Browse Questions

# Evaluate $\cot^{-1} \left ( \large \frac {{\sqrt {1+\sin x}}+{\sqrt {1-\sin x}}}{{\sqrt {1+\sin x}}-{\sqrt {1-\sin x}}} \right ) = \large \frac {x}{2}$$, x \in \left ( 0, \frac{\pi}{4} \right ) \begin{array}{1 1} 0 \\ 1 \\ \large\frac{x}{2} \\ \large\frac{-x}{2}\end{array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • $sin^2\frac{x}{2}+cos^2\frac{x}{2}=1$ • $(a+b)^2=a^2+b^2+2ab$ • $sinx=2sin\frac{x}{2}.cos\frac{x}{2}$ • $(a-b)^2=a^2+b^2-2ab$ • If $x \in \bigg( 0 \frac{\pi}{4} \bigg)$ then take $\sqrt{1-sinx} = cos\frac{x}{2}-sin\frac{x}{2}$ • If $x \in \bigg( \frac{\pi}{4} \frac{\pi}{2} \bigg)$ thentake $\sqrt{1-sinx}=sin\frac{x}{2}-cos\frac{x}{2}$ Given cot^{-1} \left ( \Large \frac {{\sqrt {1+sinx}}+{\sqrt {1-sinx}}}{{\sqrt {1+sinx}}-{\sqrt {1-sinx}}} \right ) = \large \frac {x}{2}$$ , x \in \left ( 0, \frac{\pi}{4} \right )$
We know that $sin^2\frac{x}{2}+cos^2\frac{x}{2} = 1$ and $sinx=2\sin\frac{x}{2}.cos\frac{x}{2}$
$\Rightarrow 1+sinx=sin^2\frac{x}{2}+cos^2\frac{x}{2}+2sin\frac{x}{2}cos\frac{x}{2} = (cos\frac{x}{2}+sin\frac{x}{2})^2$
$\Rightarrow 1-sinx=sin^2\frac{x}{2}+cos^2\frac{x}{2}-2sin\frac{x}{2}cos\frac{x}{2} = (cos\frac{x}{2}-sin\frac{x}{2})^2$
$\sqrt{1+sinx}=\sqrt{(cos\frac{x}{2}+sin\frac{x}{2})^2}=cos\frac{x}{2}+sin\frac{x}{2}$
$\sqrt{1-sinx}=\sqrt{(cos\frac{x}{2}-sin\frac{x}{2})^2}=cos\frac{x}{2}-sin\frac{x}{2}$
$\sqrt{1+sinx} + \sqrt{1-sinx}= cos\frac{x}{2}+sin\frac{x}{2}+cos\frac{x}{2}-sin\frac{x}{2} = 2 cos \frac{x}{2}$
$\sqrt{1+sinx} - \sqrt{1-sinx}= cos\frac{x}{2}+sin\frac{x}{2}- (cos\frac{x}{2}-sin\frac{x}{2}) = 2 sin \frac{x}{2}$