If a sin theta = b cos theta,where a, b neq 0 | vedclass

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(B) Given $a \sin 	heta = b \cos 	heta$,we have $	an 	heta = \frac{b}{a}$.We need to evaluate $a \cos 2 	heta + b \sin 2 	heta$.Using the double

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$a$

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(B) Given $a \sin 	heta = b \cos 	heta$,we have $	an 	heta = \frac{b}{a}$.We need to evaluate $a \cos 2 	heta + b \sin 2 	heta$.Using the double angle formulas in terms of $	an 	heta$:$a \cos 2 	heta + b \sin 2 	heta = a \left( \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight) + b \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight)$Substitute $	an 	heta = \frac{b}{a}$:$= a \left( \frac{1 - \frac{b^2}{a^2}}{1 + \frac{b^2}{a^2}} ight) + b \left( \frac{2 \left( \frac{b}{a} ight)}{1 + \frac{b^2}{a^2}} ight)$$= a \left( \frac{a^2 - b^2}{a^2 + b^2} ight) + b \left( \frac{2b}{a} \cdot \frac{a^2}{a^2 + b^2} ight)$$= \frac{a(a^2 - b^2)}{a^2 + b^2} + \frac{2ab^2}{a^2 + b^2}$$= \frac{a^3 - ab^2 + 2ab^2}{a^2 + b^2} = \frac{a^3 + ab^2}{a^2 + b^2}$$= \frac{a(a^2 + b^2)}{a^2 + b^2} = a$.

If $a \sin \theta = b \cos \theta$,where $a, b \neq 0$,then $a \cos 2 \theta + b \sin 2 \theta = $

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