JEE MAIN - Mathematics (2002 - No. 15)
$${\cot ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x,$$ then sin x is equal to :
$${\tan ^2}\left( {{\alpha \over 2}} \right)$$
$${\cot ^2}\left( {{\alpha \over 2}} \right)$$
$$\tan \alpha $$
$$cot\left( {{\alpha \over 2}} \right)$$
Explanation
Given that,
$${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) - {\tan ^{ - 1}}\left( \sqrt{\cos x} \right) = x\,\,\,\,...\left( 1 \right)$$
We know,
$${\cot ^{ - 1}}x + {\tan ^{ - 1}}x = {\pi \over 2}$$
$$\therefore$$ $$\,\,\,$$ $${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) + {\tan ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {\pi \over 2}\,\,\,...\left( 2 \right)$$
Adding $$(1)$$ and $$(2),$$ we get,
$$2{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = x + {\pi \over 2}$$
$$ \Rightarrow \,\,{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {x \over 2} + {\pi \over 4}$$
$$ \Rightarrow \,\,\sqrt {\cos x} = \cot \left( {{x \over 2} + {\pi \over 4}} \right)$$
$$ \Rightarrow \,\,\sqrt {\cos x} = {{\cot {x \over 2} - 1} \over {1 + \cot {x \over 2}}}$$
$$ \Rightarrow \,\,\sqrt {\cos x} = {{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}}$$
Squaring both sides we get,
$$ \Rightarrow \,\,\cos x = {{1 - 2\sin {x \over 2}\cos {x \over 2}} \over {1 + 2\sin {x \over 2}\cos {x \over 2}}}$$
$$ \Rightarrow \,\,\cos x = {{1 - \sin x} \over {1 + \sin x}}$$
$$ \Rightarrow \,\,{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}} = {{1 - \sin x} \over {1 + \sin x}}$$
Applying compounds and dividendo rule,
$$ \Rightarrow \,\,{{2\sin x} \over 2} = {{2{{\tan }^2}{x \over 2}} \over 2}$$
$$ \Rightarrow \,\,\sin x = {\tan ^2}{x \over 2}$$
Other Method :
$$ \begin{aligned} & \text { Since, } \cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}\right)=x \\\\ & \Rightarrow \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=\tan x \\\\ & \Rightarrow \cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha} \\\\ & \because \operatorname{cosec} x=\sqrt{1+\cot { }^2 x} \\\\ & \therefore \operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha} \\\\ & \Rightarrow \sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\tan ^2 \frac{\alpha}{2} \end{aligned} $$
$${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) - {\tan ^{ - 1}}\left( \sqrt{\cos x} \right) = x\,\,\,\,...\left( 1 \right)$$
We know,
$${\cot ^{ - 1}}x + {\tan ^{ - 1}}x = {\pi \over 2}$$
$$\therefore$$ $$\,\,\,$$ $${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) + {\tan ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {\pi \over 2}\,\,\,...\left( 2 \right)$$
Adding $$(1)$$ and $$(2),$$ we get,
$$2{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = x + {\pi \over 2}$$
$$ \Rightarrow \,\,{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {x \over 2} + {\pi \over 4}$$
$$ \Rightarrow \,\,\sqrt {\cos x} = \cot \left( {{x \over 2} + {\pi \over 4}} \right)$$
$$ \Rightarrow \,\,\sqrt {\cos x} = {{\cot {x \over 2} - 1} \over {1 + \cot {x \over 2}}}$$
$$ \Rightarrow \,\,\sqrt {\cos x} = {{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}}$$
Squaring both sides we get,
$$ \Rightarrow \,\,\cos x = {{1 - 2\sin {x \over 2}\cos {x \over 2}} \over {1 + 2\sin {x \over 2}\cos {x \over 2}}}$$
$$ \Rightarrow \,\,\cos x = {{1 - \sin x} \over {1 + \sin x}}$$
$$ \Rightarrow \,\,{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}} = {{1 - \sin x} \over {1 + \sin x}}$$
Applying compounds and dividendo rule,
$$ \Rightarrow \,\,{{2\sin x} \over 2} = {{2{{\tan }^2}{x \over 2}} \over 2}$$
$$ \Rightarrow \,\,\sin x = {\tan ^2}{x \over 2}$$
Other Method :
$$ \begin{aligned} & \text { Since, } \cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}\right)=x \\\\ & \Rightarrow \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=\tan x \\\\ & \Rightarrow \cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha} \\\\ & \because \operatorname{cosec} x=\sqrt{1+\cot { }^2 x} \\\\ & \therefore \operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha} \\\\ & \Rightarrow \sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\tan ^2 \frac{\alpha}{2} \end{aligned} $$
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