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I have the following expression:

$$\left( \frac{\cos \beta \sin \beta \cos\phi +\sin \beta}{\sin^2 \phi + \sin^2\beta \cos^2 \phi} \right)^2\cos \phi$$

Wolfram Alpha yields a simplified form of this as, $$\frac{\sin^2 \beta \cos \phi}{(\cos \beta \cos \phi -1)^2}$$

I can multiply the first expression out and get,

$$\frac{\cos^2\beta \sin^2\beta \cos^2 \phi +\sin^2 \beta +2 \cos \beta \sin^2 \beta \cos \phi}{\sin^4 \phi + \sin^4\beta \cos^4 \phi + 2 \sin^2 \beta \sin^2 \phi \cos^2 \phi}$$

but after that I can't see the way forward to get the simplified W-A expression.

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  • $\begingroup$ Factor-out $\sin\beta$ from the numerator of the squared quantity to get $\sin^2\beta$. That and the $\cos\phi$ give you the numerator of the result, so all you have to do is show that the rest simplifies to the final denominator. Since there are only cosines involved there, try expressing everything in terms of those. $\endgroup$
    – Blue
    Commented Feb 21 at 0:32

2 Answers 2

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So instead of multiplying the expression out, work with the first equation. Factoring out $\sin \beta$ in the numerator and substituting for sines in the denominator yields,

$$\left( \frac{\sin\beta(\cos\beta\cos\phi+1)}{1-\cos^2\phi +(1-\cos^2\beta)\cos^2\phi} \right)^2\cos\phi$$

Multiplying out in the denominator,

$$\left( \frac{\sin\beta(\cos\beta\cos\phi+1)}{1-\cos^2\phi +\cos^2\phi-\cos^2\beta\cos^2\phi} \right)^2\cos\phi$$

Simplifying, $$\left( \frac{\sin\beta(1+\cos\beta\cos\phi)}{1-\cos^2\beta\cos^2\phi} \right)^2\cos\phi$$

Breaking out the denominator yields

$$\left( \frac{\sin\beta(1+\cos\beta\cos\phi)}{(1+\cos\beta\cos\phi)(1-\cos\beta\cos\phi)} \right)^2\cos\phi$$

Dividing out common expression in numerator and denominator produces

$$\left( \frac{\sin\beta}{(1-\cos\beta\cos\phi)} \right)^2\cos\phi$$

which results in a final solution of

$$ \frac{\sin^2\beta\cos\phi}{(1-\cos\beta\cos\phi)^2}$$

which is the same as the Wolfram Alpha solution.

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$$\left( \frac{\cos \beta \sin \beta \cos\phi +\sin \beta}{\sin^2 \phi + \sin^2\beta \cos^2 \phi} \right)^2\cos \phi=\sin^2\beta\cos\phi\left( \frac{\cos\beta\cos\phi+1}{\sin^2 \phi + \sin^2\beta\cos^2\phi}\right)^2$$ It remains to show that (up to sign, because this is in a square) $$\frac{\cos\beta\cos\phi+1}{\sin^2 \phi +\sin^2\beta\cos^2\phi}=\frac1{\cos\beta\cos\phi-1}$$ This can be established as follows: $$\frac{\cos\beta\cos\phi+1}{\sin^2 \phi +\sin^2\beta\cos^2\phi}=\frac{\cos\beta\cos\phi+1}{1-\cos^2\phi+\sin^2\beta\cos^2\phi}=\frac{\cos\beta\cos\phi+1}{1-(1-\sin^2\beta)\cos^2\phi}=\frac{\cos\beta\cos\phi+1}{1-\cos^2\beta\cos^2\phi}=\frac1{1-\cos\beta\cos\phi}$$ where we have used the difference of two squares in the last step.

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