Suppose $ heta$ and $\phi(
eq 0)$ are such that $\sec ( heta+\phi)$, sec $ heta$ and $\sec ( heta-\phi)$ are in A.P. If $\cos heta=k \cos \left(\frac{\phi}{2} ight)$ for some $k$, then $k$ is equal to

Question

Suppose $	heta$ and $\phi(
eq 0)$ are such that $\sec (	heta+\phi)$, sec $	heta$ and $\sec (	heta-\phi)$ are in A.P. If $\cos 	heta=k \cos \left(\frac{\phi}{2}ight)$ for some $k$, then $k$ is equal to

Aicharya · Accepted Answer

Since, $\sec (	heta-\phi), \sec 	heta$ and $\sec (	heta+\phi)$ are in A.P., $ \begin{aligned} & 	herefore 2 \sec 	heta=\sec (	heta-\phi)+\sec (	heta+\phi) \ & \Rightarrow \frac{2}{\cos 	heta}=\frac{\cos (	heta+\phi)+\cos (	heta-\phi)}{\cos (	heta-\phi) \cos (	heta+\phi)} \end{aligned} $ $ \begin{aligned} & \Rightarrow 2\left(\cos ^2 	heta-\sin ^2 \phiight)=\cos 	heta[2 \cos 	heta \cos \phi \ & \Rightarrow \cos ^2 	heta(1-\cos \phi)=\sin ^2 \phi=1-\cos ^2 \phi \ & \Rightarrow \quad \cos ^2 	heta=1+\cos \phi=2 \cos ^2 \frac{\phi}{2} \ & 	herefore \quad \cos 	heta=\pm \sqrt{2} \cos \frac{\phi}{2} \ & 	ext { But given } \cos 	heta=\mathrm{k} \cos \frac{\phi}{2} \ & 	herefore \quad \mathrm{k}=\pm \sqrt{2} \end{aligned} $

Suppose $\theta$ and $\phi(\neq 0)$ are such that $\sec (\theta+\phi)$, sec $\theta$ and $\sec…

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