If tan theta = frac{a}{b},then b cos 2 theta | vedclass

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(A) We know that $\cos 2 	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$ and $\sin 2 	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$.Substitut

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$b$

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(A) We know that $\cos 2 	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$ and $\sin 2 	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$.Substituting these into the expression $b \cos 2 	heta + a \sin 2 	heta$:$= b \left( \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight) + a \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight)$Given $	an 	heta = \frac{a}{b}$,we substitute this value:$= b \left( \frac{1 - \frac{a^2}{b^2}}{1 + \frac{a^2}{b^2}} ight) + a \left( \frac{2 \frac{a}{b}}{1 + \frac{a^2}{b^2}} ight)$$= b \left( \frac{\frac{b^2 - a^2}{b^2}}{\frac{b^2 + a^2}{b^2}} ight) + \frac{2a^2}{b \left( \frac{b^2 + a^2}{b^2} ight)}$$= b \left( \frac{b^2 - a^2}{b^2 + a^2} ight) + \frac{2a^2 b}{b^2 + a^2}$$= \frac{b^3 - a^2 b + 2a^2 b}{b^2 + a^2}$$= \frac{b^3 + a^2 b}{b^2 + a^2} = \frac{b(b^2 + a^2)}{b^2 + a^2} = b$.

If $\tan \theta = \frac{a}{b}$,then $b \cos 2 \theta + a \sin 2 \theta = $

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