# Trig proof

1. Apr 11, 2013

### trollcast

1. The problem statement, all variables and given/known data

Show that:

$$\tan x\sec^4x\equiv\tan x\sec^2x + \tan^3x\sec^2x$$

2. Relevant equations

Trig identities / formulae

3. The attempt at a solution

I've got 2 different starts for it but I'm stuck after a few steps with both of them:

Attempt 1:

$$\tan x \sec^4 x$$
$$\frac{\tan x}{\cos^4 x}$$
$$\frac{\frac{\sin x}{\cos x}}{\cos^4 x}$$
$$\frac{\sin x \cos^4 x}{\cos x}$$
$$\sin x \cos^3 x$$

And then I can't think on anything else for this one.

Attempt 2:

$$\tan x \sec^4 x$$
$$\tan x (\tan^2 x + 1)^2$$
$$\tan x (\tan^4 x + 2\tan^2 x + 1)$$
$$\tan^5 x + 2\tan^3 x + \tan x$$

This one looks a bit closer since its got the higher power tans in it but I can't see where to get the sec terms from?

2. Apr 11, 2013

### Curious3141

Fastest way: split the second term up to $\displaystyle \tan x\tan^2x\sec^2x$, then use the identity $\displaystyle \tan^2x = \sec^2x - 1$.

3. Apr 11, 2013

### trollcast

How do you get that for the second term as theres nothing like it in our formula books?

4. Apr 11, 2013

### Staff: Mentor

If it's not in there, it should be. Certainly you know this one:
sin2x + cos2x = 1

If you divide both sides of this equation by cos2x, you get:
tan2x + 1 = sec2x

If you divide both sides of the first identity by sin2x, you get:
1 + cot2x = csc2x

You should have the first of these memorized. The latter two you can derive quickly.

5. Apr 11, 2013

### Curious3141

$\displaystyle \sin^2x + \cos^2x = 1$. Divide throughout by $\displaystyle \cos^2x$. Rearrange.

In fact, the tan-sec identify is a well-known one in its own right. So is the cot-cosec one, which you can derive by dividing the above equation by $\displaystyle \sin^2x$ instead of $\displaystyle \cos^2x$.

6. Apr 12, 2013

### HallsofIvy

This is wrong. It should be
$$\frac{sin(x)}{cos^5(x)}$$

Have you tried doing the same thing to the right side?