If (θ-α), θ and (θ+α) are in H.P., then the value of 2θ is

Correct Answer is : 1−4cos⁡2α21-4 ² α/21−4cos22α

Correct Answer is : 1−4cos⁡2α21-4 ² α/21−4cos22α

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MHTCET 14 May 2023 Shift 1 Solved Paper

Section: Mathematics

Question : 66 of 150

Marks: +1, -0

\sin (\theta-\alpha), \sin \theta

\sin (\theta+\alpha)

\cos 2\theta

Solution:

\sin (\theta-\alpha), \sin \theta

and

\sin (\theta+\alpha)

are in H.P.

\Rightarrow \frac{1}{\sin (\theta-\alpha)}, \frac{1}{\sin \theta}, \frac{1}{\sin (\theta+\alpha)}

will be in A.P.

\therefore \frac{2}{\sin \theta} = \frac{1}{\sin (\theta-\alpha)} + \frac{1}{\sin (\theta+\alpha)}

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

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

\Rightarrow \sin^2 \theta - \sin^2 \alpha = \sin^2 \theta \cos \alpha

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

\Rightarrow \sin^2 \theta \left(2 \sin^2 \frac{\alpha}{2}\right) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}

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

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

\Rightarrow 2 \cos^2 \theta -1 = 1-4 \cos^2 \frac{\alpha}{2}

\Rightarrow \cos^2 \theta = 1-4 \cos^2 \frac{\alpha}{2}

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