If frac{cos x}{a} = frac{cos (x + theta)}{b} | vedclass

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(C) Given,$\frac{\cos x}{a} = \frac{\cos (x + 	heta)}{b} = \frac{\cos (x + 2	heta)}{c} = \frac{\cos (x + 3	heta)}{d} = k$ (let).Then,$a = \frac{\co

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$\frac{b}{c}$

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(C) Given,$\frac{\cos x}{a} = \frac{\cos (x + 	heta)}{b} = \frac{\cos (x + 2	heta)}{c} = \frac{\cos (x + 3	heta)}{d} = k$ (let).Then,$a = \frac{\cos x}{k}, b = \frac{\cos (x + 	heta)}{k}, c = \frac{\cos (x + 2	heta)}{k}, d = \frac{\cos (x + 3	heta)}{k}$.Now,consider the expression $\frac{a + c}{b + d}$:$\frac{a + c}{b + d} = \frac{\frac{1}{k} [\cos x + \cos (x + 2	heta)]}{\frac{1}{k} [\cos (x + 	heta) + \cos (x + 3	heta)]} = \frac{\cos x + \cos (x + 2	heta)}{\cos (x + 	heta) + \cos (x + 3	heta)}$.Using the formula $\cos C + \cos D = 2 \cos \left( \frac{C + D}{2} ight) \cos \left( \frac{C - D}{2} ight)$:Numerator: $\cos x + \cos (x + 2	heta) = 2 \cos \left( \frac{2x + 2	heta}{2} ight) \cos \left( \frac{-2	heta}{2} ight) = 2 \cos (x + 	heta) \cos 	heta$.Denominator: $\cos (x + 	heta) + \cos (x + 3	heta) = 2 \cos \left( \frac{2x + 4	heta}{2} ight) \cos \left( \frac{-2	heta}{2} ight) = 2 \cos (x + 2	heta) \cos 	heta$.Therefore,$\frac{a + c}{b + d} = \frac{2 \cos (x + 	heta) \cos 	heta}{2 \cos (x + 2	heta) \cos 	heta} = \frac{\cos (x + 	heta)}{\cos (x + 2	heta)} = \frac{b}{c}$.

If $\frac{\cos x}{a} = \frac{\cos (x + \theta)}{b} = \frac{\cos (x + 2\theta)}{c} = \frac{\cos (x + 3\theta)}{d}$,then $\left( \frac{a + c}{b + d} \right)$ is equal to :-

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