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KCET · Maths · Inverse Trigonometric Functions

The value of \(\cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]\), where \(x \in\left(0, \frac{\pi}{4}\right)\) is

  1. A \(\frac{x}{2}-\pi\)
  2. B \(\pi-\frac{x}{3}\)
  3. C \(\pi-\frac{x}{2}\)
  4. D \(\frac{x}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\pi-\frac{x}{2}\)

Step-by-step Solution

Detailed explanation

Here,
\(\cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1-\sin x}}\right]\)
\(\begin{aligned} & =\cot ^{-1}\left[\frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})^2}{(1-\sin x)-(1+\sin x)}\right] \\ & =\cot ^{-1}\left(\frac{1-\sin x+1+\sin x+2 \sqrt{1-\sin ^2 x}}{1-\sin x-1-\sin x}\right) \\ & \Rightarrow \cot ^{-1}\left(\frac{2+2 \sqrt{\cos ^2 x}}{-2 \sin x}\right) \\ & \Rightarrow \cot ^1\left(\frac{1+\cos x}{-\sin x}\right)\end{aligned}\)
\(\begin{gathered}\Rightarrow \cot ^{-1}\left(\frac{1+2 \cos ^2 \frac{x}{2}-1}{-2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) \\ \Rightarrow \cot ^1\left(\frac{-\cos \frac{x}{2}}{\sin \frac{x}{2}}\right) \\ \Rightarrow \cot ^{-1}\left(-\cot \frac{x}{2}\right)=\cot ^{-1}\left(\cot \left(\pi-\frac{x}{2}\right)\right) \\ =\pi-\frac{x}{2}\end{gathered}\)