Question

If $\tan \left(x\right)=\frac{b}{a}$ , then the value of $a\cos \left(2x\right)+b\sin \left(2x\right)$ is

• A. $a$
• B. $a–b$
• C. $a+b$
• D. $b$

Answer 1

The correct option is A

$a$

Explanation for correct option

Given $\tan \left(x\right)=\frac{b}{a}$

$$\begin{gathered} \therefore a\cos \left(2x\right)+b\sin \left(2x\right) \\ =a\left(\frac{1-{\tan }^2\left(x\right)}{1+{\tan }^2\left(x\right)}\right)+b\left(\frac{2\tan \left(x\right)}{1+{\tan }^2\left(x\right)}\right)\left[\because \cos \left(2x\right)=\frac{1-{\tan }^2\left(x\right)}{1+{\tan }^2\left(x\right)}\mathrm{and}\sin \left(2\mathrm{x}\right)=\frac{2\tan \left(\mathrm{x}\right)}{1+{\tan }^2\left(\mathrm{x}\right)}\right] \\ =\frac{a\left(1-{\tan }^2\left(x\right)\right)+b\left(2\tan \left(x\right)\right)}{1+{\tan }^2\left(x\right)} \\ =\frac{a\left(1-{\left({\displaystyle \frac{b}{a}}\right)}^2\right)+2b\left({\displaystyle \frac{b}{a}}\right)}{1+{\left({\displaystyle \frac{b}{a}}\right)}^2}\left[\because \tan \left(x\right)=\frac{b}{a}\right] \\ =\frac{{\displaystyle \frac{a^2-b^2}{a}}+{\displaystyle \frac{2b^2}{a}}}{{\displaystyle \frac{a^2+b^2}{a^2}}} \\ =\frac{{\displaystyle \frac{a^2+b^2}{a}}}{{\displaystyle \frac{a^2+b^2}{a^2}}} \\ =a \end{gathered}$$

Hence, option (A) is correct i.e. $a$