Question

If $\cos x+\cos y+\cos \alpha =0$ and $\sin x+\sin y+\sin \alpha =0$, then $cot\frac{(x+y)}{2}=$

• A. $\sin \alpha$
• B. $\cos \alpha$
• C. $cot\alpha$
• D. $\sin \frac{(x+y)}{2}$

Answer 1

The correct option is C

$cot\alpha$

Explanation for the correct option:

Step 1. Find the value of $cot\frac{(x+y)}{2}$:

Given, $\cos x+\cos y+\cos \alpha =0$

$\Rightarrow$ $\cos x+\cos y=-\cos \alpha$

$\Rightarrow$$2\cos \frac{\left(x+y\right)}{2}\cos \frac{\left(x-y\right)}{2}=-\cos \alpha$ …..(i)

Also, $\sin x+\sin y+\sin \alpha =0$

$\Rightarrow$ $\sin x+\sin y=-\sin \alpha$

$\Rightarrow$$2\sin \frac{\left(x+y\right)}{2}\cos \frac{\left(x-y\right)}{2}=-\sin \alpha$ …..(ii)

Step 2. Divide equation (i) by (ii), we get

$\frac{2\cos \frac{\left(x+y\right)}{2}\cos \frac{\left(x-y\right)}{2}}{2\sin \frac{\left(x+y\right)}{2}\cos \frac{\left(x-y\right)}{2}}=\frac{-\cos \alpha }{-\sin \alpha }$

$\mathbf{\therefore }\mathit{c}\mathit{o}\mathit{t}\mathbf{}\frac{\left(x+y\right)}{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathit{\alpha }$

Hence, Option ‘C’ is Correct.