Question

The simplified expression of sin ($ta{n}^{−1}x$), for any real number $x$ is given by

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

Answer 1

Answer: (B)

Let tan${}^{−1}$ x = $θ$
$⇒x=tanθ=\frac{sinθ}{cosθ}=\frac{sinθ}{\sqrt{1−si{n}^2θ}}$
$⇒x=\sqrt{1−si{n}^2θ}=sinθ$
$⇒x^2\left(1−si{n}^2θ\right)=si{n}^2θ$
$⇒x^2=si{n}^2θ\left(1+x^2\right)$
$⇒si{n}^2θ=\frac{x^2}{1+x^2}⇒sinθ=\frac{x}{\sqrt{1+x^2}}$
$⇒θ=si{n}^{−1}\frac{x}{\sqrt{1+x^2}}$
$⇒ta{n}^{−1}x=si{n}^{−1}\frac{x}{\sqrt{1+x^2}}$
Now, sin $\left(ta{n}^{−1}x\right)=sin\left(si{n}^{−1}\frac{x}{\sqrt{1+x^2}}\right)$
$=\frac{x}{\sqrt{1+x^2}}$