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If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$.

HINT: The answer is $-2\pm \sqrt5$.

What I have tried so far: $$\tan{\frac{x}{2}} = \frac{1}{\sin x}-\sin x$$ $$\tan{\frac{x}{2}} = \frac{1-\sin^2 x}{\sin x}$$ $$\tan{\frac{x}{2}} = \frac{\cos^2 x}{\sin x}$$

I don't know how to solve this problem. Pls help. Thank you :)

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  • $\begingroup$ Let $u=\frac x2$. $\endgroup$ – Szeto Feb 14 at 9:49
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Notice that $$ \cos(x) = \frac{1 -\tan^2 (x/2)}{1+\tan^2(x/2)} $$ And that $$ \tan^2(x/2) = \frac{\cos^4(x)}{1-\cos^2(x)} $$ Let $t = \cos(x)$. Plug the 2nd equation into the first one and after some algebra, we get $$ (1-t)(1-t^2-t^4) = 0 $$

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$$\cos^2x=\tan\dfrac x2\sin x=\dfrac{\sin\dfrac x2}{\cos\dfrac x2}\cdot2\sin\dfrac x2\cos\dfrac x2=2\sin^2\dfrac x2=1-\cos x$$

$$\implies\cos x=\dfrac{-1\pm\sqrt5}2$$

As for real $x,\cos x\ge-1,\cos x\ne\dfrac{-1-\sqrt5}2<-1$

Using Weierstrass Substitution $$\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}=\cos x=\dfrac{-1+\sqrt5}2$$

Now apply Componendo et Dividendo

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  • $\begingroup$ Please find the updated answer. Please feel free to pinpoint any mistake/doubt $\endgroup$ – lab bhattacharjee Feb 15 at 5:57
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Hint: Use that $$\frac{1}{\sin(x)}-\sin(x)=1/2\,{\frac {1+ \left( \tan \left( x/2 \right) \right) ^{2}}{\tan \left( x/2 \right) }}-2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$

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