Trig Proof

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!!!

Related Introductory Physics Homework Help News on Phys.org

AKG

Homework Helper
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

Homework Helper
Gill said:
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).

it works! thanks

AKG

Homework Helper
Gill said:
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 cancelled 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).

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