Question

The value of $cot22{\frac{1}{2}}^{\circ }$is equal to

• A. $1+\frac{1}{\sqrt{2}}$
• B. $1+\sqrt{2}$
• C. $\sqrt{2}-1$
• D. None of these

Answer 1

The correct option is B

$1+\sqrt{2}$

Explanation for the correct option:

Find the required value

Given: $cot22{\frac{1}{2}}^{\circ }$

Let $$\theta ={45}^{\circ }$$
And$${\displaystyle \frac{\theta }{2}}={\displaystyle \frac{45}{2}}=22{{\displaystyle \frac{1}{2}}}^{\circ }$$
As we know that $\cot x=\frac{\cos x}{\sin x}$
So, $$\cot {\displaystyle \frac{\theta }{2}}={\displaystyle \frac{\cos {\displaystyle \frac{\theta }{2}}}{\sin {\displaystyle \frac{\theta }{2}}}}$$
And we can write it as,
$\Rightarrow \cot \frac{\theta }{2}=\frac{[2\cos {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}]}{[2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}]}$

$\Rightarrow \cot \frac{\theta }{2}=\frac{\left(2{\cos }^2{\displaystyle \frac{\theta }{2}}\right)}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}$

$\Rightarrow \cot \frac{\theta }{2}=\frac{1+\cos \theta }{\sin \theta }$ $[\because \sin 2\theta =2\sin \theta \cos \theta$and $\cos 2\theta =2{\cos }^2\theta -1]$
Now, substituting the values of $$\theta$$ and $${\displaystyle \frac{\theta }{2}}$$ , we get,
$$\begin{gathered} \cot 22{\frac{1}{2}}^{\circ }=\frac{(1+\cos {45}^{\circ })}{\sin {45}^{\circ }} \\ =\frac{\left(1+\frac{1}{\sqrt{2}}\right)}{\frac{1}{\sqrt{2}}} \\ =\frac{\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)}{\frac{1}{\sqrt{2}}} \\ =1+\sqrt{2} \end{gathered}$$

Therefore, the value of is $cot22{\frac{1}{2}}^{\circ }$equal to $1+\sqrt{2}$.

Hence, option (B), $1+\sqrt{2}$ is the correct answer.