Does (tanΘ)(cosΘ) always equal sinΘ?

Yes. Tan(x) = sin(x)/cos(x)

So that would be like multiplying (sin(x)/cos(x))*cos(x). The cos(x) will cancel, leaving you with just sin(x). Of course, that is assuming that cos(x) does NOT equal 0.

Not always. If cos(t) = 0, then tan(t) * cos(t) is undefined.

Yes, except when theta is an odd multiple of 90 degrees (an odd multiple of pi/2 radians). At those points, cos(theta) = 0 and sin(theta) = 1 and tan(theta) is undefined.

Yes, since tan theta = sin theta over cos theta
and sin theta over cos theta times cos theta is sin theta

Yes, as long as cosΘ is not 0, making tanΘ ± infinity.

YES!!!
Because
Tan X = Sin X / CosX
Substituting
(Sin X / Cos X) (Cos X)
Cancel down by 'Cos X'.
Hence
Sin X

Yes

yes

Well,

the answer is yes

(tanΘ)(cosΘ) = (sinΘ)/(cosΘ) * (cosΘ) = sinΘ

hope it' ll help !!

Yes.

Think about how they were originally defined using right triangles:

sin() = opp/hyp
cos() = adj/hyp
tan() = opp/adj

So if we have tan() times cos(), we get:

tan() * cos()
(opp/adj) * (adj/hyp)

The two adjacents cancel out leaving:

opp/hyp, which is the definition of sine.