Question

prove cotx/(cscx-sinx)=secx

Answer 1

Then simplify the last expression

Answer 2

\[\frac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x}-\sin x}=\frac{\cos x}{1-\sin ^2x}=\frac{\cos x}{\cos ^2x}=\frac{1}{\cos x}=\sec x\]

Answer 3

LHS = cotx /[(1-sin^2 x)/sinx]
= cotx / (cos^2 x/sinx)
= cotc /(cotxcosx)
= 1/cosx
= secx
= RHS

Answer 4

for the third line, i meant
= cotx /(cotxcosx)

Answer 5

\[(\frac{\cos}{\sin})\div(\frac{1}{\sin}-\sin)\]\[(\frac{\cos}{\sin})\div(\frac{1-\sin^2}{\sin})\]\[(\frac{\cos}{\sin})\div(\frac{\cos^2}{\sin})=(\frac{\cos.\sin}{\sin.\cos^2})=\frac{\cos}{\cos^2}\]\[\frac{\cos}{\cos}\times\frac{1}{\cos}=1\times\frac{1}{\cos}\]\[secx=secx\]I forget x