TS EAMCET · Maths · Inverse Trigonometric Functions
Fог \(\theta \in\left(0, \frac{\pi}{2}\right)\), \(\operatorname{sech}^{-1}(\cos \theta)\) is equal to
- A \(\log \left|\tan \left(\frac{\pi}{6}+\frac{\theta}{2}\right)\right|\)
- B \(\log \left|\tan \left(\frac{\pi}{3}+\frac{\theta}{2}\right)\right|\)
- C \(\log \left|\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right|\)
- D \(\log \left|\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right|\)
Answer & Solution
Correct Answer
(C) \(\log \left|\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right|\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \because \sec h^{-1}(x)=\log \left(\frac{1+\sqrt{1-x^2}}{x}\right) \\ \therefore \sec h^{-1}(\cos \theta) & =\log \left(\frac{1+\sqrt{1-\cos ^2 \theta}}{\cos \theta}\right) \\ & =\log \left(\frac{1+\sin \theta}{\cos \theta}\right) \end{aligned}\)…
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