Prove problem involving sec cos sin

**MathRaven** · July 8th 2012, 11:16 AM

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))

**HallsofIvy** · July 8th 2012, 11:45 AM

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

**skeeter** · July 8th 2012, 11:47 AM

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 ...

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

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

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

common denominator ...

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

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

use the Pythagorean identity in the numerator ...

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

divide the fractions ...

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

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