Question

In its domain of the function $f\left(x\right)=\frac{{\sin }^{-1}x}{{\cos }^{-1}x}$ is

• A. an increasing function
• B. a strictly decreasing function
• C. a decreasing function
• D. a strictly increasing function

Answer 1

The correct option is D a strictly increasing function

$f\left(x\right)=\frac{{\sin }^{-1}x}{{\cos }^{-1}x}$

$D\left({\sin }^{-1}x\right)=[-1,1]$

$D\left({\cos }^{-1}x\right)=[-1,1]$

$D_f=(-1,1)({\cos }^{-1}x\ne 0)$

$f^{\prime }\left(x\right)={\cos }^{-1}x\times \frac{1}{\sqrt{1-x^2}}+{\sin }^{-1}x\times \frac{1}{\sqrt{1-x^2}}$

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${\left({\cos }^{-1}x\right)}^2$

$f^{\prime }\left(x\right)=\frac{1}{\sqrt{1-x^2}}\times \frac{({\cos }^{-1}x+{\sin }^{-1}x)}{{\left({\cos }^{-1}x\right)}^2}$

$=\frac{\pi }{2\sqrt{1-x^2}\times {\left({\cos }^{-1}x\right)}^2}>0$

Hence,

$f\left(x\right)$ is a strictly increasing

function.