Question

If $a\tan \theta =b$, then $a\cos 2\theta +b\sin 2\theta =$

• A. $a$
• B. $b$
• C. $-a$
• D. $-b$

Answer 1

The correct option is A

$a$

Step1. Find the value of $\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{\theta }$and $\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{\theta }$:

Since $a\tan \theta =b$

$\tan \theta =b/a\dots .....\left(i\right)$

$\cos 2\theta =\frac{(1–{\tan }^2\theta )}{(1+{\tan }^2\theta )}$

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

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

$$\begin{gathered} =\frac{\left({\displaystyle \frac{2b}{a}}\right)}{\left(1+{\displaystyle \frac{b^2}{a^2}}\right)} \\ =\frac{2ab}{(a^2+b^2)} \end{gathered}$$

Step2. Find the value of $\mathit{a}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{b}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{}$:

Now,

$a\cos 2\theta +b\sin 2\theta =a\frac{(a^2-b^2)}{(a^2+b^2)}+b\frac{2ab}{(a^2+b^2)}$

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

Hence, the correct option is (A).