Q65 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

If $\theta=\sec ^{-1}(\cos h u)$. then $u=$

  • A
    $\log _e\left(\cot \left(\frac{\theta}{2}-\frac{\pi}{4}\right)\right)$
  • B
    $\log _e\left(\tan \left(\frac{\theta}{2}-\frac{\pi}{4}\right)\right)$
  • C
    $\log _e\left(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)$
  • Correct answer
    D
    $\log _e\left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)$

Solution

$\because \theta=\sec ^{-1}(\cos h u) \Rightarrow \sec \theta=\cos h u$ $\begin{aligned} & \Rightarrow u=\cosh h^{-1}(\sec \theta)=\log _e\left[\sec \theta+\sqrt{\sec ^2 \theta-1}\right] \\ & =\log _e[\sec \theta+\tan \theta]=\log _e\left[\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right]\end{aligned}$ $\begin{aligned} & =\log _e\left[\frac{1+\sin \theta}{\cos \theta}\right]=\log _e\left[\frac{1+\cos \left(\frac{\pi}{2}+\theta\right)}{\sin \left(\frac{\pi}{2}+\theta\right)}\right] \\ & =\log _e\left[\frac{2 \sin ^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)}{2 \sin \left(\frac{\pi}{4}+\frac{\theta}{2}\right) \cdot \cos \left(\frac{\pi}{4}+\frac{\theta}{2}\right)}\right] \\ & =\log _e\left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)\end{aligned}$

Hence, the correct answer is option: 'D'.

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