If sin (theta + alpha ) = a and sin (theta + | vedclass

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(B) Given: $\sin(	heta + \alpha) = a$ ... $(i)$ and $\sin(	heta + \beta) = b$ ... $(ii)$.Let $x = 	heta + \alpha$ and $y = 	heta + \beta$. Then $\

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$1 - 2a^2 - 2b^2$

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(B) Given: $\sin(	heta + \alpha) = a$ ... $(i)$ and $\sin(	heta + \beta) = b$ ... $(ii)$.Let $x = 	heta + \alpha$ and $y = 	heta + \beta$. Then $\sin x = a$ and $\sin y = b$.We know that $\cos(x - y) = \cos x \cos y + \sin x \sin y$.Since $\sin x = a$,$\cos x = \sqrt{1 - a^2}$. Since $\sin y = b$,$\cos y = \sqrt{1 - b^2}$.Thus,$\cos(\alpha - \beta) = \cos(x - y) = \sqrt{1 - a^2}\sqrt{1 - b^2} + ab$.Let $C = \cos(\alpha - \beta) = \sqrt{1 - a^2}\sqrt{1 - b^2} + ab$.Then $C - ab = \sqrt{1 - a^2}\sqrt{1 - b^2}$.Squaring both sides: $(C - ab)^2 = (1 - a^2)(1 - b^2) = 1 - a^2 - b^2 + a^2b^2$.$C^2 - 2abC + a^2b^2 = 1 - a^2 - b^2 + a^2b^2$.$C^2 - 2abC = 1 - a^2 - b^2$.We need to find $\cos 2(\alpha - \beta) - 4ab \cos(\alpha - \beta) = 2\cos^2(\alpha - \beta) - 1 - 4ab \cos(\alpha - \beta)$.$= 2C^2 - 1 - 4abC = 2(C^2 - 2abC) - 1$.Substituting $C^2 - 2abC = 1 - a^2 - b^2$:$= 2(1 - a^2 - b^2) - 1 = 2 - 2a^2 - 2b^2 - 1 = 1 - 2a^2 - 2b^2$.

If $\sin (\theta + \alpha ) = a$ and $\sin (\theta + \beta ) = b,$ then $\cos 2(\alpha - \beta ) - 4ab\cos (\alpha - \beta )$ is equal to

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