Simplifying trig Functions

\(\displaystyle \L \frac{(\tan{q}+\sec{q})^2+1}{2\sec{q}} =\)

\(\displaystyle \L \frac{\tan^2{q}+2\tan{q}\sec{q}+\sec^2{q}+1}{2\sec{q}}\)

now, a hint to help you finish up ... \(\displaystyle \L \tan^2{q} + 1 = \sec^2{q}\)
 
Here's one way. Allow you to work with some identities.

\(\displaystyle \L\\\frac{(tan(x)+sec(x))^{2}+1}{2sec(x)}\)

Expand:

\(\displaystyle \L\\\frac{tan^{2}(x)}{2sec(x)}+tan(x)+\frac{sec(x)}{2}+\frac{1}{2sec(x)}\)

Factor:

\(\displaystyle \L\\\frac{1}{2sec(x)}(tan^{2}(x)+1)+tan(x)+\frac{sec(x)}{2}\)

Now, can you whittle down to lowest terms?.

Remember, \(\displaystyle tan^{2}(x)+1=sec^{2}(x)\)
 
Re: reply

Math wiz ya rite 09 said:
i got down to:


[2(sec^2 q) + 2tan q * sec q] / [2secq]


would that simplify to :


sec^2 q + tan q * sec q]
Click to expand...

no ...

\(\displaystyle \L \frac{2\sec^2{q} + 2\tan{q}\sec{q}}{2\sec{q}} =\)

\(\displaystyle \L \frac{2\sec{q}(\sec{q} + \tan{q})}{2\sec{q}} =\)

now can you finish?
 
the sec q's would cross out and then i got

sec q + tan q

which would be

1 / cos q + sin q / cos q


which would b

(1+ sin q) / cos q


is taht as far as it goes. or does 1 + sinq have something that it can be simplified into. Cause I know that sin q + cos q =1, but I don't know if I have to incorporate this anywhere..
 
Math wiz ya rite 09 said:
the sec q's would cross out and then i got

sec q + tan q

which would be

1 / cos q + sin q / cos q


which would b

(1+ sin q) / cos q


is taht as far as it goes. or does 1 + sinq have something that it can be simplified into. Cause I know that sin q + cos q =1 no ... it's sin<sup>2</sup>q + cos<sup>2</sup>q = 1 but I don't know if I have to incorporate this anywhere..
Click to expand...

sec(q) + tan(q) or [1 + sin(q)]/cos(q) is fine ... stop there, don't make something easy into something more difficult.
 
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