Proving Trig Identities: Tan x Sec^4x = Tan x Sec^2x + Tan^3x Sec^2x

[trollcast]

Homework Statement

Show that:

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

Homework Equations

Trig identities / formulae

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?

 

[Curious3141]

Homework Statement

Show that:

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

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

 

[trollcast]

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

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

 

[Mark44]

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

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.

 

[Curious3141]

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

$\displaystyle \sin^2x + \cos^2x = 1$. 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 $\displaystyle \cos^2x$.

 

[HallsofIvy]

Homework Statement

Show that:

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

Homework Equations

Trig identities / formulae

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:

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

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

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