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Question Number 10955 by Joel576 last updated on 04/Mar/17

$$\mathrm{If}\frac{\cos \theta }{1-\sin \theta }=aa\ne \frac{\pi }{2}+2k\pi$$$$\mathrm{So},\tan \frac{\theta }{2}=...$$$$\left(\mathrm{A}\right)\frac{a}{a+1}\left(\mathrm{D}\right)\frac{a+1}{a-1}$$$$\left(\mathrm{B}\right)\frac{1}{a+1}\left(\mathrm{E}\right)\frac{a}{a-1}$$$$\left(\mathrm{C}\right)\frac{a-1}{a+1}$$

Answered by ajfour last updated on 04/Mar/17

$$$$$$\frac{\sin (\frac{\pi }{2}-\theta )}{1-\cos (\frac{\pi }{2}-\theta )}=\frac{2\sin (\frac{\pi }{4}-\frac{\theta }{2})\cos (\frac{\pi }{4}-\frac{\theta }{2})}{2\sin {}^2(\frac{\pi }{4}-\frac{\theta }{2})}$$$$=\cot (\frac{\pi }{4}-\frac{\theta }{2})=a$$$$\tan (\frac{\pi }{4}-\frac{\theta }{2})=\frac{1}{a};let\tan \left(\frac{\theta }{2}\right)=p$$$$\frac{1-p}{1+p}=\frac{1}{a}\Rightarrow a-ap=1+p$$$$p=\frac{a-1}{a+1}.$$

Commented by Joel576 last updated on 05/Mar/17

$$thankyouverymuch$$

Answered by ridwan balatif last updated on 04/Mar/17

$$\mathrm{another}\mathrm{way}$$$${\left(\frac{\cos \theta }{1-\sin \theta }\right)}^2={\left(\mathrm{a}\right)}^2$$$$\frac{{\cos }^2\theta }{{(1-\sin \theta )}^2}={\mathrm{a}}^2$$$$\frac{(1-{\sin }^2\theta )}{{(1-\sin \theta )}^2}={\mathrm{a}}^2$$$$\frac{(1-\sin \theta )(1+\sin \theta )}{(1-\sin \theta )(1-\sin \theta )}={\mathrm{a}}^2$$$$\frac{1+\sin \theta }{1-\sin \theta }={\mathrm{a}}^2$$$$\sin \theta =\frac{{\mathrm{a}}^2-1}{{\mathrm{a}}^2+1}$$$$\cos \theta =\frac{2\mathrm{a}}{{\mathrm{a}}^2+1}$$$$\mathrm{remember}:\tan \frac{\theta }{2}=\frac{\sin \theta }{1+\cos \theta }$$$$\tan \frac{\theta }{2}=\frac{\frac{{\mathrm{a}}^2-1}{{\mathrm{a}}^2+1}}{1+\frac{2\mathrm{a}}{{\mathrm{a}}^2+1}}$$$$\tan \frac{\theta }{2}=\frac{\frac{{\mathrm{a}}^2-1}{{\mathrm{a}}^2+1}}{\frac{{\mathrm{a}}^2+2\mathrm{a}+1}{{\mathrm{a}}^2+1}}$$$$\tan \frac{\theta }{2}=\frac{{\mathrm{a}}^2-1}{{\mathrm{a}}^2+2\mathrm{a}+1}$$$$\tan \frac{\theta }{2}=\frac{(\mathrm{a}-1)(\mathrm{a}+1)}{{(\mathrm{a}+1)}^2}$$$$\tan \frac{\theta }{2}=\frac{\mathrm{a}-1}{\mathrm{a}+1}$$$$\mathrm{Answer}:\left(\mathrm{C}\right)$$

Commented by Joel576 last updated on 05/Mar/17

$$thankyouverymuch$$

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