(A) Given the equation: $\sin (A+B) \sin (A-B)+\cos (A+B) \cos (A-B)=\frac{1}{2}$ Using the trigonometric identity $\cos (x-y) = \cos x \cos y + \sin

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(A) Given the equation: $\sin (A+B) \sin (A-B)+\cos (A+B) \cos (A-B)=\frac{1}{2}$ Using the trigonometric identity $\cos (x-y) = \cos x \cos y + \sin x \sin y$,where $x = A+B$ and $y = A-B$: $\cos ((A+B) - (A-B)) = \frac{1}{2}$ $\cos (A+B-A+B) = \frac{1}{2}$ $\cos (2B) = \frac{1}{2}$ Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have $2B = \frac{\pi}{3}$ Therefore,$B = \frac{\pi}{6}$ Thus,option $A$ is correct.

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Trigonometrical ratios of sum and difference of two and three angles

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If $\sin (A+B) \sin (A-B)+\cos (A+B) \cos (A-B)=\frac{1}{2}$ and $0 < B < \frac{\pi}{2}$,then $B=$

TS EAMCET 2022 All TS EAMCET

A
$\frac{\pi}{6}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{3}$
D
$\frac{5 \pi}{12}$

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Trigonometrical Ratios, Functions and Identities

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Trigonometrical ratios of sum and difference of two and three angles

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If $\sin (A+B) \sin (A-B)+\cos (A+B) \cos (A-B)=\frac{1}{2}$ and $0 < B < \frac{\pi}{2}$,then $B=$

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