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 $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$
Try: let $\tan^2\alpha=a,\tan^2\beta=b,,\tan^2\gamma=c$. Then given $ab+bc+ca+2abc=1$
Then how I calculate $\sum\sin^2\alpha$. Could some help me to solve it, Thanks.
\begin{align*} \cot^2\alpha+\cot^2\beta+\cot^2\gamma+2&=\cot^2\alpha\cot^2\beta\cot^2\gamma\\ \csc^2\alpha+\csc^2\beta +\csc^2\gamma-1&=(\csc^2\alpha-1)(\csc^2\beta-1)(\csc^2\gamma-1)\\ \csc^2\alpha+\csc^2\beta +\csc^2\gamma-1&=\csc^2\alpha\csc^2\beta\csc^2\gamma\\&\qquad-\csc^2\alpha\csc^2\beta-\csc^2\beta\csc^2\gamma-\csc^2\gamma\csc^2\alpha\\&\qquad+\csc^2\alpha+\csc^2\beta+\csc^2\gamma-1\\ \csc^2\alpha\csc^2\beta\csc^2\gamma&=\csc^2\alpha\csc^2\beta+\csc^2\beta\csc^2\gamma+\csc^2\gamma\csc^2\alpha\\ 1&=\sin^2\gamma+\sin^2\alpha+\sin^2\beta \end{align*}
Hint
Let $\sin \alpha =a$,$\sin \beta =b$ and $\sin \gamma =c$
And then use $$\cos^2 x=1-\sin^2 x$$
Substitute this in original equation to get $$a^2b^2(1-c^2)+b^2c^2(1-a^2)+a^2c^2(1-b^2)+2a^2b^2c^2=(1-a^2)(1-b^2)(1-c^2)$$ Which simplifies to
$$a^2+b^2+c^2=1$$ Hope you can solve it to get
$$ \tan^2\alpha\tan^2\beta+\tan^2\beta\tan^2\gamma+\tan^2\gamma\tan^2\alpha+2\tan^2\alpha\tan^2\beta\tan^2\gamma=1 $$ Multiplying both sides of this by $\cos^2\alpha\cos^2\beta\cos^2\gamma,$ one gets \begin{align} & \sin^2\alpha\sin^2\beta\cos^2\gamma + \cos^2\alpha\sin^2\beta\sin^2\gamma + \sin^2\alpha\sin^2\gamma\cos^2\beta + 2\sin^2\alpha\sin^2\beta\sin^2\gamma \\[10pt] = {} & \cos^2\alpha\cos^2\beta\cos^2\gamma. \end{align} Then replacing $\cos^2\theta$ with $1-\sin^2\theta$ for $\theta=\alpha,\beta,\gamma,$ we get \begin{align} & (\sin^2\alpha\sin^2\beta)( 1-\sin^2\gamma) + (\sin^2\beta\sin^2\gamma)(1-\sin^2\alpha) + (\sin^2\alpha\sin^2\gamma)(1-\cos^2\beta) \\ & {} + 2\sin^2\alpha\sin^2\beta\sin^2\gamma = (1-\sin^2\alpha)(1-\sin^2\beta)(1-\sin^2\gamma). \end{align} Expanding both sides we get \begin{align} & (\sin^2\alpha\sin^2\beta - \sin^2\alpha\sin^2\beta\sin^2\gamma) + (\sin^2\beta\sin^2\gamma - \sin^2\alpha\sin^2\beta\sin^2\gamma) \\ & {} + (\sin^2\alpha\sin^2\gamma - \sin^2\alpha\sin^2\beta\sin^2\gamma) + 2\sin^2\alpha\sin^2\beta\sin^2\gamma \\[10pt] = {} & 1-\sin^2\alpha-\sin^2\beta-\sin^2\gamma + \sin^2\alpha\sin^2\beta + \sin^2\alpha\sin^2\gamma+\sin^2\beta\sin^2\gamma -\sin^2\alpha \sin^2\beta \sin^2\gamma \end{align} Collecting like terms and cancelling terms common to both sides, we get $$ 0 = 1- \sin^2\alpha-\sin^2\beta-\sin^2\gamma. $$
