Question

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

(a) a (b) b (c) $\frac{a}{b}$ (d) $\frac{b}{a}$

Answer 1

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

Now,

$$\begin{gathered} b\cos 2\theta +a\sin 2\theta \\ =b\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)+a\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right) \\ =b\left(\frac{1-{\displaystyle \frac{a^2}{b^2}}}{1+{\displaystyle \frac{a^2}{b^2}}}\right)+a\left(\frac{2\times {\displaystyle \frac{a}{b}}}{1+{\displaystyle \frac{a^2}{b^2}}}\right) \\ =\frac{b\left(b^2-a^2\right)}{a^2+b^2}+\frac{2a^{2}b}{a^2+b^2} \end{gathered}$$

$$\begin{gathered} =\frac{b^3-a^{2}b+2a^{2}b}{a^2+b^2} \\ =\frac{b^3+a^{2}b}{a^2+b^2} \\ =\frac{b\left(b^2+a^2\right)}{a^2+b^2} \\ =b \end{gathered}$$

Hence, the correct answer is option B.