1. ## simplifying the expression

I'm having trouble with this

$(1-sin^2(x)) / sin(x) - csc(x)$

2. Originally Posted by yeunju
I'm having trouble with this

$(1-sin^2(x)) / sin(x) - csc(x)$
$\frac{1-\sin^2x}{\sin x}-\csc x=\frac{1}{\sin x}-\frac{\sin^2x}{\sin x}-\csc x=\dots$

Can you continue?

3. $\frac{1-\sin^2x}{\sin x-\csc x}=\frac{1-\sin^2x}{\sin x-\frac{1}{\sin x}}=\frac{1-\sin^2x}{\frac{\sin^2x-1}{\sin x}}=-\sin x$

4. Originally Posted by yeunju
I'm having trouble with this

$(1-sin^2(x)) / sin(x) - csc(x)$
$\color{red} sin^2(x) + cos^2(x) = 1 \qquad \implies sin^2(x) - 1 = -cos^2(x)$

$\frac{(1-sin^2(x))}{sin(x)} - csc(x)$

$\frac{(1-sin^2(x))}{sin(x)} -\frac{1}{sin(x)}$

$\frac{(1-sin^2(x))-1}{sin(x)}$

$\frac{-sin^2(x)}{sin(x)}$

$-sin(x)$
------------------------------------------------------------------------

$\frac{1-sin^2(x)}{sin(x) - csc(x)}$

Denominator = $sin(x) - csc(x) = sin(x) - \frac{1}{sin(x)}$

$
= \frac{sin^2(x) - 1}{sin(x)}$

$
= \frac{-cos^2(x)}{sin(x)}$

Numerator $= 1-sin^2(x) = cos^2(x)$

So fraction is

$=\frac{cos^2(x)}{\frac{-cos^2(x)}{sin(x)}}$

$=\frac{sin(x)}{-1}$

$=~ -sin(x)$