TRIG!!!

[taylorann92]

I need helppp!

Verify that (CosX / 1-SinX)-(1+SinX/CosX)= 0

i have been trying to figure this out for two days and i just keep starting over!!

 

[BigGlenntheHeavy]

$\displaystyle \frac{cos(x)}{1-sin(x)}-\frac{1+sin(x)}{cos(x)} \ = \ 0$

$\displaystyle \frac{cos^2(x)-[1+sin(x)][1-sin(x)]}{cos(x)[1-sin(x)]} \ = \ 0$

$\displaystyle \frac{cos^2(x)-[1-sin^2(x)]}{cos(x)[1-sin(x)]} \ = \ 0$

$\displaystyle \frac{cos^2(x)-cos^2(x)}{cos(x)[1-sin(x)]} \ = \ 0$

$\displaystyle \frac{0}{cos(x)[1-sin(x)]} \ = \ 0, \ 0 \ = \ 0$

$\displaystyle Addendum: \ Any \ restrictions?$

 

[taylorann92]

Is there anyway you could explain how you got the first step? It is okay if you dont but I am just confused how the cos^2 happened..

 

[soroban]

Hello, taylorann92!

$\displaystyle \text{Verify that: }\;\frac{\cos x}{1-\sin x} - \frac{1 +\sin x}{\cos x} \;=\;0$

$\displaystyle \text{Multiply the first fraction by }\:\frac{1+\sin x}{1+\sin x}:$

. . . $\displaystyle \frac{\cos x}{1-\sin x}\cdot\frac{1+\sin x}{1+\sin x} - \frac{1+\sin x}{\cos x}$

. . . $\displaystyle =\;\frac{\cos x(1+\sin x)}{1-\sin^2\!x} - \frac{1+\sin x}{\cos x}$

. . . $\displaystyle =\;\frac{\cos(1+\sin x)}{\cos^2\!x} - \frac{1+\sin x}{\cos x}$

. . . $\displaystyle =\;\frac{1+\sin x}{\cos x} - \frac{1+\sin x}{\cos x}$

. . . $\displaystyle =\;\qquad 0$

 

[BigGlenntheHeavy]

$\displaystyle cos(x) \ times \ cos(x) \ = \ [cos(x)]^2 \ = \ cos^2(x).$

$\displaystyle Note: \ If \ you \ don't \ have \ a \ firm \ grasp \ of \ algebra, \ then \ doing \ trig. \ identities \ can$

$\displaystyle become \ quite \ burdensome.$

 

[taylorann92]

thank you both so much! it was so helpful