Question

Help me simplify this Identity?

Answer 1

\[\frac{ 1-\sin^2 x }{ \sin x - \csc x }\]

Answer 2

Would it just be cos^2 x because the bottom two are opposites? o/h - h/o?

Answer 3

\[\huge \frac{1-\sin^2x}{\sin x - cscx} =\frac{1-\sin^2x}{\sin x - \frac{1}{\sin x}}\]
\[\huge =\frac{1-\sin^2x}{\frac{\sin^2 x -1}{sinx}}\]
\[\huge =\frac{\cos^2x}{\frac{-(1-\sin^2 x )}{sinx}}\]
\[\huge =\frac{\cos^2x}{-(\cos^2 x )} \times \sin x = -\sin x\]

@This_Is_Batman

Answer 4

Do you think you could explain to me what you did? I'm confused on this part:
\[\frac{ 1 - \sin^2 x }{ \sin x - 1/\sin x } = \frac{ 1 -\sin^2 x }{ \frac{ \sin^2 x-1 }{ \sin x } }\]

Answer 5

Yes off-course. please stand by..

Answer 6

\[\huge \frac{1-\sin^2x}{\frac{\sin^2x-1}{sinx}} = (1-\sin^2x) \times \frac{sinx}{\sin^2x-1}\]
\[\huge = \cos^2 x \times \frac{sinx}{-(1-\sin^2x)}\]
\[\huge = \cos^2 x \times \frac{sinx}{-\cos^2x} = - \sin x\]
@This_Is_Batman

Answer 7

Where did you get the sin^2 x on the bottom?

Answer 8

looking at just the bottom
\[ \sin x - \csc x \\ \sin x - \frac{1}{\sin x }
\]
multiply the sin x by sin x / sin x (to put it over a common denominator)
\[ \frac{ \sin x}{ \sin x} \cdot \sin x - \frac{1}{\sin x } \\
\frac{ \sin^2 x}{ \sin x} - \frac{1}{\sin x } \\
\frac{ \sin^2 x-1}{ \sin x}
\]

at this point I would "flip" the bottom and multiply times the top in the original equation
\[ \frac{ 1-\sin^2 x }{ \sin x - \csc x } = (1-\sin^2 x) \cdot \frac{ \sin x} { \sin^2 x-1}\]
notice that if we factor -1 out (1 - sin^2 x) we have
\[ -(\sin^2 x -1 ) \cdot \frac{ \sin x} { \sin^2 x-1}\]
which simplifies to - sin x

Answer 9

Ah! See,that's all I needed. I'm a little loopy right now. Forgot you had to do the common denominator. Duh. Thanks guys! God bless!