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Simplify: $$\tan\alpha \sin^2 \beta + \tan\beta \sin^2 \beta$$ to $$\tan\beta - \tan\alpha$$

I have played around with all the trigonometric relations and just can't same to figure this out.

Any pointers?

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  • $\begingroup$ The equality $\tan\alpha \sin^2 \beta + \tan\beta \sin^2 \beta=\tan\beta - \tan\alpha$ is patently false for $0<\beta<\alpha<\pi/2$, because the left-hand side is positive and the right-hand side is negative. $\endgroup$ – egreg Jul 8 '17 at 9:23
  • $\begingroup$ Maybe the second sine factor is $\sin^2 \alpha$? $\endgroup$ – Oscar Lanzi Jul 8 '17 at 11:40
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If, say, $\alpha=\beta=\dfrac\pi4$ this is false. Therefore, it is not true in general.

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The best it can be simplified to is $\sin^2\beta \ (\tan \alpha + \tan \beta)$. No other manipulation would lead to the second equation.

Using Jose Carlos Santos' example, if we use $\pi/4$ for each side, we would have $(1/2 + 1/2) = 1$ for the first equation and $1 - 1 = 0$ for the second, which is a contradiction.

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