Solving sqrt of ((1-sinx)/(1+sinx)) = |secx-tanx|

[Gill]

yea ok this problem has been making me mad for the past 2 weeks or so (it's a summer project due next week) any help would b great

sqrt of ((1-sinx)/(1+sinx))= |secx-tanx|

i tried:
=|secx-tanx|
=|(1/cosx)-(sinx/cosx)|
=|(1-sinx)/cosx|
and i need it to equal the other side without touching the other side... please HELP!

 

[AKG]

Hint: |A| = \sqrt{A^2}
Hint: \cos ^2x + \sin ^2x = 1

 

[Gill]

sorry but that still seems to not work, wat can i do? am i doing something wrong?

 

[lurflurf]

yea ok this problem has been making me mad for the past 2 weeks or so (it's a summer project due next week) any help would b great

sqrt of ((1-sinx)/(1+sinx))= |secx-tanx|

i tried:
=|secx-tanx|
=|(1/cosx)-(sinx/cosx)|
=|(1-sinx)/cosx|
and i need it to equal the other side without touching the other side... please HELP!

on the sqrt side you could multiply the numerator and denominator by 1-sin(x)
and use (sin(x))^2=1-(cos(x))^2
on the other side write it in terms of sin and cos as you have but then writ cos in terms of sin (hit use sqrt).

 

[Gill]

it works! thanks

 

[AKG]

sorry but that still seems to not work, wat can i do? am i doing something wrong?

You must have been doing something wrong, because it does indeed work:

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

Note that I canceled a factor of (1 - \sin x) without checking that it wasn't 0. But you can check for yourself that if it is zero, then sin(x) = 1, which implies that cos(x) = 0, and so sec(x) and tan(x) are undefined, and I would assume that you're only asked to show that this identity holds for those values of x that don't lead to us having anything undefined (this happens when sec(x) or tan(x) are undefined, or when (1 + sin(x)) = 0).