First slide

Trigonometric transformations

Question

If $\cos (x - y), \cos x and \cos (x + y)$ are in H.P., then $|\cos x \sec \frac{y}{2}|$ equal to

Moderate

A

1
B

2
C

$\sqrt{2}$
D

none of these

Solution

It is given that
$\cos (x - y), \cos x and \cos (x + y)$ are in H.P
$∴ \frac{2}{\cos x} = \frac{1}{\cos (x - y)} + \frac{1}{\cos (x + y)}$
$\begin{array}{l} \Rightarrow \frac{2}{\cos x} = \frac{2 \cos x \cos y}{\cos^{2} x - \sin^{2} y} \\ \Rightarrow \cos^{2} x \cos y = \cos^{2} x - \sin^{2} y \\ \Rightarrow \cos^{2} x (1 - \cos y) = \sin^{2} y \\ \Rightarrow 2 \cos^{2} x \sin^{2} \frac{y}{2} = 4 \sin^{2} \frac{y}{2} \cos^{2} \frac{y}{2} \\ \Rightarrow \cos^{2} x \sec^{2} \frac{y}{2} = 2 \\ \Rightarrow \cos x \sec \frac{y}{2} ∣= \sqrt{2} \end{array}$

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