If a tan theta = b,then a cos 2theta + b sin | vedclass

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(A) Given that $a 	an 	heta = b$,which implies $	an 	heta = \frac{b}{a}$.We need to evaluate the expression $a \cos 2	heta + b \sin 2	heta$.Usin

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$a$

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(A) Given that $a 	an 	heta = b$,which implies $	an 	heta = \frac{b}{a}$.We need to evaluate the expression $a \cos 2	heta + b \sin 2	heta$.Using the double angle formulas in terms of $	an 	heta$:$\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:$a \left( \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight) + b \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight)$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 \frac{b}{a}}{1 + \frac{b^2}{a^2}} ight)$$= a \left( \frac{\frac{a^2 - b^2}{a^2}}{\frac{a^2 + b^2}{a^2}} ight) + b \left( \frac{\frac{2b}{a}}{\frac{a^2 + b^2}{a^2}} ight)$$= a \left( \frac{a^2 - b^2}{a^2 + b^2} ight) + b \left( \frac{2ab}{a^2 + b^2} ight)$$= \frac{a^3 - ab^2 + 2ab^2}{a^2 + b^2}$$= \frac{a^3 + ab^2}{a^2 + b^2} = \frac{a(a^2 + b^2)}{a^2 + b^2} = a$.

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

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