Can someone please prove this equation:

Cos x+Cos x= 2Sec x

1 - Sin x 1 + Sin x

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- Sep 8th 2010, 03:33 PMGAVREED2Trigonometry Proof
Can someone please prove this equation:

__Cos x__+__Cos x__= 2Sec x

1 - Sin x 1 + Sin x - Sep 8th 2010, 03:39 PMskeeter
- Sep 8th 2010, 09:48 PMGAVREED2
I'm not sure if i completed the question correctly

Is the answer 2/cos x = 2/cos x - Sep 8th 2010, 10:36 PMUnknown008
You should consider only one hand side and should end up with 2sec(x).

$\displaystyle \frac{cos(x)}{1-sin(x)} + \frac{cos(x)}{1+sin(x)} = \frac{cos(x)(1+sin(x) + cos(x)(1-sin(x))}{(1-sin(x))(1+sin(x))}$

Expand, simplify.

Ultimately, you should get;

$\displaystyle \frac{cos(x)}{1-sin(x)} + \frac{cos(x)}{1+sin(x)} = 2sec(x)$ - Sep 9th 2010, 01:23 AMEducated
Yes, the answer is $\displaystyle \frac{2}{cos(x)}$ but that is not how you solve a prove question.

There is no point in saying 2/cos x = 2/cos x, because of course that equation is true, and what you have basicly done is rewritten 2sec(x) in the form 2/cos(x).

You should start off on the left, ignoring the right hand side, having only the equation $\displaystyle \frac{cos(x)}{1-sin(x)} + \frac{cos(x)}{1+sin(x)}$ and simplifying/rearranging it to match what is stated.

You must show the steps of how you got that, without the answer of 2sec(x) in the equation:

$\displaystyle \dfrac{cos(x)}{1-sin(x)} + \dfrac{cos(x)}{1+sin(x)}$

$\displaystyle = \dfrac{cos(x)(1+sin(x) + cos(x)(1-sin(x))}{(1-sin(x))(1+sin(x))}$

$\displaystyle = \dfrac{cos(x) + cos(x)sin(x) + cos(x) - cos(x)sin(x)}{1-sin^2(x)}$

Continue simplifying/rearranging it to get 2sec(x).

Hint: ($\displaystyle cos^2(x) + sin^2(x) = 1$)