Question

If ${\tan }^2\alpha +2\tan \alpha .\tan 2\beta ={\tan }^2\beta +2\tan \beta .\tan 2\alpha$, then

• A. ${\tan }^2\alpha +2\tan \alpha .\tan 2\beta =0$
• B. $\tan \alpha +\tan \beta =0$
• C. ${\tan }^2\beta +2\tan \beta .\tan 2\alpha =0$
• D. $\tan \alpha =\tan \beta$

Answer 1

The correct option is D $\tan \alpha =\tan \beta$

${\tan }^2\alpha +2\tan \alpha .\tan 2\beta ={\tan }^2\beta +2\tan \beta .\tan 2\alpha$

$\Rightarrow {\tan }^2\alpha +2\tan \alpha .\left(\frac{2\tan \beta }{1-{\tan }^2\beta }\right)={\tan }^2\beta +2\tan \beta .\left(\frac{2\tan \alpha }{1-{\tan }^2\alpha }\right)[\because \tan 2\theta =\frac{2\tan \theta }{1-ta{n}^2\theta }]$

Now , ${\tan }^2\alpha +\frac{4\tan \alpha .\tan \beta }{1-{\tan }^2\beta }={\tan }^2\beta +\frac{4\tan \alpha .\tan \beta }{1-{\tan }^2\alpha }$

$\Rightarrow {\tan }^2\alpha -{\tan }^2\beta =4\tan \alpha \tan \beta (\frac{1}{1-{\tan }^2\alpha }-\frac{1}{1-{\tan }^2\beta })$

$\Rightarrow {\tan }^2\alpha -{\tan }^2\beta =4\tan \alpha \tan \beta \left(\frac{1-{\tan }^2\beta -1+{\tan }^2\alpha }{(1-{\tan }^2\alpha )(1-{\tan }^2\beta )}\right)$

$\Rightarrow {\tan }^2\alpha -{\tan }^2\beta =4\tan \alpha \tan \beta \left(\frac{{\tan }^2\alpha -{\tan }^2\beta }{(1-{\tan }^2\alpha )(1-{\tan }^2\beta )}\right)$

$\Rightarrow {\tan }^2\alpha -{\tan }^2\beta -4\tan \alpha \tan \beta \left(\frac{{\tan }^2\alpha -{\tan }^2\beta }{(1-{\tan }^2\alpha )(1-{\tan }^2\beta )}\right)=0$

$\Rightarrow ({\tan }^2\alpha -{\tan }^2\beta )[1-\frac{4\tan \alpha \tan \beta }{(1-{\tan }^2\alpha )(1-{\tan }^2\beta )}]=0$

either

$\Rightarrow ({\tan }^2\alpha -{\tan }^2\beta )=0\Rightarrow \tan \alpha =\pm \tan \beta$ --- eq.1

or, $\Rightarrow [1-\frac{4\tan \alpha \tan \beta }{(1-{\tan }^2\alpha )(1-{\tan }^2\beta )}]=0$

or, $\left[\frac{4\tan \alpha \tan \beta }{(1-{\tan }^2\alpha )(1-{\tan }^2\beta )}\right]=1$

$4\tan \alpha \tan \beta =(1-{\tan }^2\alpha )(1-{\tan }^2\beta )$

$4\tan \alpha \tan \beta =1-{\tan }^2\alpha -{\tan }^2\beta +{\tan }^2\alpha .{\tan }^2\beta$

$2\tan \alpha \tan \beta +{\tan }^2\alpha +{\tan }^2\beta =1+{\tan }^2\alpha .{\tan }^2\beta -2\tan \alpha \tan \beta$

or, $(\tan \alpha +\tan \beta {)}^2=(1-\tan \alpha .\tan \beta {)}^2$

or, $(\tan \alpha +\tan \beta )=\pm (1-\tan \alpha .\tan \beta )$

or, $\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }=\pm 1$

or, $\tan (\alpha +\beta )=\pm 1$

or, $\alpha +\beta =\frac{\pi }{4}\text{or},\alpha +\beta =-\frac{\pi }{4}$

$[\therefore \tan \frac{\pi }{4}=1]$

Again from 1, we get,

$(\tan \alpha +\tan \beta )(\tan \alpha -\tan \beta )=0$

$\Rightarrow$ either $\tan \alpha +\tan \beta =0$ or, $\tan \alpha -\tan \beta =0$

Hence option B & D are correct.