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UPSC NDA Math Model Paper 9

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Question : 85 of 120

Marks: +1, -0

The value of the expression

\tan\left(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\right)

is

$2-\sqrt{5}$
$\sqrt{5}-2$
$(\sqrt{5}-2)2$
$5-\sqrt{2}$

Solution:

Given expression is

\tan\left(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\right)

Consider,

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

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

\Rightarrow\cos 2\theta=\frac{2}{\sqrt{5}}

\Rightarrow2\cos^{2}\theta-1=\frac{2}{\sqrt{5}}

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

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

Similarly,

\sin\theta=\sqrt{\frac{1}{2}-\frac{2}{\sqrt{5}}}

We know that

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

\Rightarrow\tan\theta=\frac{\sqrt{\frac{1}{2}-\frac{2}{\sqrt{5}}}}{\sqrt{\frac{1}{2}+\frac{2}{\sqrt{5}}}}

\Rightarrow\tan\theta=\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}

=\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}\times\frac{\sqrt{5}-2}{\sqrt{5}-2}}

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

\Rightarrow\tan\theta=\sqrt{5}-2

Hence, The value of the expression

\tan\left(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\right)

is

\sqrt{5}-2

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