Question

If $a\cos 2\theta +b\sin 2\theta =c$ has α and β as its roots, then prove that

(i) $\tan \alpha +\tan \beta =\frac{2b}{a+c}$ [NCERT EXEMPLAR]

(ii) $\tan \alpha \tan \beta =\frac{c-a}{c+a}$

(iii) $\tan \left(\alpha +\beta \right)=\frac{b}{a}$ [NCERT EXEMPLAR]

Answer 1

Given: $a\cos 2\theta +b\sin 2\theta =c$

$$\begin{gathered} \Rightarrow a\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)+b\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)-c=0 \\ \Rightarrow a\left(1-{\tan }^2\theta \right)+2b\tan \theta -c\left(1+{\tan }^2\theta \right)=0 \\ \Rightarrow a-a{\tan }^2\theta +2b\tan \theta -c-c{\tan }^2\theta =0 \\ \Rightarrow \left(a+c\right){\tan }^2\theta -2b\tan \theta +\left(c-a\right)=0.....\left(1\right) \end{gathered}$$

This a quadratic equation in terms of ${\tan }^2\theta$.

It is given that α and β are the roots of the given equation, so tanα and tanβ are the roots of (1).

Since tanα and tanβ are the roots of the equation $\left(a+c\right){\tan }^2\theta -2b\tan \theta +\left(c-a\right)=0$. Therefore,

(i)
$$\begin{gathered} \tan \alpha +\tan \beta =-\frac{\left(-2b\right)}{a+c}\left(\mathrm{Sum}\mathrm{of}\mathrm{roots}=-\frac{b}{a}\right) \\ \Rightarrow \tan \alpha +\tan \beta =\frac{2b}{a+c} \end{gathered}$$

(ii)

$$\begin{gathered} \tan \alpha \tan \beta =\frac{c-a}{a+c}\left(\mathrm{Product}\mathrm{of}\mathrm{roots}=\frac{c}{a}\right) \\ \mathrm{Or}\tan \alpha \tan \beta =\frac{c-a}{c+a} \end{gathered}$$

(iii)
$$\begin{gathered} \tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ =\frac{{\displaystyle \frac{2b}{a+c}}}{1-\left({\displaystyle \frac{c-a}{c+a}}\right)}\left[\mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\right] \\ =\frac{{\displaystyle \frac{2b}{a+c}}}{{\displaystyle \frac{c+a-c+a}{c+a}}} \\ =\frac{2b}{2a} \\ =\frac{b}{a} \end{gathered}$$