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Question Number 114100 by deleteduser12 last updated on 17/Sep/20

$$$$$$If\cos =\frac{a\cos \alpha -b}{a-b\cos \alpha },provethat,$$$$\frac{\tan \frac{1}{2}\theta }{\sqrt{a+b}}=\frac{\tan \frac{1}{2}\alpha }{\sqrt{a+b}}$$

Commented by Henri Boucatchou last updated on 17/Sep/20

$$cos?=\frac{acos\alpha ....}{}$$

Commented by som(math1967) last updated on 17/Sep/20

$$\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{should}\mathrm{be}\cos \theta$$

Answered by som(math1967) last updated on 17/Sep/20

$$\frac{1}{\cos \theta }=\frac{\mathrm{a}-\mathrm{bcos}\alpha }{\mathrm{acos}\alpha -\mathrm{b}}$$$$\frac{1+\cos \theta }{1-\cos \theta }=\frac{\mathrm{a}-\mathrm{b}+(\mathrm{a}-\mathrm{b})\cos \alpha }{(\mathrm{a}+\mathrm{b})-(\mathrm{a}+\mathrm{b})\cos \alpha }$$$$\left[\mathrm{using}\mathrm{componendo}\mathrm{and}\mathrm{dividendo}\right]$$$$\frac{1}{{\tan }^2\frac{\theta }{2}}=\frac{(\mathrm{a}-\mathrm{b})(1+\cos \alpha )}{(\mathrm{a}+\mathrm{b})(1-\cos \alpha )}$$$${\tan }^2\frac{\theta }{2}=\frac{(\mathrm{a}+\mathrm{b})}{(\mathrm{a}-\mathrm{b})}{\tan }^2\frac{\alpha }{2}$$$$\tan \frac{\theta }{2}=\sqrt{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}-\mathrm{b}}}\tan \frac{\alpha }{2}$$$$\frac{\tan \frac{\theta }{2}}{\sqrt{\mathrm{a}+\mathrm{b}}}=\frac{\tan \frac{\alpha }{2}}{\sqrt{\mathrm{a}-\mathrm{b}}}\left(\mathrm{proved}\right)$$

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