Q65 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Mathematics

If $\theta=\frac{\pi}{6}$ and $x=\log \left[\cot \left(\frac{\pi}{4}+\theta\right)\right]$, then $\sin h(x)=$

  • A
    $\sqrt{3}$
  • B
    $\frac{1}{\sqrt{3}}$
  • Correct answer
    C
    $-\sqrt{3}$
  • D
    $-\frac{1}{\sqrt{3}}$

Solution

$$ \begin{aligned} & \text { At } \theta=\frac{\pi}{6}, x=\log \left[\cot \left(\frac{\pi}{4}+\theta\right)\right] \\ & \Rightarrow \quad x=\log \left[\cot \left(\frac{\pi}{4}+\frac{\pi}{6}\right)\right] \\ & \Rightarrow \quad e^x=\frac{\cot \pi / 4 \cot \pi / 6-1}{\cot \pi / 6+\cot \pi / 4} \\ & \Rightarrow \quad e^x=\frac{\sqrt{3}-1}{\sqrt{3}+1} \\ & \therefore \quad e^{-x}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \\ & \because \quad \sin h(x)=\frac{e^x-e^{-x}}{2}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}+1}-\frac{\sqrt{3}+1}{\sqrt{3}-1}}{2} \\ & =\frac{(3+1-2 \sqrt{3})-(3+1+2 \sqrt{3})}{4}=-\frac{4 \sqrt{3}}{4}=-\sqrt{3} \\ & \end{aligned} $$ Hence, option (c) is correct.

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

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