x\right)$ is $\frac{\pi }{2}$
• D. $f\left(x\right)$ is many one function
• E. All of these

Answer 1

The correct option is E All of these

$f\left(x\right)=si{n}^{-1}\left(tanx\right)+co{s}^{-1}\left(cotx\right)$

$=ta{n}^{-1}\left(\frac{tanx}{\sqrt{1-ta{n}^{2}x}}\right)+ta{n}^{-1}\left(\frac{\sqrt{-1-co{t}^{2}x}}{cotx}\right)$

$=ta{n}^{-1}\left(\frac{tanx}{\sqrt{1-ta{n}^{2}x}}\right)+ta{n}^{-1}\left(\frac{\sqrt{ta{n}^{2}x-1}}{tanx}\right)$

As inside square root, quantity $\ge 0$

$\Rightarrow 1-ta{n}^{2}x>0$ and $ta{n}^{2}x-1>0$

$\Rightarrow ta{n}^{2}x\le 1$ and $ta{n}^{2}x\ge 1$

$\Rightarrow ta{n}^{2}x=1\Rightarrow tanx=\pm 1$

$\Rightarrow x=n\pi \pm \pi /4$

$f\left(x\right)=\pi /2+0=\pi /2$

As $f\left(x\right)=\pi /2$ at $\pi /4,3\pi /4,5\pi /\mathrm{4...}\Rightarrow$ Period $=\pi /2$

As $f\left(\pi /4\right)=f\left(3\pi /4\right)=\pi /2\Rightarrow f$ is many - one

$\Rightarrow$ All are correct