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If $$\tan^2x\tan^2y + \tan^2y\tan^2z + \tan^2z\tan^2x + 2\tan^2x\tan^2y\tan^2z = 1$$ then find $$\sin^2x + \sin^2y + \sin^2z$$

Blue
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Deepak
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    Hint: Multiply-through by $\cos^2x \cos^2y \cos^2z$ to get an equation involving only sines and cosines. Then re-write all the cosines in terms of sines, and simplify. – Blue Aug 09 '16 at 15:40

1 Answers1

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Make the every term in $\sin $ $$\sin^2 x \sin^2 y (1-\sin^2 z)+\sin^2 x\sin^2 z(1-\sin^2 y)+\sin^2 z\sin^2 x(1-\sin^2 y)+2\sin^2 x\sin^2 y \sin^2 z =(1-\sin^2 x )(1-\sin^2 y)(1-\sin^2 z)$$

Aakash Kumar
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