Question

$1+co{t}^2\left({\sin }^{-1}x\right)=$

• A. $\frac{1}{2x}$
• B. $x^2$
• C. $\frac{1}{x^2}$
• D. $\frac{2}{x}$

Answer 1

The correct option is C

$\frac{1}{x^2}$

Explanation for the correct option:

Simplify using trigonometric identities

Let ${\sin }^{-1}x=\theta$

$\Rightarrow \sin \theta =x$

Put the value of ${\sin }^{-1}x$ in the given expression:

$1+co{t}^2\left({\sin }^{-1}x\right)=1+co{t}^2\theta$

$=\cos e{c}^2\theta =\frac{1}{{\sin }^2\theta }$ $\left[Since,1+co{t}^2\theta =\cos e{c}^2\theta ;\cos e{c}^2\theta =\frac{1}{{\sin }^2\theta }\right]$

Put the value of $\sin \theta$:

$\Rightarrow 1+co{t}^2\left({\sin }^{-1}x\right)=\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}$

Hence, Option ‘C’ is Correct.