Question

If $\cos (\theta -\alpha ),\cos \theta ,\cos (\theta +\alpha )$ are in H.P., Then prove that ${\cos }^2\theta =1+\cos \alpha$

Answer 1

Given:$\cos (\theta -\alpha ),\cos \theta ,\cos (\theta +\alpha )$ are in H.P

$\Rightarrow \frac{1}{\cos (\theta -\alpha )},\frac{1}{\cos \theta },\frac{1}{\cos (\theta +\alpha )}$ are in A.P

$\frac{1}{\cos \theta }-\frac{1}{\cos (\theta -\alpha )}=\frac{1}{\cos (\theta +\alpha )}-\frac{1}{\cos \theta }$

$\frac{2}{\cos \theta }=\frac{1}{\cos (\theta +\alpha )}+\frac{1}{\cos (\theta -\alpha )}$

$\Rightarrow \frac{2}{\cos \theta }=\frac{\cos (\theta -\alpha )+\cos (\theta +\alpha )}{\cos (\theta +\alpha )\cos (\theta -\alpha )}$

$\Rightarrow \frac{2}{\cos \theta }=\frac{\cos \theta \cos \alpha -\sin \theta \sin \alpha +\cos \theta \cos \alpha +\sin \theta \sin \alpha }{{\cos }^2\theta -{\sin }^2\alpha }$

$\Rightarrow \frac{2}{\cos \theta }=\frac{2\cos \theta \cos \alpha }{{\cos }^2\theta -{\sin }^2\alpha }$

$\Rightarrow \frac{1}{\cos \theta }=\frac{\cos \theta \cos \alpha }{{\cos }^2\theta -{\sin }^2\alpha }$

$\Rightarrow {\cos }^2\theta -{\sin }^2\alpha ={\cos }^2\theta \cos \alpha$

$\Rightarrow {\sin }^2\alpha ={\cos }^2\theta -{\cos }^2\theta \cos \alpha$

$\Rightarrow {\sin }^2\alpha ={\cos }^2\theta (1-\cos \alpha )$

$\Rightarrow \frac{{\sin }^2\alpha }{(1-\cos \alpha )}={\cos }^2\theta$

$\Rightarrow \frac{1-{\cos }^2\alpha }{(1-\cos \alpha )}={\cos }^2\theta$

$\Rightarrow \frac{(1-\cos \alpha )(1+\cos \alpha )}{(1-\cos \alpha )}={\cos }^2\theta$

$\Rightarrow 1+\cos \alpha ={\cos }^2\theta$

Hence ${\cos }^2\theta =1+\cos \alpha$

Hence proved.