Hi,
Can anyone give me a hint/tip to get started with this question. I've been at it for an hour and just get a mish mash of sin and cos:
Prove:
$\displaystyle
\frac{cos(A)}{1 - tan(A)} + \frac{sin(A)}{1-cot(A)} = sin(A) + sin(A)
$
Thank you
Hi,
Can anyone give me a hint/tip to get started with this question. I've been at it for an hour and just get a mish mash of sin and cos:
Prove:
$\displaystyle
\frac{cos(A)}{1 - tan(A)} + \frac{sin(A)}{1-cot(A)} = sin(A) + sin(A)
$
Thank you
correction
you mean you want to prove this right ??
$\displaystyle \frac{cos(A)}{1 - tan(A)} + \frac{sin(A)}{1-cot(A)} = cos(A) + sin(A)$
to prove it first you know that
$\displaystyle \tan A = \frac{\sin A }{\cos A } $
$\displaystyle \cot A = \frac{\cos A}{\sin A } $
$\displaystyle 1- \frac{\sin A }{\cos A } = \frac{\cos A - \sin A }{\cos A } $
make the same thing to the other denominator
simplify you will have to fraction have the same denominator
use
$\displaystyle (a^2 - b^2 ) = (a-b)(a+b) $
something will cancel then you will get the answer