Question

(cot^2x - 1)/(csc^2x) = cos2x​

Answer 1

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):

$cos(2x)=cos^2x-sin^2x$,

$cos(2x)=1-2sin^2x$, and

$cos(2x)=2cos^2x-1$

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:

$\frac{\frac{cos^2x}{sin^2x} -\frac{sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$

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:

$\frac{\frac{cos^2x-sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$

The rule is bring up the lower fraction and flip it to multiply, so that will give us:

$\frac{cos^2x-sin^2x}{sin^2x} *\frac{sin^2x}{1}$

And canceling out the sin^2 x leaves us with just

$cos^2x-sin^2x$ which is one of our identities.