TS EAMCET · Maths · Differentiation
\(\frac{d}{d x} \tan ^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right]\) is equal to
- A 1
- B \(-\frac{1}{2}\)
- C \(\frac{1}{2}\)
- D -1
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Let } y=\tan ^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right] \\ & =\tan ^{-1}\left[\frac{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)-\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\sin \frac{x}{2}+\cos…
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