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Question Number 187330 by Humble last updated on 16/Feb/23

$$$$$$showthat\frac{1-cos\theta }{1+cos\theta }={tan}^2\left(\frac{1}{2}\theta \right)$$

Answered by Frix last updated on 16/Feb/23

$$\mathrm{What}\mathrm{are}\mathrm{we}\mathrm{allowed}\mathrm{to}\mathrm{use}?$$$${\tan }^{2}x=\frac{{\sin }^{2}x}{{\cos }^{2}x}=\frac{\frac{1-\cos 2x}{2}}{\frac{1+\cos 2x}{2}}=\frac{1-\cos 2x}{1+\cos 2x}$$$$\Rightarrow$$$$\frac{1-\cos \theta }{1+\cos \theta }=\frac{\frac{1-\cos \theta }{2}}{\frac{1+\cos \theta }{2}}=\frac{{\sin }^2\frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}}={\tan }^2\frac{\theta }{2}$$$$\mathrm{or}$$$$\cos \theta =\frac{{\mathrm{e}}^{\mathrm{i}\theta }+{\mathrm{e}}^{-\mathrm{i}\theta }}{2}=\frac{1+{\mathrm{e}}^{2\mathrm{i}\theta }}{{2\mathrm{e}}^{\mathrm{i}\theta }}$$$$\sin \theta =\frac{1-{\mathrm{e}}^{2\mathrm{i}\theta }}{{2\mathrm{e}}^{\mathrm{i}\theta }}\mathrm{i}$$$$\frac{1-\cos \theta }{1+\cos \theta }=\frac{-\frac{{(1-{\mathrm{e}}^{\mathrm{i}\theta })}^2}{{2\mathrm{e}}^{\mathrm{i}\theta }}}{\frac{{(1+{\mathrm{e}}^{\mathrm{i}\theta })}^2}{{2\mathrm{e}}^{\mathrm{i}\theta }}}=-{\left(\frac{1-{\mathrm{e}}^{\mathrm{i}\theta }}{1+{\mathrm{e}}^{\mathrm{i}\theta }}\right)}^2=$$$$={\left(\frac{1-{\mathrm{e}}^{\mathrm{i}\theta }}{1+{\mathrm{e}}^{\mathrm{i}\theta }}\mathrm{i}\right)}^2={\left(\frac{1-{\mathrm{e}}^{2\mathrm{i}\frac{\theta }{2}}}{1+{\mathrm{e}}^{2\mathrm{i}\frac{\theta }{2}}}\mathrm{i}\right)}^2={\left(\frac{\frac{1-{\mathrm{e}}^{2\mathrm{i}\frac{\theta }{2}}}{{2\mathrm{e}}^{i\frac{\theta }{2}}}\mathrm{i}}{\frac{1+{\mathrm{e}}^{2\mathrm{i}\frac{\theta }{2}}}{{2\mathrm{e}}^{i\frac{\theta }{2}}}}\right)}^2=$$$$={\left(\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\right)}^2={\tan }^2\frac{\theta }{2}$$

Commented by Humble last updated on 16/Feb/23

$$awesome!!.thankyou,sir$$

Commented by Frix last updated on 16/Feb/23

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