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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

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

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

5 Answers 5

up vote 3 down vote accepted

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

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?

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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

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.

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

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$$ \ tan A-\ cosecA\ secA+ \ 2CosecA\ cosA $$ $$ \dfrac{\ sinA}{\ cosA}-\dfrac{1}{\ sinA.\ cosA}+2\ cotA $$ $$ \dfrac {\ sin^2A-1}{\ sinA \ cosA}+2\ cotA $$ $$ -\ cotA + 2\ cotA $$ $$\ cotA$$

proved....

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

I just want to comment on how you could take your work so far a little further.

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

What you need to do is to get a common denominator $$\frac{\sin A}{\cos A}\cdot\frac{\sin A}{\sin A}-\frac{1}{\sin A\cos A}(1- \cos^2 A- \cos^2 A)$$ $$=\frac{\sin^2 A}{\sin A\cos A}-\frac{1- \cos^2 A- \cos^2 A}{\sin A\cos A}$$ $$=\frac{\sin^2 A}{\sin A\cos A}-\frac{\sin^2 A- \cos^2 A}{\sin A\cos A}$$ $$=\frac{\cos^2 A}{\sin A\cos A}$$ $$=\frac{\cos A}{\sin A}$$ $$=\cot A$$

But in general I do think thatit would be nice to use more weapons than just breaking everything into sines and cosines. As a starting point we have $$\cos^2\theta + \sin^2\theta = 1$$ $$\tan^2\theta + 1 = \sec^2\theta$$ $$\cot^2\theta + 1 = \csc^2\theta$$ $$\tan\theta = \frac{\sin\theta}{\cos\theta}, \cot\theta = \frac{\cos\theta}{\sin\theta}$$ $$\csc\theta = \frac{1}{\sin\theta}, \sec\theta = \frac{1}{\cos\theta}, \cot\theta = \frac{1}{\tan\theta}$$ But we could use the above identities to come up with these identities $$\cot\theta\tan\theta = \sin\theta\csc\theta = \cos\theta\sec\theta = 1$$ $$\tan\theta = \frac{\sec\theta}{\csc\theta}, \cot\theta = \frac{\csc\theta}{\sec\theta}$$

So lets try the same problem again, but this time deploying these new weapons. $$\tan A - \csc A \sec A (1-2\cos^2 A)$$ $$=\cot A \tan A(\tan A - \csc A \sec A (1-2\cos^2 A))$$ $$=\cot A( \tan^2 A - \frac{\sec A}{\csc A}\csc A \sec A (1-\frac{2}{\sec^2 A})))$$ $$=\cot A( \tan^2 A - \sec^2 A (1-\frac{2}{\sec^2 A})))$$ $$=\cot A( \tan^2 A - \sec^2 A + 2)$$ $$=\cot A( (\sec^2 A - 1) - \sec^2 A + 2)$$ $$=\cot A$$

The point that I'm trying to make is that sometimes it's more convenient to express everything in terms of sins and coss, but at other times it's better to express everything in terms of tans and secs (or cots and cscs). These other identities will make your trigonometric simplifications all the more powerful.

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