If $\cos (\theta-\alpha), \cos \theta$ and $\cos (\theta+\alpha)$ are in harmonic progressions, then $2 \tan…
If $\cos (\theta-\alpha), \cos \theta$ and $\cos (\theta+\alpha)$ are in harmonic progressions, then $2 \tan ^2 \theta=$
- $\tan ^2 \frac{\alpha}{2}-1$
- $1+\tan ^2 \frac{\alpha}{2}$
- $1+\cot ^2 \frac{\alpha}{2}$
- $1-\cot ^2 \frac{\alpha}{2}$
Solution
Given, $\cos (\theta-\alpha), \cos \theta$ and $\cos (\theta+\alpha$ ) are in harmonic progression
$\therefore \quad \frac{2}{\cos \theta}=\frac{\cos (\theta+\alpha)+\cos (\theta-\alpha)}{\cos (\theta-\alpha) \cos (\theta+\alpha)}$
$\begin{aligned} & \Rightarrow \quad \cos \theta=\frac{2 \cos (\theta-\alpha) \cos (\theta+\alpha)}{\cos (\theta-\alpha)+\cos (\theta+\alpha)} \\ & \frac{2 \cos (\theta-\alpha) \cos (\theta+\alpha)}{2 \cos \theta \cos \alpha}\end{aligned}$
$\begin{aligned} & =\frac{\cos ^2 \theta \cos ^2 \alpha-\sin ^2 \theta \sin ^2 \alpha}{\cos \theta \cos \alpha} \\ & =\frac{\cos ^2 \theta \cos ^2 \alpha-\left(1-\cos ^2 \theta\right) \sin ^2 \alpha}{\cos \theta \cos \alpha} \\ & =\frac{\cos ^2 \theta \cos ^2 \alpha+\cos ^2 \theta \sin ^2 \alpha-\sin ^2 \alpha}{\cos \theta \cos \alpha} \\ & \Rightarrow \cos ^2 \theta \cos \alpha=\cos ^2 \theta-\sin ^2 \alpha \\ & \Rightarrow \cos ^2 \theta=\frac{\sin ^2 \alpha}{1-\cos \alpha}=\frac{4 \sin ^2 \frac{\alpha}{2}}{2 \sin ^2 \frac{\alpha}{2}} \cos ^2 \frac{\alpha}{2}=2 \cos ^2 \frac{\alpha}{2} \\ & \Rightarrow \sin ^2 \theta=1-2 \cos ^2 \frac{\alpha}{2} \\ & \text { Now, } \tan ^2 \theta=\frac{1-2 \cos ^2 \frac{\alpha}{2}}{2 \cos ^2 \frac{\alpha}{2}}=\frac{1}{2} \sec ^2 \frac{\alpha}{2}-1 \\ & \Rightarrow 2 \tan ^2 \theta=\sec ^2 \frac{\alpha}{2}-2 \\ & \Rightarrow 2 \tan ^2 \theta=\tan ^2 \frac{\alpha}{2}-1 .\end{aligned}$
Asked in: AP EAMCET 2023 (16 May Shift 1)
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