Question

If $\sec (\theta +\alpha )+\sec (\theta -\alpha )=2\sec \theta$, then ${\cos }^2\theta =a+\cos \alpha$ Find a

Answer 1

$\sec (\theta +\alpha )+\sec (\theta -\alpha )=2\sec \theta$
or $\frac{1}{\cos (\theta +\alpha )}+\frac{1}{\cos (\theta -\alpha )}\frac{2}{\cos \theta }$
or $\frac{\cos (\theta -\alpha )+\cos (\theta +\alpha )}{\cos (\theta +\alpha )\cos (\theta -\alpha )}=\frac{2}{\cos \theta }$
or $\frac{2\cos \theta \cos \alpha }{{\cos }^2\theta -{\sin }^2\alpha }=\frac{2}{\cos \theta }$
or $2{\cos }^2\theta \cos \alpha ={\cos }^2\theta -{\sin }^2\alpha$
or ${\sin }^2\alpha ={\cos }^2\theta (1-\cos \alpha )$
or $1-{\cos }^2\alpha ={\cos }^2\theta (1-\cos \alpha )$ or $1+\cos \alpha ={\cos }^2\theta$
Ans: 1