Help with proving identities

**Croton** · February 10th 2010, 12:46 PM

I've been studying for an upcoming math test and have come across two problems that I can't figure out how to solve, help is appreciated.

prove:

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

and question 2:

$\frac{\ tan^2(x)}{1 - \cos x}=\frac{1}{\cos(x)} + \frac{1}{\cos^2(x)}$

**e^(i*pi)** · February 10th 2010, 12:52 PM

Originally Posted by Croton

I've been studying for an upcoming math test and have come across two problems that I can't figure out how to solve, help is appreciated.

prove:

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

and question 2:

$\frac{\ tan^2(x)}{1 - \cos x}=\frac{1}{\cos(x)} + \frac{1}{\cos^2(x)}$

1. $1+tan(x) = \frac{cos(x)+sin(x)}{cos(x)}<br />$

Sub that expression into the LHS and do some cancelling

2.

$<br /> \frac{sin^2(x)}{cos^2(x)(1-cos(x))}$

$sin^2(x) = 1-cos^2(x)$

$\frac{1-cos^2(x)}{cos^2(x)(1-cos(x)}$

Use the difference of two squares on the numerator and cancel

**icemanfan** · February 10th 2010, 12:54 PM

For the first problem, write:

$1 + \tan x = \frac{\cos x}{\cos x} + \frac{\sin x}{\cos x} = \frac{\sin x + \cos x}{\cos x}$

For the second problem, write:

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

$= \frac{1 - \cos^2 x}{(1 - \cos x)(\cos^2 x)} = \frac{(1 - \cos x)(1 + \cos x)}{(1 - \cos x)(\cos^2 x)} = \frac{1 + \cos x}{\cos^2 x}$

**Croton** · February 11th 2010, 01:05 AM

Ah so that how you do it, Thanks!

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