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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$b$

Vedclass · Answer

(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 = $

Similar Questions

If $\cos ^3 x \sin 4 x = \sum_{r=0}^{n} a_{r} \sin rx$ for all $x \in R$,then $a_3+a_5 : a_1+a_7 = $

If $\theta$ is any angle,then $\sin^2 \theta \cos^2 \theta =$

Which of the following number$(s)$ is/are rational?

If $3 \sin 2 \theta = 2 \sin 3 \theta$ and $0 < \theta < \pi$,then the value of $\sin \theta$ is equal to

If $\tan \frac{\theta}{2} = \operatorname{cosec} \theta - \sin \theta$,then $\tan^2 \frac{\theta}{2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module