Simplgying trig functions

**Mr_Green** · November 12th 2006, 06:30 AM

Simplify the following:

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

**ThePerfectHacker** · November 12th 2006, 06:47 AM

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,
$\frac{\sin x(\cos x+1)}{(\cos x+1)}=\sin x$

**Soroban** · November 12th 2006, 06:48 AM

Hello, Mr_Green!

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

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