Verify trigonometry equation $\tan A - \csc A \sec A (1-2\cos^2 A)= \cot A$

How would I verify the following trigonometry identity?

$$\tan A - \csc A \sec A (1-2\cos^2 A)= \cot A$$

My work so far is

$$\frac{\sin A}{\cos A}-\frac{1}{\sin A}\frac{1}{\cos A}(1- \cos^2 A- \cos^2 A)$$

-
Remember $1=\sin^2+\cos^2$. Can you move further with that? –  anon Jul 15 '12 at 22:21
add comment

4 Answers

$$\frac{\sin A}{\cos A}-\frac{1-2\cos^2 A}{\sin A \cos A}=\cot A$$

By the pythagorean identity, $1-2\cos^2 A=\sin^2 A-\cos^2 A$.

$$\frac{\sin A}{\cos A}-\frac{\sin^2 A -\cos^2 A}{\sin A \cos A}=\cot A$$

If I told you to split up the fraction, could you get it from there?

-
In splitting it up would it end up being (sin/cos)-(sin-cos). –  Fernando Martinez Jul 15 '12 at 22:34
@Rakishi the first one is right, the second one is not. Physically write out both separate fractions, then cancel them individually. –  Robert Mastragostino Jul 15 '12 at 22:41
Oh I see thank you for your help. –  Fernando Martinez Jul 15 '12 at 22:44
add comment

First of all, I would not separate out $2\cos^2A$ as you have (at least not yet). On the other hand, I see a fractions being multiplied and added. If it helps, replace $\sin A$ with $s$ and $\cos A$ with $c$ so that you can do the algebraic manipulations for the fractions.

-
add comment

$$\tan A - \csc A \sec A (1-2\cos^2 A)= \cot A$$ As we can write $\sin^2A+\cos^2A$ in place of $1$. So,

$$\tan A - \csc A \sec A ((\sin^2A+\cos^2A)-2\cos^2 A)$$ $$\tan A - \csc A \sec A (\sin^2A-\cos^2 A)$$ $$\tan A - \csc A \sec A\ \sin^2 A+\csc A \sec A\ \cos^2 A$$ As $\csc A \sec A\ \sin^2 A$ will reduce to $\tan A$ and $\csc A \sec A\ \cos^2 A$ wil reduce to $\cot A$ $$\tan A - \tan A + \cot A$$ $$\cot A$$

-
add comment

$$\ tanA-\ cosecAsecA+ \ 2CosecAcosA$$ $$\frac{\ sinA}{\ cosA}-\frac{1}{\ sinA.\ cosA}+2\ cotA$$ $$\frac {\ sin^2A-1}{\ sinA*cosA}+2\ cotA$$ $$-\ cotA + 2\ cotA$$ $$\ cotA$$

proved....

-
$sin^2A-1$=$-cos^2A$ –  iostream007 May 10 '13 at 4:52
add comment