Find the value of ≤ft(221/2^{})

Correct Answer is : 1+21+{2}1+2

Correct Answer is : 1+21+{2}1+2

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UPSC NDA Math Model Paper 9

Question : 5 of 120

Marks: +1, -0

\cot\left(22\frac{1}{2}^{\circ}\right)

Solution:

As we know that, cot x = cos x / sin x

\Rightarrow \cot\left(22\frac{1}{2}^{\circ}\right) = \frac{\cos\left(22\frac{1}{2}^{\circ}\right)}{\sin\left(22\frac{1}{2}^{\circ}\right)}

As we know that,

\sin\frac{A}{2} = \pm\sqrt{\frac{1-\cos A}{2}}

\Rightarrow \sin\left(\frac{45^{\circ}}{2}\right) = \pm\sqrt{\frac{1-\cos(45^{\circ})}{2}}

As we know that, 0° < A / 2 < 90° where all trigonometric ratios are positive.

\Rightarrow \sin\left(\frac{45^{\circ}}{2}\right) = \sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}

........(1)

As we know that,

\cos\frac{A}{2} = \pm\sqrt{\frac{1+\cos A}{2}}

\Rightarrow \cos\left(\frac{45^{\circ}}{2}\right) = \pm\sqrt{\frac{1+\cos(45^{\circ})}{2}}

.........(2)

As we know that, 0° < A / 2 < 90° where all trigonometric ratio’s are positive.

\Rightarrow \cos\left(\frac{45^{\circ}}{2}\right) = \sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}

So, from equation (1) and (2), we get

\Rightarrow \cot\left(22\frac{1}{2}^{\circ}\right) = \frac{\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}}{\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}}

=1+\sqrt{2}

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