Question

Would someone please help me break down the steps of proving the following:?

1+cos^2(-x)/1-csc^2(-x) = -sin^2x-tan^2x

Answer 1

Is this \[\frac{1+cos^2{x}}{1-\csc^2{x}}=\sin^2{x}-\tan^2{x}\]
?
Change everything into sin and cos and then start mass reducing if you can

Answer 2

He has negative x's in the parentheses

Answer 3

I've tried but I get stuck... And yes those are negative x's

Answer 4

\[\frac{1+cos^2{(-x)}}{1-\csc^2{(-x)}}\]
\[\frac{1+cos^2{(-x)}}{1-\frac{1}{\sin^2{(-x)}}}\]
\[\frac{1+cos^2{(-x)}}{1-\frac{1}{\sin^2{(-x)}}}\cdot \frac{\sin^2{(-x)}}{\sin^2{(-x)}}\]

\[\frac{\sin^2{(-x)}+\sin^2{(-x)}cos^2{(-x)}}{\sin^2{(-x)}-1}\]
\[\frac{\sin^2{(-x)}+\sin^2{(-x)}cos^2{(-x)}}{-\cos^2{(-x)}}\]

\[-\frac{\sin^2{(-x)}}{\cos^2{(-x)}}-\sin^2{(-x)}\]
\[-\tan^2{(-x)}-\sin^2{(-x)}\]
Note that sin(-x)=-sin(x)
(-sin(x))^2 = sin(x)

so
\[-\tan^2{(x)}-\sin^2{(x)}\]

Answer 5

oh and
\[1=\sin^2{x} + \cos^2{x}\]
so
\[\sin^2{x} - 1=-\cos^2{x}\]

Answer 6

and
\[\tan{(-x)} = -\tan{(x)}\]
so \[\tan^2{(-x)} = \tan^2{(x})\]

Answer 7

Oh and above that should be
\[\sin^2{(-x)}=\sin^2{(x)}\]
I omitted the square term

Answer 8

Thank you