Question

If $\sin \theta =\frac{1}{2},\cos \varphi =1$, where $0<\theta <\frac{\pi }{2}$ and $0<\varphi \le \frac{\pi }{2}$, then $\left(\cot \right(\theta +2\varphi ){)}^2$ is equal to:

Answer 1

$\sin \theta =\frac{1}{2}$

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

$=\sqrt{1-\frac{1}{4}}$

$=\frac{\sqrt{3}}{2}$

$\cos \varphi =1$

$\Rightarrow sin\varphi =\sqrt{1-{\cos }^2\varphi }$

$=\sqrt{1-1}$

$=0$

Now, $\cot \left(\theta +2\varphi \right)$$=\frac{\cos \left(\theta +2\varphi \right)}{\sin \left(\theta +2\varphi \right)}$

$=\frac{\cos \theta \cos 2\varphi -\sin \theta \sin 2\varphi }{\sin \theta \cos 2\varphi +\cos \theta \sin 2\varphi }$

$=\frac{\frac{\sqrt{3}}{2}\left(2{\cos }^2\varphi -1\right)-\frac{1}{2}\times 2\sin \varphi \cos \varphi }{\frac{1}{2}\left(2{\cos }^2\varphi -1\right)+\frac{\sqrt{3}}{2}\times 2\sin \varphi \cos \varphi }$

$=\frac{\frac{\sqrt{3}}{2}\left(2-1\right)-0}{\frac{1}{2}\left(2-1\right)+0}$

$=\sqrt{3}$

Now, ${\left(\cot \left(\theta +2\varphi \right)\right)}^2={\left(\sqrt{3}\right)}^2=3$