If : $\tan^2\alpha \tan^2\beta +\tan^2\beta \tan^2\gamma + \tan^2\gamma \tan^2\alpha + 2\tan^2\alpha \tan^2\beta \tan^2\gamma =1\dots$
Problem :
If $\tan^2\alpha \tan^2\beta +\tan^2\beta \tan^2\gamma + \tan^2\gamma \tan^2\alpha + 2\tan^2\alpha \tan^2\beta \tan^2\gamma =1$
Then find the value of $\sin^2\alpha + \sin^2\beta +\sin^2\gamma$
Please suggest how to proceed in such problem.
Solution 1:
HINT:
If $\sin^2\alpha=x$ etc,
$\displaystyle\tan^2\alpha=\frac{\sin^2\alpha}{\cos^2\alpha}=\frac{\sin^2\alpha}{1-\sin^2\alpha}=\frac x{1-x}$
Just simplify to find $x+y+z=1$ assuming $(1-x)(1-y)(1-z)\ne0$ i.e, $\tan\alpha$ etc. are finite