Suppose θ and φ(≠ 0) are such that sec (θ+φ), sec θ and sec (θ-φ) are in A.P. If cos θ=k cos ( φ 2 ) for some k, then k is equal to

Since, $\sec (\theta-\phi), \sec \theta$sec(θ−ϕ),secθ and $\sec (\theta+\phi)$sec(θ+ϕ) are in A.P., $$\begin{aligned} & \therefore 2 \sec \theta=\sec (\theta-\phi)+\sec (\theta+\phi) \\ & \Rightarrow \frac{2}{\cos \theta}=\frac{\cos (\theta+\phi)+\cos (\theta-\phi)}{\cos (\theta-\phi) \cos (\theta+\phi)} \end{aligned}$$​∴2secθ=sec(θ−ϕ)+sec(θ+ϕ)⇒cosθ2​=cos(θ−ϕ)cos(θ+ϕ)cos(θ+ϕ)+cos(θ−ϕ)​​ $$\begin{aligned} & \Rightarrow 2\left(\cos ^2 \theta-\sin ^2 \phi\right)=\cos \theta[2 \cos \theta \cos \phi \\ & \Rightarrow \cos ^2 \theta(1-\cos \phi)=\sin ^2 \phi=1-\cos ^2 \phi \\ & \Rightarrow \quad \cos ^2 \theta=1+\cos \phi=2 \cos ^2 \frac{\phi}{2} \\ & \therefore \quad \cos \theta=\pm \sqrt{2} \cos \frac{\phi}{2} \\ & \text { But given } \cos \theta=\mathrm{k} \cos \frac{\phi}{2} \\ & \therefore \quad \mathrm{k}=\pm \sqrt{2} \end{aligned}$$​⇒2(cos2θ−sin2ϕ)=cosθ[2cosθcosϕ⇒cos2θ(1−cosϕ)=sin2ϕ=1−cos2ϕ⇒cos2θ=1+cosϕ=2cos22ϕ​∴cosθ=±2​cos2ϕ​ But given cosθ=kcos2ϕ​∴k=±2​​