Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Since, $\sec (\theta -\varphi ),\sec \theta$ and $\sec (\theta +\varphi )$ are in A.P., $$\begin{array}{rl}& \therefore 2\sec \theta =\sec (\theta -\varphi )+\sec (\theta +\varphi )\\ & \Rightarrow \frac{2}{\cos \theta }=\frac{\cos (\theta +\varphi )+\cos (\theta -\varphi )}{\cos (\theta -\varphi )\cos (\theta +\varphi )}\end{array}$$ $$\begin{array}{rl}& \Rightarrow 2\left({\cos }^2\theta -{\sin }^2\varphi \right)=\cos \theta [2\cos \theta \cos \varphi \\ & \Rightarrow {\cos }^2\theta (1-\cos \varphi )={\sin }^2\varphi =1-{\cos }^2\varphi \\ & \Rightarrow {\cos }^2\theta =1+\cos \varphi =2{\cos }^2\frac{\varphi }{2}\\ & \therefore \cos \theta =\pm \sqrt{2}\cos \frac{\varphi }{2}\\ & \text{ But given }\cos \theta =\mathrm{k}\cos \frac{\varphi }{2}\\ & \therefore \mathrm{k}=\pm \sqrt{2}\end{array}$$