Question

If ${\cot }^{-1}\left[\sqrt{\cos \alpha }\right]-{\tan }^{-1}\left[\sqrt{\cos \alpha }\right]=x$, then $\sin x$ is equal to

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

Answer 1

Answer: (A)

${\tan }^2\left(\frac{\alpha }{2}\right)$

Given equation: ${cot}^{-1}\left(\sqrt{cos\alpha }\right)-{tan}^{-1}\left(\sqrt{cos\alpha }\right)=x$

$let{cot}^{-1}\left(\sqrt{cos\alpha }\right)=P$

$⟹cotP=\sqrt{cos\alpha }$

$⟹sinP=\frac{1}{\sqrt{1+cos\alpha }}$

Also ${tan}^{-1}\left(\sqrt{cos\alpha }\right)=Q$

$⟹tanQ=\sqrt{cos\alpha }$

$⟹sinQ=\frac{\sqrt{cos\alpha }}{\sqrt{1+cos\alpha }}$

Now the given equation can be written as

$x=P-Q$

$sinx=sin(P-Q)$

$sinx=sinPcosQ-cosPsinQ$

$sinx=\left(\frac{1}{\sqrt{1+cos\alpha }}\right)\left(\frac{1}{\sqrt{1+cos\alpha }}\right)-\left(\frac{\sqrt{cos\alpha }}{\sqrt{1+cos\alpha }}\right)\left(\frac{\sqrt{cos\alpha }}{\sqrt{1+cos\alpha }}\right)$

$sinx=\frac{1-cos\alpha }{1+cos\alpha }=\frac{2{sin}^2\frac{\alpha }{2}}{2{cos}^2\frac{\alpha }{2}}sinx={tan}^2\frac{\alpha }{2}$

Hence option (a) is correct.