Q62 / 170 · Manipal (MET) 2015 · Mathematics

If $\tan \left(\sec ^{-1} x\right)=\sin \left(\cos ^{-1} \frac{1}{\sqrt{5}}\right)$, then $x$ is equal to

  • Correct answer
    A
    $\pm \frac{3}{\sqrt{5}}$
  • B
    $\pm \frac{\sqrt{5}}{3}$
  • C
    $\pm \sqrt{\frac{3}{5}}$
  • D
    None of these

Solution

We have, $\begin{array}{rlrl} & & \tan \left(\sec ^{-1} x\right) & =\sin \left(\cos ^{-1} \frac{1}{\sqrt{5}}\right) \\ \Rightarrow & & \sqrt{\sec ^2\left(\sec ^{-1} x\right)-1}=\sqrt{1-\cos ^2\left(\cos ^{-1} \frac{1}{\sqrt{5}}\right)} \\ \Rightarrow & & \sqrt{x^2-1}=\sqrt{1-\frac{1}{5}} \\ \Rightarrow & & x^2-1=\frac{4}{5}\end{array}$ $\begin{array}{ll}\Rightarrow & x^2=\frac{9}{5} \\ \Rightarrow & x= \pm \frac{3}{\sqrt{5}}\end{array}$

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

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