Question

If ${\text{cot}}^{-1}\left[{\left(\text{cos}\alpha \right)}^{1/2}\right]-{\text{tan}}^{-1}\left[{\left(\text{cos}\alpha \right)}^{1/2}\right]=x,$ then $\sin x$ equals

• A. ${\tan }^2\left(\frac{\alpha }{2}\right)$
• B. ${\text{cot}}^2\left(\frac{\alpha }{2}\right)$
• C. $\text{tan}\alpha$
• D. $\text{cot}\left(\frac{\alpha }{2}\right)$

Answer 1

The correct option is A ${\tan }^2\left(\frac{\alpha }{2}\right)$

${\text{cot}}^{-1}\left[{\left(\text{cos}\alpha \right)}^{1/2}\right]-{\text{tan}}^{-1}\left[{\left(\text{cos}\alpha \right)}^{1/2}\right]=x$

$\Rightarrow {\text{tan}}^{-1}\left[\frac{1}{\sqrt{\text{cos}\alpha }}\right]-{\text{tan}}^{-1}\left[\sqrt{\text{cos}\alpha }\right]=x$

$\Rightarrow {\text{tan}}^{-1}\left[\frac{\frac{1}{\sqrt{\text{cos}\alpha }}-\sqrt{\text{cos}\alpha }}{1+\frac{\sqrt{\cos \alpha }}{\sqrt{\cos \alpha }}}\right]=x$

$\Rightarrow {\tan }^{-1}\left[\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}\right]=x$

$\Rightarrow \text{tan}x=\frac{1-\text{cos}\alpha }{2\sqrt{\text{cos}\alpha }}$

$\therefore \text{sin}x=\frac{1-\text{cos}\alpha }{1+\text{cos}\alpha }$

$=\frac{2{\text{sin}}^2\frac{\alpha }{2}}{2{\text{cos}}^2\frac{\alpha }{2}}={\text{tan}}^2\left(\frac{\alpha }{2}\right)$