Trig Identities question for test tomorrow Help!!

[Johnny Neutron]

Need some help with these two problems:

Thanks in advance if you could answer them:

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

[AKG]

sin (X) / Cos (x) - 1 = show workWhat?Sec ^2/ cot (x) - Tan ^3x = Tan X
\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.

Sec x - Cos x/tanx = sinx

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).

[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.

[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- cos² x)/cos x)(cos x/sin x)
= (sin² x/cos x)(cos x/sin x)
= sin x