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INVERSE TRIGNOMETRIC FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINVERSE TRIGNOMETRIC FUNCTIONS1 Mark
MCQ
Choose the correct answer from the given four options. The value of the expression $\tan\Big(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\Big)$ is: Hint: $\bigg[\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\bigg]$
  • A
    $2+\sqrt{5}$
  • ✓
    $\sqrt{5}-2$
  • C
    $\frac{\sqrt{5}+2}{2}$
  • D
    $5+\sqrt{2}$
✓

Answer

Correct option: B.
$\sqrt{5}-2$
We have, $\tan\Big(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\Big)$

Let $\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}=\theta$

$\Rightarrow\ \cos^{-1}\frac{2}{\sqrt{5}}=2\theta$

$\Rightarrow\ \cos2\theta=\frac{2}{\sqrt{5}}$

$\therefore\ 2\cos^{2}\theta-1=\frac{2}{\sqrt{5}}$

$\Rightarrow\ \cos^2\theta=\frac{1}{2}+\frac{1}{\sqrt{5}}$

$\Rightarrow\ \cos\theta=\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}}$

$\therefore\ \tan\theta=\frac{\sin\theta}{\cos\theta}$

$=\sqrt{\frac{\frac{1}{2}-\frac{1}{\sqrt{5}}}{\frac{1}{2}+\frac{1}{\sqrt{5}}}}=\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}$

$=\sqrt{\frac{(\sqrt{5}-2)^2}{(\sqrt{5}+2)(\sqrt{5}-2)}}=\sqrt{5}-2$

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