Question

If ${\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=ax$, $xϵ(0,\frac{\pi }{4})$.
Find the value of $a$.

• A. $a=0$
• B. $a=\frac{1}{2}$
• C. $a=1$
• D. $a=2$

Answer 1

The correct option is B $a=\frac{1}{2}$

Consider $\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$

$=\frac{{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}^2}{{\left(\sqrt{1+\sin x}\right)}^2-{\left(\sqrt{1-\sin x}\right)}^2}$ (by rationalizing)

$=\frac{(1+\sin x)+(1-\sin x)+2\sqrt{(1+\sin x)(1-\sin x)}}{1+\sin x-1+\sin x}$

$=\frac{2(1+\sqrt{1-{\sin }^{2}x})}{2\sin x}$

$=\frac{1+\cos x}{\sin x}$

$=\frac{2{\cos }^2\frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}$ $=\cot \frac{x}{2}$

So, ${\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)={\cot }^{-1}\left(\cot \frac{x}{2}\right)=\frac{x}{2}$

Also, $\frac{x}{2}=ax$ ...Given

Hence, $a=\frac{1}{2}$