Answer:
Step-by-step explanation:
The idea here is to get the left side simplified down so it is the same as the right side. Consequently, there are 3 identities for cos(2x):
[tex]cos(2x)=cos^2x-sin^2x[/tex],
[tex]cos(2x)=1-2sin^2x[/tex], and
[tex]cos(2x)=2cos^2x-1[/tex]
We begin by rewriting the left side in terms of sin and cos, since all the identities deal with sines and cosines and no cotangents or cosecants. Rewriting gives you:
[tex]\frac{\frac{cos^2x}{sin^2x} -\frac{sin^2x}{sin^2x} }{\frac{1}{sin^2x} }[/tex]
Notice I also wrote the 1 in terms of sin^2(x).
Now we will put the numerator of the bigger fraction over the common denominator:
[tex]\frac{\frac{cos^2x-sin^2x}{sin^2x} }{\frac{1}{sin^2x} }[/tex]
The rule is bring up the lower fraction and flip it to multiply, so that will give us:
[tex]\frac{cos^2x-sin^2x}{sin^2x} *\frac{sin^2x}{1}[/tex]
And canceling out the sin^2 x leaves us with just
[tex]cos^2x-sin^2x[/tex] which is one of our identities.
