# Thread: Verifying an Identity

1. ## Verifying an Identity

(tan x )/(1 + cos x) + (sin x)/(1 - cos x) = cot x + sec x csc x

Starting with the left side:

Get the LCD and add the fractions to get:

[tan x (1 - cos x) + sin x (1 + cos x)] / (1 + cos x)(1 - cos x)

Distributive property:

(tan x - tan x cos x + sin x + sin x cos x) /( 1 - cos2x)

Simplify.

tan x - sin x + sin x + sin x cos x / sin2x

(tan x + sin x cos x) / (sin2x)

Now working with the right side.

cot x + sec x csc x

cos x/sin x + 1/(cos x sin x)

Find the LCD and add to get:

(cos2x + 1) / (sin x cos x)

(1 - sin2x + 1)/ (sin x cos x)

(2 - sin2x) / (sin x cos x)

From here, I cannot find how to relate the two sides of the equation. Please help!

2. ## Re: Verifying an Identity

I think what you meant to wrirte is this:

$\frac {\tan x}{1 + \cos x} + \frac {\sin x }{1 - \cos x} = \cot x \ +\ \sec x \ \csc x$

As usual with these types of problems its best to start by converting the various trig functions into t heir sine and cosine equivalents, then simplify fractions while looking for ways to apply a simple identy such as $\sin^2 x + \cos^2 x = 1$. It works right out. Try it - and post back with your attempt.

3. ## Re: Verifying an Identity

That's very hard to read but I think you want to show that
$\frac{tan(x)}{1+ cos(x)}+ \frac{sin(x)}{1- cos(x)}= cot(x)+ sec(x)csc(x)$

It's often easiest to immediately put everything in terms of sine and cosine:
$\frac{\frac{sin(x)}{cos(x)}}{1+ cos(x)}+ \frac{sin(x)}{1- cos(x)}= \frac{cos(x)}{sin(x)}+ \frac{1}{cos(x)}\frac{1}{sin(x)}$

On the right we have a sum of fractions so get the "common denominator" $(1+ cos(x))(1- cos(x))= 1- cos^2(x)= sin^2(x)$ by multiplying numerator and denominator of the first fraction by 1- cos(x) and the second by 1+ cos(x):
$\frac{sin(x)}{cos(x)}- sin(x)}{sin^2(x)}+ \frac{sin(x)+ sin(x)cos(x)}{sin^2(x)}= \frac{\frac{sin(x)}{cos(x)}+ sin(x)cos(x)}{sin^2(x)}= \frac{\frac{1}{cos(x)}+ cos(x)}{sin(x)}= \left(\frac{1}{cos(x)}+ cos(x)\right)\frac{1}{sin(x)}$

5. ## Re: Verifying an Identity

Hello, cdbowman42!

$\text{Prove: }\:\frac{\tan x }{1 + \cos x} + \frac{\sin x}{1 - \cos x} \;=\; \cot x + \sec x \csc x$

Starting with the left side:

Get the LCD and add the fractions to get:

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

Distributive property:

$\frac{\tan x - \tan x\cos x + \sin x + \sin x \cos x}{1 - \cos^2\!x}$

Simplify.

$\frac{\tan x - \sin x + \sin x + \sin x\cos x}{\sin^2x} \;=\;\frac{\tan x + \sin x\cos x}{\sin^2\!x}$

All this is correct . . . Keep going!

$\frac{\tan x + \sin x\cos x}{\sin^2\!x}$

. . $=\;\frac{\tan x}{\sin^2\!x} + \frac{\sin x\cos x}{\sin^2\!x}$

. . $=\;\frac{\frac{\sin x}{\cos x}}{\sin^2\!x} + \frac{\cos x}{\sin x}$

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

. . $=\;\sec x\csc x + \cot x$