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

6 years agocould you put it in latex?

6 years agowhat do you mean?

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

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

6 years agoso \[ctg(\theta)-tg(\theta)\]

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

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

6 years agoThank you so much!

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

6 years ago