# Trig Identities question for test tomorrow Help

1. Jun 20, 2004

### Johnny Neutron

Trig Identities question for test tomorrow Help!!

Need some help with these two problems:

sin (X) / Cos (x) - 1 = show work

Sec ^2/ cot (x) - Tan ^3x = Tan X

show work to prove

last one

Sec x - Cos x/tanx = sinx

show work to prove

2. Jun 20, 2004

### AKG

What?
$$\frac{\sec ^2 x}{\cot x} - \tan ^3 x = \tan x$$

Note, $\cot x$ must not be zero. Now, multiply by $\cot x$:

$$\sec ^2 x - \tan ^2 x = 1 \ \dots \ (1)$$

Note, if $\cot x$ were zero, then $\cos x$ would have to be zero (since $\cot x = \frac{\cos x}{\sin x}$), but since it's not, then $\cos x \neq 0$. So, we can multiply both sides by $\cos ^2 x$:

$$1 - \sin ^2 x = \cos ^2 x$$

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

This is a basic identity you should know. In fact (1) is a commonly used identity too, but I figured I'd get you down at least this far. I assume you won't have to prove this. If you do, then you know that $\sin x$ is the ratio of the side opposite the angle x in a right triangle to the hypoteneuse. You should also know the definition for $\cos x$. With these two definitions and the Pythagorean Theorem, you should be able to prove those two identities.

As a general approach to any of these kinds of problems, express everything in terms of sine and cosine. Mutiplying both sides by $\sin x \cos x$:

$$\sin x - \cos ^3 x = \sin ^2 x \cos x$$

$$\sin x = \cos x(\cos ^2 x + \sin ^2 x)$$

$$\sin x = \cos x$$

This is wrong. Try x = 32 degrees. It doesn't work. I guess it's a trick question or you mistyped (or I made a mistake).

3. Jun 20, 2004

### Parth Dave

Yea i'm almost positive that last one doesn't work. If you make that tanx, sinx/cosx you are left with secx = 2sinx, which is not true.

4. Jun 21, 2004

### HallsofIvy

But (sec x- cos x)/tan x= sin x is true.

As AKG suggested change everything to sin and cos:

(1/cos x- cos x)/(sin x/cos x)

= ((1- cos<sup>2</sup> x)/cos x)(cos x/sin x)
= (sin<sup>2</sup> x/cos x)(cos x/sin x)
= sin x