Question

If $\sec \left(x+\alpha \right)+\sec \left(x-\alpha \right)=2\sec x$, prove that $\cos x=\pm \sqrt{2}\cos \frac{\alpha }{2}$

Answer 1

Equation $\sec \left(x+\alpha \right)+\sec \left(x-\alpha \right)=2\sec x$ can be written as

$$\begin{gathered} \frac{1}{\cos \left(x+\mathrm{\alpha }\right)}+\frac{1}{\cos \left(x-\mathrm{\alpha }\right)}=\frac{2}{\cos x} \\ \Rightarrow \frac{1}{\cos x\times \mathrm{cos\alpha }-\sin x\times \mathrm{sin\alpha }}+\frac{1}{\cos x\times \mathrm{cos\alpha }+\sin x\times \mathrm{sin\alpha }}=\frac{2}{\cos x}\left[\because \cos \left(A+B\right)=\cos A\times \cos B-\sin A\times \sin B\mathrm{and}\cos \left(A-B\right)=\cos A\times \cos B+\sin A\times \sin B\right] \\ \Rightarrow \frac{2\cos x\times \mathrm{cos\alpha }}{{\cos }^{2}x\times {\cos }^2\mathrm{\alpha }-{\sin }^{2}x\times {\sin }^2\mathrm{\alpha }}=\frac{2}{\cos x} \\ \Rightarrow \frac{\cos x\times \mathrm{cos\alpha }}{{\cos }^{2}x\times {\cos }^2\mathrm{\alpha }-\left(1-{\cos }^{2}x\right)\times {\sin }^2\mathrm{\alpha }}=\frac{1}{\cos x} \end{gathered}$$

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

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

$$\begin{gathered} \Rightarrow {\cos }^2\mathrm{x}=\frac{4{\sin }^2{\displaystyle \frac{\mathrm{\alpha }}{2}}\times {\cos }^2{\displaystyle \frac{\mathrm{\alpha }}{2}}}{2{\sin }^2{\displaystyle \frac{\mathrm{\alpha }}{2}}}\left(\because {\sin }^{2}x=4{\sin }^2\frac{x}{2}\times {\cos }^2\frac{x}{2}\right) \\ \Rightarrow \cos x=\pm \sqrt{2}\cos \frac{\mathrm{\alpha }}{2} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$