If (x+ )+ (x- )=2 , prove that - 2 2

Question

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

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We have, $\sec(\theta+\alpha)+\sec(\theta-\alpha)=2\sec\theta.$ $\Rightarrow\frac{1}{\cos^2\theta.\cos^2\alpha-\sin^\theta\sin^2\alpha}+\frac{1}{\cos\theta.\cos\alpha-\sin\theta\sin\alpha}=\frac{2}{\cos\theta}$ $\Rightarrow\frac{2\cos\theta\cos\alpha}{\cos^2\cos^2\alpha-\sin^2\sin^2\alpha}=\frac{2}{\cos\theta}$ $\Rightarrow\frac{2\cos\theta\cos\alpha}{\cos^2\theta\cos^2\alpha-(\cos^2\alpha+\sin^2\alpha)}=\frac{1}{\cos\theta}$ $\Rightarrow\cos^2\theta\cos\alpha=\cos^2\theta(\cos^2\alpha+\sin^2\alpha)-\sin^2\alpha$ $\Rightarrow\cos^2\theta(1-\cos\alpha)=\sin^2\alpha$ $\Rightarrow\cos^2\theta=\frac{\sin^2\alpha}{2\sin^2\frac{\alpha}{2}}$ $=\frac{1\sin^2\frac{\alpha}{2}.\cos^2\frac{\alpha}{2}}{2\sin^2\frac{\alpha}{2}}$ $\Rightarrow\cos\theta=\pm\sqrt{2}\cos\frac{\alpha}{2}$

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