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Question Number 189468 by cortano12 last updated on 17/Mar/23

$$\mathrm{If}\tan \left(\frac{\mathrm{x}}{2}\right)=\csc \mathrm{x}-\sin \mathrm{x},\mathrm{then}$$$$\tan {}^2\left(\frac{\mathrm{x}}{2}\right)=?$$

Commented by mehdee42 last updated on 17/Mar/23

$$thegivenrelationiswrong.$$$$tan\left(\frac{x}{2}\right)=cscx-cotx$$

Commented by cortano12 last updated on 17/Mar/23

$$\mathrm{why}\mathrm{wrong}?$$

Commented by Frix last updated on 17/Mar/23

$$\mathrm{Why}\mathrm{do}\mathrm{you}\mathrm{think}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{relation}?$$

Commented by mehdee42 last updated on 17/Mar/23

$$sorry.itwasmymistake.thisisanequation.$$

Answered by horsebrand11 last updated on 17/Mar/23

$$\tan \left(\frac{x}{2}\right)=\frac{1+\tan {}^2\left(\frac{x}{2}\right)}{2\tan \left(\frac{x}{2}\right)}-\frac{2\tan \left(\frac{x}{2}\right)}{1+\tan {}^2\left(\frac{x}{2}\right)}$$$$let\tan \left(\frac{x}{2}\right)=p\Rightarrow p^2=\tan {}^2\left(\frac{x}{2}\right)$$$$\Rightarrow p=\frac{1+p^2}{2p}-\frac{2p}{1+p^2}$$$$\Rightarrow p=\frac{{(1+p^2)}^2-{\left(2p\right)}^2}{2p(1+p^2)}$$$$\Rightarrow 2p^2(1+p^2)=p^4-2p^2+1$$$$letk=p^2,k>0$$$$\Rightarrow 2k(1+k)=k^2-2k+1$$$$\Rightarrow 2k+2k^2=k^2-2k+1$$$$\Rightarrow k^2+4k-1=0$$$$\Rightarrow k=p^2=\frac{-4+\sqrt{20}}{2}=\sqrt{5}-2$$$$$$

Answered by Frix last updated on 17/Mar/23

$$\tan \frac{x}{2}=\csc x-\sin x$$$$t=\frac{t^2+1}{2t}-\frac{2t}{t^2+1}$$$$t=\frac{{(t^2-1)}^2}{2t(t^2+1)}$$$$t^4+4t^2-1=0$$$$t^2=-2+\sqrt{5}(\mathrm{for}t\in \mathbb{R})$$

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