Question 383 of 519HardMCQ
Suppose θ and φ (=0) are such that sec(θ+φ),secθ,sec(θ−φ) are in A.P. If cosθ=kcos2φ for some k, then k is equal to: Solutions · 1
Official solution
A.P. condition: 2secθ=sec(θ−φ)+sec(θ+φ), i.e.
\[\frac{2}{\cos\theta}=\frac{\cos(\theta+\varphi)+\cos(\theta-\varphi)}{\cos(\theta-\varphi)\cos(\theta+\varphi)}=\frac{2\cos\theta\cos\varphi}{\cos^2\theta-\sin^2\varphi}.\]
\[\Rightarrow \cos^2\theta(1-\cos\varphi)=\sin^2\varphi=1-\cos^2\varphi \Rightarrow \cos^2\theta=1+\cos\varphi=2\cos^2\frac{\varphi}{2}.\]
\[\cos\theta=\pm\sqrt{2}\cos\frac{\varphi}{2} \Rightarrow k=\pm\sqrt{2}.\] Answer the question to unlock