# Prove trig identity: $\csc \tan - \cos = \frac{\sin^2}{\cos}$

\begin{align*} \csc\ x \tan\ x - \cos\ x &= \left(\frac{1}{\sin\ x}\right)\left(\frac{\sin\ x}{\cos\ x}\right) - \cos\ x \\ &= \frac{\sin\ x}{(\sin\ x)(\cos\ x)} - \cos\ x \\ &= \frac{\sin\ x}{(\sin\ x)(\cos\ x)} - \frac{(\cos\ x)(\sin\ x)(\cos\ x)}{(\sin\ x)(\cos\ x)} \\ &= \frac{(\cos^2 x)(\sin\ x)}{(\sin\ x)(\cos\ x)} \\ &= \cos\ x \end{align*}

Thank you!

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Your last line is wrong. You made an error in subtraction from your second last line. –  Peter Feb 3 '14 at 1:40

$$\cos^2 x + \sin^2 x = 1\\\frac{a^2}{h^2}+\frac{o^2}{h^2} = 1\\a^2 + o^2 = h^2$$ In other words pythagorean theorem –  Flowers Feb 3 '14 at 2:35
@user125736 Yes, you just use a Pythagorean identity. The fact that $\sin^2 t + \cos^2 t = 1$ is fundamentally important, and should always be one of the first things you think about. –  user61527 Feb 3 '14 at 2:35