Answer:
[tex]\frac{cos^2(\frac{\pi}{2}-x)}{\sqrt{1-sin^2(x)}}=\frac{sin^2(x)}{|cos(x)|}[/tex]
Step-by-step explanation:
To simplify this expression you must use the following trigonometric identities
[tex]cos(\frac{\pi}{2}-x) = sinx[/tex] I
[tex]1-sin (x) ^ 2 = cos ^ 2(x)[/tex] II
Remember that
[tex]\sqrt{f(x)^2} =f(x)[/tex]
Only if [tex]f(x)> 0[/tex] for all x
If f(x) is not greater than 0 for all x then
[tex]\sqrt{f(x)^2} =|f(x)|[/tex]
Now we have the expression:
[tex]\frac{cos^2(\frac{\pi}{2}-x)}{\sqrt{1-sin^2(x)}}[/tex]
then using the trigonometric identities I and II we have to:
[tex]\frac{cos^2(\frac{\pi}{2}-x)}{\sqrt{1-sin^2(x)}}=\frac{sin^2(x)}{\sqrt{1-sin^2(x)}}\\\\\\\frac{sin^2(x)}{\sqrt{1-sin^2(x)}}= \frac{sin^2(x)}{\sqrt{cos^2(x)}}[/tex]
[tex]cos(x)[/tex] is not greater or equal than 0 for all x. So.
[tex]\frac{sin^2(x)}{\sqrt{cos^2(x)}}=\frac{sin^2(x)}{|cos(x)|}[/tex]
Finally
[tex]\frac{cos^2(\frac{\pi}{2}-x)}{\sqrt{1-sin^2(x)}}=\frac{sin^2(x)}{|cos(x)|}[/tex]
