Question

If $\alpha ,\beta$ and $\gamma$ are connected by the relation $2{\tan }^2\alpha {\tan }^2\beta {\tan }^2\gamma +{\tan }^2\alpha {\tan }^2\beta +{\tan }^2\beta {\tan }^2\gamma +{\tan }^2\gamma {\tan }^2\alpha =1$ then

• A. ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =1$
• B. ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =2$
• C. $\cos 2\alpha +\cos 2\beta +\cos 2\gamma =1$
• D. $\cos (\alpha +\beta )\cos (\alpha -\beta )={\cos }^2\gamma$

Answer 1

Answer: (A), (B), (C)

The correct options are
A ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =2$
B ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =1$
C $\cos 2\alpha +\cos 2\beta +\cos 2\gamma =1$

Given, $2{\tan }^2\alpha {\tan }^2\beta {\tan }^2\gamma +{\tan }^2\alpha {\tan }^2\beta +{\tan }^2\beta {\tan }^2\gamma +{\tan }^2\gamma {\tan }^2\alpha =1$
${\tan }^2\alpha {\tan }^2\beta +{\tan }^2\beta {\tan }^2\gamma +{\tan }^2\gamma {\tan }^2\alpha =1-2{\tan }^2\alpha {\tan }^2\beta {\tan }^2\gamma$ ...(1)
Now,
${\sin }^2\gamma +{\sin }^2\beta +{\sin }^2\alpha =\frac{{\tan }^2\gamma }{1+{\tan }^2\gamma }+\frac{{\tan }^2\beta }{1+{\tan }^2\beta }+\frac{{\tan }^2\alpha }{1+{\tan }^2\alpha }=\frac{{\tan }^2\gamma (1+{\tan }^2\beta )(1+{\tan }^2\alpha )+{\tan }^2\beta (1+{\tan }^2\gamma )(1+{\tan }^2\alpha )+{\tan }^2\alpha (1+ta{n}^2\beta )(1+ta{n}^2\gamma )}{(1+{\tan }^2\gamma )(1+{\tan }^2\beta )(1+{\tan }^2\alpha )}$
Using (1), we get
${\sin }^2\gamma +{\sin }^2\beta +{\sin }^2\alpha =\frac{\sum {\tan }^2\alpha +2\sum {\tan }^2\beta {\tan }^2\gamma +3{\tan }^2\alpha {\tan }^2\beta {\tan }^2\gamma }{1+\sum {\tan }^2\alpha +1-{\tan }^2\alpha {\tan }^2\beta {\tan }^2\gamma }=\frac{2+\sum {\tan }^2\alpha {\tan }^2\beta {\tan }^2\gamma }{2+\sum {\tan }^2\alpha -{\tan }^2\alpha {\tan }^2\beta {\tan }^2\gamma }=1$
$\Rightarrow {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =2\Rightarrow \cos 2\alpha +\cos 2\beta +\cos 2\gamma =1$
Now if $\cos (\alpha +\beta )\cos (\alpha -\beta )-{\cos }^2\gamma$ is true
${\cos }^2\alpha -{\sin }^2\beta =-{\cos }^2\gamma \Rightarrow {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$