A) \[{{\tan }^{2}}\left( \frac{\alpha }{2} \right)\]
B) \[{{\cot }^{2}}\left( \frac{\alpha }{2} \right)\]
C) \[\tan \alpha \]
D) \[\cot \left( \frac{\alpha }{2} \right)\]
Correct Answer: A
Solution :
We know that, \[{{\cot }^{-1}}(\sqrt{\cos \alpha })+{{\tan }^{-1}}(\sqrt{\cos \alpha })=\frac{\pi }{2}\] ... (i) and given that \[{{\cot }^{-1}}\sqrt{\cos \alpha }-{{\tan }^{-1}}\sqrt{\cos \alpha }=x\] ... (ii) On adding Eqs. (i) and (ii), we get \[2{{\cot }^{-1}}(\sqrt{\cos \alpha })=\frac{\pi }{2}+x\] \[\Rightarrow \] \[\sqrt{\cos \alpha }=\cot \left( \frac{\pi }{4}+\frac{x}{2} \right)\] \[\Rightarrow \] \[\sqrt{\cos \alpha }=\frac{\cot \frac{x}{2}-1}{1+\cot \frac{x}{2}}\] \[\Rightarrow \] \[\sqrt{\cos \alpha }=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\] On squaring both sides, we get \[\cos \alpha =\frac{1-\sin x}{1+\sin x}\Rightarrow \frac{1-{{\tan }^{2}}\frac{\alpha }{2}}{1+{{\tan }^{2}}\frac{\alpha }{2}}=\frac{1-\sin x}{1+\sin x}\] Applying componendo and dividendo \[\sin x={{\tan }^{2}}\frac{\alpha }{2}\] Alternate Solution Since, \[{{\cot }^{-1}}(\sqrt{\cos \alpha })-{{\tan }^{-1}}(\sqrt{\cos \alpha })=x\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{1}{\sqrt{\cos \alpha }} \right)-{{\tan }^{-1}}(\sqrt{\cos \alpha })=x\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{\frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}{1+\frac{1}{\sqrt{\cos \alpha }}.\sqrt{\cos \alpha }} \right)=x\] \[\Rightarrow \frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}=\tan x\Rightarrow \cot x=\frac{2\sqrt{\cos \alpha }}{1-\cos \alpha }\] \[\because \] \[\cos ecx=\sqrt{1+{{\cot }^{2}}x}\] \[\therefore \] \[\cos ecx=\frac{1+\cos \alpha }{1-\cos \alpha }\] \[\Rightarrow \] \[\sin x=\frac{1-\cos \alpha }{1+\cos \alpha }={{\tan }^{2}}\frac{\alpha }{2}\]
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