Sequences And Series Ques 103
If $\cos (x-y), \cos x$ and $\cos (x+y)$ are in HP. Then $\cos x \cdot \sec \frac{y}{2}=\ldots$.
(1997C, 2M)
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Answer:
Correct Answer: 103.$(\pm \sqrt{2})$
Solution:
Formula:
HARMONICAL PROGRESSION (H.P.)](/important-formula/mathematics/sequence_and_series)
Since, $\cos (x-y), \cos x$ and $\cos (x+y)$ are in AP.
$ \begin{aligned} & \therefore \quad \cos x=\frac{2 \cos (x-y) \cos (x+y)}{\cos (x-y)+\cos (x+y)} \\ & \Rightarrow \quad \cos x(2 \cos x \cdot \cos y)=2\{\cos ^{2} x+\sin ^{2} y \} \\ & \Rightarrow \quad \cos ^{2} x \cdot \cos y=\cos ^{2} x \cos y-\sin ^{2} y \ & \Rightarrow \quad \cos ^{2} x(1-\cos y)=\sin ^{2} y \\ & \Rightarrow \quad \cos ^{2} x \cdot 2 \sin ^{2} \frac{y}{2}=2 \sin ^{2} \frac{y}{2} \cdot \cos ^{2} \frac{y}{2} \\ & \Rightarrow \quad \cos ^{2} x \cdot \sec ^{2} \frac{y}{2}=2 \\ & \therefore \quad \cos x \cdot \sec \frac{y}{2}= \pm \sqrt{2} \end{aligned} $
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