Question

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

Answer 1

We have,

$sec(\theta +\alpha )+sec(\theta -\alpha )=2sec\theta$

$\Rightarrow \frac{1}{cos\theta .cos\alpha -sin\theta sin\alpha }+\frac{1}{cos\theta .cos\alpha +sin\theta sin\alpha }=\frac{2}{cos\theta }\Rightarrow \frac{2cos\theta cos\alpha }{co{s}^2\theta co{s}^2\alpha -si{n}^2\theta si{n}^2\alpha }=\frac{2}{cos\theta }\Rightarrow \frac{cos\theta cos\alpha }{co{s}^2\theta co{s}^2\alpha -(1-co{s}^2\theta )si{n}^2\alpha }=\frac{1}{cos\theta }\Rightarrow co{s}^2\theta cos\alpha =co{s}^2\theta (co{s}^2\alpha +si{n}^2\alpha )=si{n}^2\alpha \Rightarrow co{s}^2\theta (1-cos\alpha )=si{n}^2\alpha \Rightarrow co{s}^2\theta =\frac{si{n}^2\alpha }{2si{n}^2\frac{\alpha }{2}}=\frac{4si{n}^2\frac{\alpha }{2}.co{s}^2\frac{\alpha }{2}}{2si{n}^2\frac{\alpha }{2}}\Rightarrow cos\theta =\pm \sqrt{2}cos\frac{\alpha }{2}$