how to show that $\tan^2 y=-\csc 2x$ If $\csc y=\sin x -\cos x$ how to show that $\tan^2 y=-\csc 2x$
Can anyone explain to me? What identity I should use?

Question: Show $tan^2(y) = -csc(2x)$, given that $csc(y) = sin(x) - cos(x)$.
Identities/Knowledge: 


*

*$csc(\theta) = \frac {1}{sin(\theta)}$

*$sin(2\theta) = 2sin(\theta)cos(\theta)$

*$cos^2(\theta) + sin^2(\theta) = 1$

*$tan(\theta) = \frac {1}{csc(\theta)cos(\theta)}$


Working Out:
Lets' begin with what the right hand side is equal too:
$$R.H.S = -csc(2x) = -\frac {1}{sin(2x)}$$
That was simple, now we move onto the left hand side (the fun part, I'll annotate as I go on):
$$L.H.S = \frac {sin^2(y)}{cos^2(y)} = (\frac {1}{csc(y)cos(y)})^2 $$
$$= \frac {1}{csc^2(y)cos^2(y)}$$
But $csc(y) = sin(x) - cos(x)$, so (equation 1);
$$ \frac {1}{(sin(x) - cos(x))^2 cos^2(y)}$$
Note that $cos^2(\theta) = 1 - sin^2(\theta)$
$$cos^2(y) = ? = 1 - (\frac {1}{sin(y)})^{-2} = 1 - (\frac {1}{\frac {1}{sin(y)}})^2$$
But $\frac {1}{sin(y)} = csc(y)$
$$ cos^2(y) = 1 - (\frac {1}{sin(x) - cos(x)})^2$$
Sub $cos^2(y)$ into equation 1:
$$\frac {1}{(sin(x) - cos(x))^2 (1 - \frac {1}{(sin(x) - cos(x))^2})}$$
$$ = \frac {1}{(sin(x) - cos(x))^2 (\frac {(sin(x) - cos(x))^2 -1}{(sin(x) - cos(x))^2})}$$
Let $a = sin(x) - cos(x)$
$$\frac {1}{\frac {a^2(a^2-1)}{a^2}}$$
The $a^2$ cancels out, so you're left with:
$$\frac {1}{a^2 - 1}$$
But $a = sin(x) - cos(x)$
$$\frac {1}{(sin(x) - cos(x))^2 - 1}$$
Note that $(sin(x) - cos(x))^2 = cos^2(x) + sin^2(x) - 2sin(x)cos(x)$, but because of the identities listed above we can simplify it to; $1 - sin(2x)$.
$$\therefore \frac {1}{1 - sin(2x) - 1} = - \frac {1}{sin(2x)} = R.H.S$$
