If sec (x+α)+sec (x−α)=2 sec x,

Solution:

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

$\frac{1}{\cos (x+\alpha)}+\frac{1}{\cos (x-\alpha)}=\frac{2}{\cos x}$

$\Rightarrow \frac{1}{\cos x \times \cos \alpha-\sin x \times \sin \alpha}+\frac{1}{\cos x \times \cos \alpha+\sin x \times \sin \alpha}=\frac{2}{\cos x}$

$[\because \cos (A+B)=\cos A \times \cos B-\sin A \times \sin B$ and $\cos (A-B)=\cos A \times \cos B+\sin A \times \sin B]$

$\Rightarrow \frac{2 \cos x \times \cos \alpha}{\cos ^{2} x \times \cos ^{2} \alpha-\sin ^{2} x \times \sin ^{2} \alpha}=\frac{2}{\cos x}$

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

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

$\Rightarrow \frac{\cos ^{2} x \times \cos \alpha}{\cos ^{2} x \times \cos ^{2} \alpha-\sin ^{2} \alpha+\cos ^{2} x \sin ^{2} \alpha}=1$

$\Rightarrow \cos ^{2} x \times \cos \alpha=\cos ^{2} x \times \cos ^{2} \alpha-\sin ^{2} \alpha+\cos ^{2} x \sin ^{2} \alpha$

$\Rightarrow \cos ^{2} x \times \cos \alpha=\cos ^{2} x\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)-\sin ^{2} \alpha$

$\Rightarrow \cos ^{2} x \times \cos \alpha=\cos ^{2} x-\sin ^{2} \alpha$

$\Rightarrow \cos ^{2} \mathrm{x} \times \cos \alpha-\cos ^{2} \mathrm{x}=-\sin ^{2} \alpha$

$\Rightarrow \cos ^{2} x(\cos \alpha-1)=-\sin ^{2} \alpha$

$\Rightarrow \cos ^{2} x(1-\cos \alpha)=\sin ^{2} \alpha$

$\Rightarrow \cos ^{2} \mathrm{x}=\frac{\sin ^{2} \alpha}{2 \sin ^{2} \frac{\alpha}{0}} \quad\left(\because 2 \sin ^{2} \frac{\mathrm{x}}{2}=1-\cos x\right)$

$\Rightarrow \cos ^{2} x=\frac{4 \sin ^{2} \frac{\alpha}{2} \times \cos ^{2} \frac{\alpha}{2}}{2 \sin ^{2} \frac{\alpha}{2}} \quad\left(\because \sin ^{2} x=4 \sin ^{2} \frac{x}{2} \times \cos ^{2} \frac{x}{2}\right)$

$\Rightarrow \cos x=\pm \sqrt{2} \cos \frac{\alpha}{2}$

Hence proved.