Question

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

(a) a

(b) b

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

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

Answer 1

Given: $\tan x=\frac{a}{b}$
Now,

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

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

Hence, the correct answer is option B.
Given: $\tan x=\frac{a}{b}$
Now,

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

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

Hence, the correct answer is option B.