Q102 / 160 · TS EAMCET 2014 · Mathematics

$\sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$ equals to

  • A
    $\log _e(3(2+\sqrt{3}))$
  • B
    $\log _e\left(\frac{1+\sqrt{3}}{3}\right)$
  • Correct answer
    C
    $\log _e\left(\frac{2+\sqrt{3}}{3}\right)$
  • D
    $\log _e\left(\frac{2-\sqrt{3}}{3}\right)$

Solution

$\begin{aligned} & \because \sec h^{-1} x=\log _e\left(\frac{1+\sqrt{1-x^2}}{x}\right) \\ & \text { and cosec } h^{-1} x=\log _e\left(\frac{1+\sqrt{1+x^2}}{x}\right) \\ & \therefore \sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h\left(\frac{3}{4}\right) \\ & =\log _e\left(\frac{1+\sqrt{1-\left(\frac{1}{2}\right)^2}}{\frac{1}{2}}\right)-\log _e\left(\frac{1+\sqrt{1+\frac{9}{16}}}{\frac{3}{4}}\right) \\ & =\log _e\left(\frac{1+\sqrt{\frac{3}{2}}}{\frac{1}{2}}\right)-\log _e\left(\frac{1+\frac{5}{4}}{\frac{3}{4}}\right) \\ & =\log _e(2+\sqrt{3})-\log _e\left(\frac{9}{3}\right) \\ & =\log _e(2+\sqrt{3})-\log _e(3) \\ & =\log _e \frac{(2+\sqrt{3})}{3}\end{aligned}$

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

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