How do you prove #cosA/(1-tanA) = (sinA)/(1-cotA)#?

Question

Emery Adkins · Accepted Answer

Since this relationship is generally untrue, it cannot be proven. For example if #A= pi/6#
#color(white)("XXXX")##(cos A)/(1-tan A) = 2.05# #color(white)("XXXX")##(sin A)/(1-cot A) = -0.68# (Maybe you should add squares to a few of the terms in your relation?)

Asher Friedman · Answer

undefined To prove $ \frac{\cos A}{1 - \tan A} = \frac{\sin A}{1 - \cot A} $, we'll start with the left-hand side (LHS) and manipulate it to match the right-hand side (RHS).

LHS:
$$ \frac{\cos A}{1 - \tan A} $$
$$ = \frac{\cos A}{1 - \frac{\sin A}{\cos A}} $$
$$ = \frac{\cos A}{\frac{\cos A - \sin A}{\cos A}} $$
$$ = \frac{\cos A \cdot \cos A}{\cos A - \sin A} $$
$$ = \frac{\cos^2 A}{\cos A - \sin A} $$

Now, let's simplify the right-hand side (RHS) and see if it matches the simplified LHS:

RHS:
$$ \frac{\sin A}{1 - \cot A} $$
$$ = \frac{\sin A}{1 - \frac{\cos A}{\sin A}} $$
$$ = \frac{\sin^2 A}{\sin A - \cos A} $$

Comparing the simplified expressions of LHS and RHS, we see that they are not equal. Therefore, the given statement is incorrect as it stands. If you intended a different statement or there's a typo, please clarify so we can provide the correct proof.