Question

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

Answer 1

Equation $\sec \left(\mathrm{\theta }+\alpha \right)+\sec \left(\mathrm{\theta }-\alpha \right)=2\sec \mathrm{\theta }$ can be written as

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

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

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

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