Question

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

Answer 1

could you put it in latex?

Answer 2

what do you mean?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

More abridged,

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

Answer 8

Thank you so much!

Answer 9

There was a typo,

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