Proving d/dx(tan(x)/sec²(x)) = cos(2x)

[QuarkCharmer]

Homework Statement

Prove that:
$$\frac{d}{dx} \frac{tan(x)}{sec^2(x)} = cos(2x)$$

Homework Equations

The Attempt at a Solution

$$\frac{d}{dx} \frac{tan(x)}{sec^2(x)} = cos(2x)$$
$$\frac{d}{dx} \frac{tan(x)}{sec^2(x)} =$$
$$\frac{sec^2(x)(sec^2(x))-(tan(x)(2sec^2(x)tan^2(x)))}{(sec^2(x))^2}=$$
$$\frac{sec^4(x)}{sec^4(x)}-\frac{2sec^2(x)tan^2(x)}{sec^4(x)}=$$
$$1-\frac{2tan^2(x)}{sec^2(x)}=$$
$$\frac{cos(x)cos(x)}{cos^2(x)}-\frac{2sin^2(x)cos^2(x)}{cos^2(x)}=$$
$$1-2sin^2(x)(1)=$$
$$1-2sin^2(x) = cos(2x)$$
Is that right?

 

[vela]

Other than numerous typos (I assume they're typos), it looks okay.

It would have been easier if you simplified first and then differentiated.

 

[QuarkCharmer]

I'm not seeing these typos, but I am horrible with latex so I don't doubt you. What should I correct?

I did it this way because it was for a problem that wanted me to find the derivative of the initial function using the quotient rule, and then as a part B it asked to prove that it was equal to cos2x. I just wrote it here as one big thing.

 

[vela]

Where you first applied the quotient rule, check the parentheses and the exponents.

 

[QuarkCharmer]

Oh yeah, that was a typo! (fixed it)

Thanks for your help.