# Thread: trig identities, proof

1. ## trig identities, proof

prove these identities;

A) $\frac{1-cos\theta}{sin\theta}$ = $\frac{sin\theta}{1+cos\theta}$

b) $(\frac{1}{cos\theta} + tan\theta)$ $(\frac{1}{cos\theta} - tan\theta)$ = 1

is this working correct for the second one;

$y = \frac{1}{cos\theta}$

$(y+tan\theta) (y-tan\theta)$

$y^2 - tan^2\theta = 1$

$y^2 - \frac{sin^2\theta}{cos^2\theta}$

where would I go from here?

2. Originally Posted by Tweety
prove these identities;

A) $\frac{1-cos\theta}{sin\theta}$ = $\frac{sin\theta}{1+cos\theta}$

b) $(\frac{1}{cos\theta} + tan\theta)$ $(\frac{1}{cos\theta} - tan\theta)$ = 1

is this working correct for the second one;

$y = \frac{1}{cos\theta}$

$(y+tan\theta) (y-tan\theta)$

$y^2 - tan^2\theta = 1$

$y^2 - \frac{sin^2\theta}{cos^2\theta}$

where would I go from here?
Remember what you made y equal to and put it back in:

$y^2 = \frac{1}{cos^2\theta}$

This gives you

$\frac{1-sin^2\theta}{cos^2\theta} = \frac{cos^2\theta}{cos^2\theta} = 1 = RHS$

For number 1 get the same denominator by cross multiplying:

$\frac{1-cos\theta}{sin\theta}$ = $\frac{sin\theta}{1+cos\theta}$

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

The denominators cancel and the LHS is the difference of two squares:

$1-cos^2\theta = sin^2\theta$

$sin^2\theta = sin^2\theta$

3. How does getting this expression $sin^2\theta = sin^2\theta$ prove that $\frac{1-cos\theta}{sin\theta}$ = $\frac{sin\theta}{1+cos\theta}$?

Are you not meant to make the first expression look like the second one?

Thanks,

4. Follow the listed steps: Take that second equality, cross-multiply, and simplify using the listed identities. What do you get?

5. Hello, Tweety!

Prove these identities:

. . $(a)\;\;\frac{1-\cos\theta}{\sin\theta} \:=\: \frac{\sin\theta}{1+\cos\theta}$

Multiply the left side by $\frac{1+\cos\theta}{1+\cos\theta}$

$\frac{1-\cos\theta}{\sin\theta}\cdot\frac{1+\cos\theta}{1+ \cos\theta} \:=\:\frac{1-\cos^2\!\theta}{\sin\theta(1 + \cos\theta)} \:=\:\frac{\sin^2\!\theta}{\sin\theta(1 + \cos\theta)} \;=\;\frac{\sin\theta}{1 + \cos\theta}$

$(b)\;\;\left(\frac{1}{\cos\theta} + \tan\theta\right)\left(\frac{1}{\cos\theta} - \tan\theta\right) \:=\: 1$

The left side is: . $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) \:=\:\sec^2\!\theta - \tan^2\!\theta \;=\;1$

6. Originally Posted by Tweety
prove these identities;

A) $\frac{1-cos\theta}{sin\theta}$ = $\frac{sin\theta}{1+cos\theta}$

b) $(\frac{1}{cos\theta} + tan\theta)$ $(\frac{1}{cos\theta} - tan\theta)$ = 1

is this working correct for the second one;

$y = \frac{1}{cos\theta}$

$(y+tan\theta) (y-tan\theta)$

$y^2 - tan^2\theta = 1$

$y^2 - \frac{sin^2\theta}{cos^2\theta}$

where would I go from here?
1.You multiply on the diagonal and get:
$sin^2x=(1-cosx)(1+cosx) \Leftrightarrow sin^2x=1-cos^2x \Leftrightarrow sin^2x+cos^2x=1$

2.You have $tanx=\frac{sinx}{cosx}$
So
$
(\frac{1}{cosx}+tanx)(\frac{1}{cosx}-tanx)=1
$

$
(\frac{1}{cosx}+\frac{sinx}{cosx})(\frac{1}{cosx}-\frac{sinx}{cosx})=1$

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

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

$
\frac{cos^2x}{cos^2x}=1
$