If sin (alpha - beta ) = frac{1}{2} and cos ( | vedclass

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(A) Given,$\sin (\alpha - \beta ) = \frac{1}{2}$. Since $\sin 30^\circ = \frac{1}{2}$,we have $\alpha - \beta = 30^\circ$ $(i)$.Also,$\cos (\alpha + \

Vedclass · Accepted Answer

$\alpha = 45^\circ, \beta = 15^\circ$

Vedclass · Answer

(A) Given,$\sin (\alpha - \beta ) = \frac{1}{2}$. Since $\sin 30^\circ = \frac{1}{2}$,we have $\alpha - \beta = 30^\circ$ $(i)$.Also,$\cos (\alpha + \beta ) = \frac{1}{2}$. Since $\cos 60^\circ = \frac{1}{2}$,we have $\alpha + \beta = 60^\circ$ $(ii)$.Adding equations $(i)$ and $(ii)$:$(\alpha - \beta ) + (\alpha + \beta ) = 30^\circ + 60^\circ$$2\alpha = 90^\circ \Rightarrow \alpha = 45^\circ$.Substituting $\alpha = 45^\circ$ in equation $(ii)$:$45^\circ + \beta = 60^\circ \Rightarrow \beta = 15^\circ$.Thus,$\alpha = 45^\circ$ and $\beta = 15^\circ$ satisfy both conditions.

If $\sin (\alpha - \beta ) = \frac{1}{2}$ and $\cos (\alpha + \beta ) = \frac{1}{2},$ where $\alpha$ and $\beta$ are positive acute angles,then

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