TS EAMCET · Maths · Inverse Trigonometric Functions
\(\operatorname{sech}^{-1}(\sin \theta)\) is equal to
- A \(\log \tan \frac{\theta}{2}\)
- B \(\log \sin \frac{\theta}{2}\)
- C \(\log \cos \frac{\theta}{2}\)
- D \(\log \cot \frac{\theta}{2}\)
Answer & Solution
Correct Answer
(D) \(\log \cot \frac{\theta}{2}\)
Step-by-step Solution
Detailed explanation
We have, sech \({ }^{-1}(\sin \theta)\) \(\begin{aligned} & =\cos h^{-1}(\operatorname{cosec} \theta) \\ & =\log \left[\operatorname{cosec} \theta+\sqrt{\left(\operatorname{cosec}^2 \theta-1\right)}\right] \\ & =\log \cot \frac{\theta}{2}\end{aligned}\)
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