Question

How do you prove cotx/(cscx-sinx)=secx

Answer 1

$\dfrac{\cot x}{\csc x-\sin x}=\dfrac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x}-\sin x}= \dfrac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x}-\frac{\sin^2x}{\sin x}}= \dfrac{\frac{\cos x}{\sin x}}{\frac{1-\sin^2x}{\sin x}}=\\\\\\= \dfrac{\frac{\cos x}{\sin x}}{\frac{\cos^2x}{\sin x}}=\dfrac{\cos x\cdot\sin x}{\sin x\cdot\cos^2x}=\dfrac{1}{\cos x}=\boxed{\sec x}$