Question

If $a\cos 2\theta +b\sin 2\theta =c$ has $\alpha$ and $\beta$ as its solution, then the value of $\tan \alpha +\tan \beta$ is

• A. $\frac{(c+a)}{2b}$
• B. $\frac{2b}{(c+a)}$
• C. $\frac{(c–a)}{2b}$
• D. $\frac{b}{(c+a)}$

Answer 1

The correct option is B

$\frac{2b}{(c+a)}$

Explanation for the correct option:

Step 1. Find the value of $\tan \alpha +\tan \beta$:

Given,

$a\cos 2\theta +b\sin 2\theta =c$ ….(i)

$\Rightarrow$ $a\left(\frac{1–\tan 2\theta }{1+\tan 2\theta }\right)+b\left(\frac{2\tan \theta }{1+\tan 2\theta }\right)=c$

$\Rightarrow$ $a(1–\tan 2\theta )+2b\tan \theta =c(1+\tan 2\theta )$

$\Rightarrow$$c+c\tan 2\theta –a+a\tan 2\theta –2b\tan \theta =0$

$\Rightarrow$ $(c+a)\tan 2\theta –2b\tan \theta +(c–a)=0$ ….(ii)

Step 2. Given that $\alpha$ and $\beta$ are the roots of equation (i).

$\therefore$$\tan \alpha$ and $\tan \beta$ are the roots of equation (ii) since it is a quadratic equation in $\tan \theta$.

Now, Sum of the roots of equation (ii),

$\begin{array}{rcl}\therefore \tan \alpha +\tan \beta & =& \frac{-(-2b)}{(c+a)}\\ & =& \frac{\mathbf{2}\mathbf{b}}{\mathbf{(}\mathbf{c}\mathbf{}\mathbf{+}\mathbf{}\mathbf{a}\mathbf{)}}\end{array}$

Hence, Option ‘B’ is Correct.