# Thread: Prove problem involving sec cos sin

1. ## Prove problem involving sec cos sin

Hi guys been struggling to solve this
If you know how to do it, please do solve and show work so that I can follow your steps

prove (sec(x)-cos(x))/(sec(x)+cos(x))=(sin^2(x))/(1+cos^2(x))

2. ## Re: Prove problem involving sec cos sin

I would start by changing everything into terms of cosine only. What is the definition of sec(x)?

3. ## Re: Prove problem involving sec cos sin

Originally Posted by MathRaven
Hi guys been struggling to solve this
If you know how to do it, please do solve and show work so that I can follow your steps

prove (sec(x)-cos(x))/(sec(x)+cos(x))=(sin^2(x))/(1+cos^2(x))
when proving identities, beginners are advised to change all trig functions to terms involving cosine and/or sine ...

$\displaystyle \frac{\sec{x} - \cos{x}}{\sec{x}+\cos{x}} =$

change $\displaystyle \sec{x}$ to $\displaystyle \frac{1}{\cos{x}}$ ...

$\displaystyle \frac{\frac{1}{\cos{x}} - \cos{x}}{\frac{1}{\cos{x}} + \cos{x}} =$

common denominator ...

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

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

use the Pythagorean identity in the numerator ...

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

divide the fractions ...

$\displaystyle \frac{\sin^2{x}}{1+\cos^2{x}}$