$\displaystyle \frac{\sqrt{x^{2}+9}}{x^{2}} $. Help write in terms of sines and cosines can anyone give me steps on how to do this problem
Factoring out the 9, which the square root makes it a 3 is good:
$\displaystyle \frac{\sqrt{9\sec^2(\theta)+9}}{9\sec^2(\theta)}= \frac{\sqrt{9(\sec^2(\theta)+1)}}{9\sec^2(\theta)} =\frac{3\sqrt{\sec^2(\theta)+1}}{9\sec^2(\theta)}=$
$\displaystyle \frac{\sqrt{\sec^2(\theta)+1}}{3\sec^2(\theta)}$
However, $\displaystyle \sec^2(\theta)+1\ne\csc^2(\theta)$. You are further instructed to write in terms of sines and cosines. How can you do this with the secant function?
Also, I would use standard notation for the square of a trig function, i.e.:
$\displaystyle \cos^2(\theta)$
The way you have it written implies the argument (angle) is being squared.
So, we now have:
$\displaystyle \frac{\sqrt{\frac{1}{\cos^2(\theta)}+1}}{\frac{3}{ \cos^2(\theta)}}$
How can we rewrite the division as a multiplication?