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$$\tan\left(\frac{\pi}{2} -x\right) - \cot\left(\frac{3\pi}{2} -x\right) + \tan(2\pi-x) - \cot(\pi-x) = \frac{4-2\sec^{2}x}{\tan{x}}$$

L.S.

$= \cot{x} - \tan{x} - \tan{x} + \cot{x}$

$= 2\cot{x} - 2\tan{x}$

$= 2\left(\frac{\cos{x}}{\sin{x}} - \frac{\sin{x}}{\cos{x}}\right)$

$= 2\left(\frac{\cos^{2}x - \sin^{2}x}{\sin{x}\cos{x}}\right)$

$= 2\left(\frac{1-2\sin^{2}x}{\sin{x}\cos{x}}\right)$

$= \frac{4 - 2\sin^{2}x}{\sin{x}\cos{x}}$

  • Not sure where to go from here.
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You may want to look here (meta.math.stackexchange.com/questions/5020/…) on how to typeset your questions so that it is easier for people to read. –  user17762 Nov 8 '12 at 2:08

1 Answer 1

You have simplified your initial expression to $2 \cot(x) - 2 \tan(x)$. Make use of the following trigonometric identities: $$\cot(x) = \dfrac1{\tan(x)}$$ $$\sec^2(x) - \tan^2(x) = 1$$ and simplify to get what you want.

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$$2 \cot(x) - 2 \tan(x) = 2 \left( \dfrac1{\tan(x)} - \tan(x)\right) = 2 \left( \dfrac{1 - \tan^2(x)}{\tan(x)}\right)$$ Making use of this identity we have that $\tan^2(x) = \sec^2(x) - 1$. Plug this in your numerator and you that $$2 \cot(x) - 2 \tan(x) = 2 \left( \dfrac{1 - \sec^2(x) + 1}{\tan(x)}\right) = \dfrac{4 - 2 \sec^2(x)}{\tan(x)}$$

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Ok makes sense... Is sec^2(x) - tan^2(x) = 1 created off of sin^2x + cos^2x = 1 because we haven't learned that and this is a Grade 12 course. –  DavidSalib Nov 8 '12 at 2:09
    
@DavidSalib Yes. You have $\sin^2(x) + \cos^2(x) = 1$. Now divide by $\cos^2(x)$ to get $\tan^2(x) + 1 = \sec^2(x)$. –  user17762 Nov 8 '12 at 2:10
1  
Yes I was able to solve it very easily with the knowledge of sec^2x -tan^2x = 1 . Thank You. –  DavidSalib Nov 8 '12 at 2:13

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