Question

cot inverse{√ 1+ sinx+ √ 1- sinx}/{ √1+sinx-√1-sinx}

Answer 1

Solution for this question

Answer 2

Answer:

The value of the given expression is :

$\bf\frac{x}{2}$

Step-by-step explanation:

$\cot^{-1}(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-sin x}})\\\\\text{Rationalizing the above expression, We get , }\\\\\implies \cot^{-1}(\frac{1+\sin x+1-\sin x+2\sqrt{1-\sin^2x}}{1+\sin x-1+\sin x})\\\\\implies \cot^{-1}(\frac{2+2\cos x}{2\sinx})\\\\\implies \cot^{-1}(\frac{1+\cos x}{\sin x})\\\\\implies \cot^{-1}(\frac{2\cos^2\frac{x}{2}}{2\cos\frac{x}{2}\cdot\sin\frac{x}{2}})\\\\\implies \cot^{-1}(\cot\frac{x}{2})\\\\\implies \frac{x}{2}$

Hence, The value of the given expression is :

$\bf\frac{x}{2}$