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

Following your first line, just write

\begin{align*} \csc t \tan t - \cos t &= \frac{1}{\cos t} - \cos t \\ &= \frac{1 - \cos^2 t}{\cos t} \end{align*}

from a common denominator. Can you take it from here?

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From there you just replace the numerator with a Pythagorean identity, right? My problem is it's hard for me to recognize that 1-cos^2 is a variation of the Pythagorean identity – Learner Feb 3 '14 at 2:07
$$\cos^2 x + \sin^2 x = 1\\\frac{a^2}{h^2}+\frac{o^2}{h^2} = 1\\a^2 + o^2 = h^2$$ 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
So, does this look right? csc tan - cos = 1/sin*sin/cos-cos = sin - cos^2/sin(cos) = 1-cos^2/cos = sin^2/cos – Learner Feb 3 '14 at 2:39

Your method was also not wrong,

L.H.S=$\frac{sinx}{sinxcosx}$-$\frac{cosxsinxcosx}{sinxcosx}$

= $\frac{sinx-cos^2xsinx}{sinxcosx}$

= $\frac{sinx(1-cos^2)}{sinxcosx}$

= $\frac{1-cos^2}{cosx}$

= $\frac{sin^2}{cosx}$=R.H.S

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