CUET Preparation Today
For $|x|<1, \sin (\tan^{-1}x)$ equal to |
$\frac{1}{\sqrt{1+x^2}}$ $\frac{1}{\sqrt{1-x^2}}$ $\frac{x}{\sqrt{1-x^2}}$ $\frac{x}{\sqrt{1+x^2}}$ |
$\frac{x}{\sqrt{1+x^2}}$ |
The correct answer is Option (4) → $\frac{x}{\sqrt{1+x^2}}$ Let $\theta=\tan^{-1}(x)$. $\tan\theta=\frac{x}{1}$ gives a right triangle with: Opposite $=x$, Adjacent $=1$, Hypotenuse $=\sqrt{1+x^{2}}$. Hence, $\sin\theta=\frac{x}{\sqrt{1+x^{2}}}$ $\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^{2}}}$ |
