Question

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

• A. $b$
• B. $a$
• C. $\frac{a}{b}$
• D. $\frac{a}{a+b}$

Answer 1

The correct option is B $a$

Given:- $\tan \theta =\frac{b}{a}$

To find:- $a\cos 2\theta +b\sin 2\theta =?$

Sol:-

As we know that,

$\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$

$\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

Therefore,

$a\cos 2\theta +b\sin 2\theta$

$=a\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)+b\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)$

$=a\left(\frac{1-{\left(\frac{b}{a}\right)}^2}{1+{\left(\frac{b}{a}\right)}^2}\right)+b\left(\frac{2\left(\frac{b}{a}\right)}{1+{\left(\frac{b}{a}\right)}^2}\right)$

$=a\left(\frac{a^2-b^2}{a^2+b^2}\right)+b\left(\frac{2ab}{a^2+b^2}\right)$

$=\frac{a^3-a{b}^2+2a{b}^2}{a^2+b^2}$

$=\frac{a^3+a{b}^2}{a^2+b^2}$

$=\frac{a(a^2+b^2)}{a^2+b^2}$

$=a$