Verify the identity: (1 - 2 sin^2 θ)/(sin θ cos θ) = cot θ - tan θ

could you put it in latex?

what do you mean?

\[1-2.\sin(\theta)/\sin(\theta).\cos(\theta)=\sin^{2}+\cos(\theta)^{2}-2\sin(\theta)/(\sin(\theta.\cos(\theta)))\]

so: \[\cos(\theta)^{2}/(\sin(\theta).\cos(\theta))-\sin(\theta)^{2}/(\sin(\theta).\cos(\theta))\]

so \[ctg(\theta)-tg(\theta)\]

\[ \frac{(1 - 2 \sin^2 θ)}{(\sin θ \cos θ)} = \frac{2\cos 2θ}{\sin 2θ} = 2\cot 2θ = \frac {\cos^2θ-\sin^2θ}{2\sinθ\cos θ} \]\[=\frac {\cos^2θ}{2\sinθ\cos θ}-\frac {\sin^2θ}{2\sinθ\cos θ} = \cot θ-\tan θ\] QED!

More abridged, \[ \frac{(1 - 2 \sin^2 θ)}{(\sin θ \cos θ)} = \frac{2\cos 2θ}{\sin 2θ} = \frac {\cos^2θ-\sin^2θ}{2\sinθ\cos θ} \] \[ =\frac {\cos^2θ}{2\sinθ\cos θ}-\frac {\sin^2θ}{2\sinθ\cos θ} = \cot θ-\tan θ \]

Thank you so much!

There was a typo, \[\frac{(1 - 2 \sin^2 θ)}{(\sin θ \cos θ)} = \frac{2\cos 2θ}{\sin 2θ} = \frac {\cos^2θ-\sin^2θ}{\sinθ\cos θ} \]\[\frac {\cos^2θ}{\sinθ\cos θ}-\frac {\sin^2θ}{\sinθ\cos θ} = \cot θ-\tan θ\]

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