Question

If $\tan \frac{x}{2}=\frac{\sqrt{1-e}}{1+e}\tan \frac{\alpha }{2}$, then $\cos \alpha =$

(a) $1-e\cos \left(\cos x+e\right)$

(b) $\frac{1+e\cos x}{\cos x-e}$

(c) $\frac{1-e\cos x}{\cos x-e}$

(d) $\frac{\cos x-e}{1-e\cos x}$

Answer 1

(d) $\frac{\cos x-e}{1-e\cos x}$

$$\begin{gathered} \mathrm{Given}:\tan \frac{x}{2}=\sqrt{\frac{1-\mathrm{e}}{1+\mathrm{e}}}\tan \frac{\mathrm{\alpha }}{2} \\ \Rightarrow \frac{\tan {\displaystyle \frac{x}{2}}}{\tan {\displaystyle \frac{\mathrm{\alpha }}{2}}}=\sqrt{\frac{1-\mathrm{e}}{1+\mathrm{e}}} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}, \\ \frac{{\tan }^2{\displaystyle \frac{x}{2}}}{{\tan }^2{\displaystyle \frac{\mathrm{\alpha }}{2}}}=\frac{1-\mathrm{e}}{1+\mathrm{e}} \\ \Rightarrow {\tan }^2\frac{\mathrm{\alpha }}{2}\left(1-\mathrm{e}\right)={\tan }^2\frac{x}{2}\left(1+\mathrm{e}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{{\sin }^2{\displaystyle \frac{\mathrm{\alpha }}{2}}}{{\cos }^2{\displaystyle \frac{\mathrm{\alpha }}{2}}}\left(1-\mathrm{e}\right)=\frac{{\sin }^2{\displaystyle \frac{x}{2}}}{{\cos }^2{\displaystyle \frac{x}{2}}}\left(1+\mathrm{e}\right) \\ \Rightarrow \frac{{\displaystyle \frac{1}{2}}\left(1-\mathrm{cos\alpha }\right)}{{\displaystyle \frac{1}{2}}\left(1+\mathrm{cos\alpha }\right)}\left(1-\mathrm{e}\right)=\frac{{\displaystyle \frac{1}{2}}\left(1-\cos x\right)}{{\displaystyle \frac{1}{2}}\left(1+\cos x\right)}\left(1+\mathrm{e}\right) \\ \Rightarrow \left(1-\mathrm{cos\alpha }\right)\left(1+\cos x\right)\left(1-\mathrm{e}\right)=\left(1+\mathrm{cos\alpha }\right)\left(1-\cos x\right)\left(1+\mathrm{e}\right) \\ \Rightarrow \left(1+\cos x\right)\left(1-\mathrm{e}\right)-\mathrm{cos\alpha }\left(1+\cos x\right)\left(1-\mathrm{e}\right)=\left(1-\cos x\right)\left(1+\mathrm{e}\right)+\mathrm{cos\alpha }\left(1-\cos x\right)\left(1+\mathrm{e}\right) \\ \Rightarrow \mathrm{cos\alpha }\left\{\left(1+\cos x\right)\left(1-\mathrm{e}\right)+\left(1-\cos x\right)\left(1+\mathrm{e}\right)\right\}=\left(1+\cos x\right)\left(1-\mathrm{e}\right)-\left(1-\cos x\right)\left(1+\mathrm{e}\right) \\ \Rightarrow \mathrm{cos\alpha }=\frac{2\cos x-2\mathrm{e}}{2-2\mathrm{ecos}x}=\frac{\cos x-\mathrm{e}}{1-\mathrm{ecos}x} \end{gathered}$$