# Need Help Simplifying trig expression

• Dec 4th 2006, 04:17 PM
Mr_Green
Need Help Simplifying trig expression
Well first off I am not the best simplifier, yet i still am managable, but now with the whole trig function ordeal and it makes it twice as hard.

Could someone please show me step by step on how to simplify
[(tan q + sec q)^2 +1] / [2sec q]

Thanks much!
• Dec 4th 2006, 07:56 PM
CaptainBlack
Quote:

Originally Posted by Mr_Green
Well first off I am not the best simplifier, yet i still am managable, but now with the whole trig function ordeal and it makes it twice as hard.

Could someone please show me step by step on how to simplify
[(tan q + sec q)^2 +1] / [2sec q]

Thanks much!

First replace all functions by $\displaystyle \sin$s and $\displaystyle \cos$s:

$\displaystyle f(\theta)=\frac{(\tan(\theta)+\sec(\theta))^2+1)}{ 2 \sec(\theta)}= $$\displaystyle \left(\left(\frac{s}{c}+\frac{1}{c}\right)^2+1\rig ht)\frac{c}{2} , where \displaystyle c and \displaystyle s are shorthand for \displaystyle \cos(\theta) and \displaystyle \sin(\theta), so: \displaystyle f(\theta)= \left(\left(\frac{s+1}{c}\right)^2+1\right)\frac{c }{2}= \left(\left(\frac{s^2+2s+1}{c^2}\right)+1\right)\f rac{c}{2}=$$\displaystyle \left(\frac{s^2+2s+1+c^2}{c^2}\right)\frac{c}{2}= \left(\frac{2s+2}{c^2}\right)\frac{c}{2}$

.........$\displaystyle =\left(\frac{s+1}{c}\right)=\tan(\theta)+\sec(\the ta)$

RonL
• Dec 4th 2006, 11:02 PM
Soroban
Hello, Mr_Green!

Quote:

Simplify: .$\displaystyle \frac{(\tan\theta + \sec\theta)^2 +1}{2\sec\theta}$

We have: .$\displaystyle \frac{\tan^2\theta + 2\tan\theta\sec\theta + \sec^2\theta + 1}{2\sec\theta}$ $\displaystyle = \;\frac{(\tan^2\theta + 1) + \sec^2\theta + 2\sec\theta\tan\theta}{2\sec\theta}$

. . . . . .$\displaystyle = \;\frac{\sec^2\theta + \sec^2\theta + 2\sec\theta\tan\theta}{2\sec\theta}\;=\;\frac{2\se c^2\theta + 2\sec\theta\tan\theta}{2\sec\theta}$

Factor and reduce: . $\displaystyle \frac{2\sec\theta(\sec\theta + \tan\theta)}{2\sec\theta} \;= \;\sec\theta + \tan\theta$