Question

1-sin^2x/ (sinx-cscx)
a.-cos x

b. -sin x

c. sin2x

d. cos2x

Answer 1

wht do u think

Answer 2

write \(\csc x\) as \(\frac{1}{\sin x}\) and try to simplify
\[\dfrac{1-\sin^2x}{\sin x-\csc x} = \dfrac{1-\sin^2x}{\sin x-\frac{1}{\sin x}} = \dfrac{\sin x(1-\sin^2 x)}{\sin^2 x-1} = -\sin x\]

Answer 3

Oh I got Cos^2x...