Question
Assertion: If $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$,then $\sin x=\tan ^{2} \frac{\alpha}{2}$Reason: $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$
Step 1
We can rewrite this using the identity $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ as follows: \[\tan ^{-1}\left(\frac{1/\sqrt{\cos \alpha}-\sqrt{\cos \alpha}}{1+(1/\sqrt{\cos \alpha})\cdot \sqrt{\cos \alpha}}\right)=x\] Show more…
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