Question

If $\tan 2\alpha \cdot \tan \beta =1$ and $\tan \alpha \cdot \tan \gamma =1$ then

• A. $(1+\tan \alpha )(1+\tan \left(\frac{\beta }{2}\right))=2\sin (\alpha +\gamma )$
• B. $\tan \alpha +\tan \beta =\tan \gamma$
• C. $\tan (\alpha +\beta )=\tan \gamma$
• D. $\tan 2\gamma \tan \beta =-1$

Answer 1

Answer: (A), (C), (D)

$\tan 2\gamma \tan \beta =-1$

Given,
$\tan 2\alpha \times \tan \beta =1\Rightarrow 2\alpha =\frac{\pi }{2}-\beta \Rightarrow 2\alpha +\beta =\frac{\pi }{2}...\left(1\right)$

$\tan \alpha \times \tan \gamma =1\Rightarrow \alpha =\frac{\pi }{2}-\gamma \Rightarrow \alpha +\gamma =\frac{\pi }{2}...\left(2\right)$

$(1+\tan \alpha )(1+\tan (\frac{\pi }{4}-\alpha ))$
$=(1+\tan \alpha )\times \frac{2}{(1+\tan \alpha )}=2=RHS$

From $\left(1\right)$
$\alpha +\beta =\frac{\pi }{2}-\alpha$
$\therefore \tan (\alpha +\beta )=\cot \alpha =\cot (\frac{\pi }{2}-\gamma )=\tan \gamma$

$\alpha +\beta =\frac{\pi }{2}\Rightarrow 2\gamma =\pi -2\alpha \Rightarrow \tan 2\gamma =-\tan 2\alpha$
and $\beta =\frac{\pi }{2}-2\alpha \Rightarrow \tan \beta =\cot 2\alpha$
$\therefore \tan 2\gamma \cdot \tan \beta =-1=RHS$