OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

could you put it in latex?

6 years ago
OpenStudy (anonymous):

what do you mean?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

so $ctg(\theta)-tg(\theta)$

6 years ago
OpenStudy (anonymous):

$\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 ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

Thank you so much!

6 years ago
OpenStudy (anonymous):

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$

6 years ago