If the solutions of $\cos p θ + \cos q θ = 0$ are in A.P., then the common difference of A.P. is

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\frac{2 π}{p + q} or \frac{2 π}{p - q}

\frac{π}{2 (p + q)} or \frac{π}{2 (p - q)}

\frac{2 π}{p + q} or \frac{π}{2 (p - q)}

\frac{p}{p + q} or \frac{p}{p - q}

answer is B.

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Detailed Solution

\cos p θ + \cos q θ = 0

$2. \cos (\frac{p + q}{2}) θ . \cos (\frac{p - q}{2}) θ = 0$

$\cos (\frac{p + q}{2}) θ = 0 (o r) \cos (\frac{p - q}{2}) θ = 0$

$∴ (\frac{p + q}{2}) θ = (2 n + 1) \frac{π}{2} (o r) (\frac{p - q}{2}) θ = (2 n + 1) \frac{π}{2}$

When $∴ θ = \frac{(2 n + 1) π}{p + q}$

Sequence is $\frac{3 π}{p + q}, \frac{5 π}{p + q}, \frac{7 π}{p + q} ...$ common difference is $\frac{2 π}{p + q}$

When $∴ θ = \frac{(2 n + 1) π}{p - q}$

Sequence is $\frac{3 π}{p - q}, \frac{5 π}{p - q}, \frac{7 π}{p - q} ...$ common difference is $\frac{2 π}{p - q}$

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