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If we are given

$u=\cot^{-1}\sqrt{\cos 2m} - \arctan \sqrt{\cos2m}$

Then we have to prove $\sin u = \tan^2m$

I tried and it as $\tan u = \frac{(1+\cos2m)}{2\sqrt{\cos2m})}$

But got stuck after that .

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    $\begingroup$ There are quite a few proofs involving inverse trigonometric functions tonight. However, the function $\tan(\cdot)$ is not a linear transformation. In other words, you can't just take the tangent of both pieces. $\endgroup$ – Decaf-Math Dec 21 '16 at 6:50
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$$u =\cot^{-1} \sqrt{\cos 2m} -\arctan\sqrt{\cos 2m}$$ $$\Rightarrow u = \frac{\pi}{2}-2\arctan\sqrt{\cos 2m}$$ $$\Rightarrow \tan(\frac{\frac{\pi}{2}-u}{2}) = \sqrt{\cos 2m}$$ $$\Rightarrow \tan^{2} (\frac{\frac{\pi}{2}-u}{2}) = \cos 2m =\frac{1-\tan^{2}m}{1+\tan^{2}m}$$ Using componendo and dividendo, we can easily get $$\cos(\frac{\pi}{2}-u) = \tan^{2}m \Rightarrow \sin u=\tan^{2}m $$ Hope it helps.

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As $\sqrt{\cos2m}\ge0$

using Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

$u=$arccot$\sqrt{\cos2m}-\arctan\sqrt{\cos2m}=\arctan\dfrac1{\sqrt{\cos2m}}-\arctan\sqrt{\cos2m}$ $=\arctan\dfrac{1-\cos2m}{2\sqrt{\cos2m}}$

$\tan u=\dfrac{1-\cos2m}{2\sqrt{\cos2m}}$

As $\dfrac{1-\cos2m}{2\sqrt{\cos2m}}\ge0,0\le u\le\dfrac\pi2;$

$\cos u=+\dfrac1{\sqrt{1+\tan^2u}}=?$

$\sin u=\tan u\cdot\cos u=?$

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