If cos ,(theta - alpha ) = a,,,sin ,(theta - | vedclass

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Question

(c) We have $\sin (\alpha - \beta ) = \sin (	heta - \beta - \overline {	heta - \alpha } )$

$ = \sin (	heta - \beta )\cos (	heta - \alpha ) - \c

Vedclass · Accepted Answer

${a^2} + {b^2}$

Vedclass · Answer

(c) We have $\sin (\alpha - \beta ) = \sin (	heta - \beta - \overline {	heta - \alpha } )$

$ = \sin (	heta - \beta )\cos (	heta - \alpha ) - \cos (	heta - \beta )\sin (	heta - \alpha )$

$ = ba - \sqrt {1 - {b^2}} \sqrt {1 - {a^2}} $

and $\cos (\alpha - \beta ) = \cos (	heta - \beta - \overline {	heta - \alpha } )$

$ = \cos (	heta - \beta )\cos (	heta - \alpha ) + \sin (	heta - \beta )\sin (	heta - \alpha )$

$ = a\sqrt {1 - {b^2}} + b\sqrt {1 - {a^2}} $

$	herefore $ Given expression is ${\cos ^2}(\alpha - \beta ) + 2ab\sin (\alpha - \beta )$

$ = (a\sqrt {1 - {b^2}} + b\sqrt {1 - {a^2}{)^2}} + 2ab\{ ab - \sqrt {1 - {a^2}} \sqrt {1 - {b^2}} \} $

$ = {a^2} + {b^2}$.

Trick : Put $\alpha = 30^\circ ,\beta = 60^\circ $ and $	heta = 90^\circ ,$

then $a = \frac{1}{2},b = \frac{1}{2}$

$	herefore {\cos ^2}(\alpha - \beta ) + 2ab\sin (\alpha - \beta ) = \frac{3}{4} + \frac{1}{2} 	imes \left( { - \frac{1}{2}} ight) = \frac{1}{2}$

which is given by option $(c).$

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

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