# Simplgying trig functions

• Nov 12th 2006, 07:30 AM
Mr_Green
Simplgying trig functions
Simplify the following:

(sin q + tan q) / (1 + sec q)
• Nov 12th 2006, 07:47 AM
ThePerfectHacker
Quote:

Originally Posted by Mr_Green
Simplify the following:

(sin q + tan q) / (1 + sec q)

Multiply numerator and denominator by,
$\cos x$
Thus,
$\frac{\sin x\cos x+\tan x \cos x}{\cos x+\sec x\cos x}$
But,
$\tan x\cos x=\frac{\sin x}{\cos x}\cdot \cos x=\sin x$
And,
$\sec x\cos x=1$
Thus,
$\frac{\sin x\cos x+\sin x}{\cos x+1}$
Factor,
$\frac{\sin x(\cos x+1)}{(\cos x+1)}=\sin x$
• Nov 12th 2006, 07:48 AM
Soroban
Hello, Mr_Green!

Quote:

Simplify: . $\frac{\sin\theta + \tan\theta}{1 + \sec\theta}$

We have: . $\frac{\sin\theta + \frac{\sin\theta}{\cos\theta}}{1 + \frac{1}{\cos\theta}}$

Multiply by $\frac{\cos\theta}{\cos\theta}\!:\;\;\frac{\cos\the ta}{\cos\theta}\cdot\frac{\sin\theta + \frac{\sin\theta}{\cos\theta}}{1 + \frac{1}{\cos\theta}} \;= \;\frac{\sin\theta\cos\theta + \sin\theta}{\cos\theta + 1}$

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