Question

If $\cos (\theta -\alpha ),cos\theta ,cos(\theta +\alpha )$ are in H.P. and $\cos \alpha \ne 1$, then the angle $\alpha$ cannot lie in the

• A. $Iquadrant$
• B. $IIquadrant$
• C. $IIIquadrant$
• D. $IVquadrant$

Answer 1

Answer: (D)

$IVquadrant$

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

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

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

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

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

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

${\cos }^2\alpha -\cos \alpha \left({\cos }^2\theta \right)+{\cos }^2\theta =0$

$\cos \alpha =\frac{{\cos }^2\theta \pm \sqrt{{\cos }^4\theta -4{\cos }^2\theta +4}}{2}$

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

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

$\Rightarrow \cos \alpha =1$

$\cos \alpha =\cos 0$

$\alpha =0$

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

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

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

$\Rightarrow \cos \alpha$ is always negative

$\Rightarrow \cos \alpha$ does not lie in $1^{st}$ and $4^{th}$ quadrant.