Question

(Secx/sinx)-(sinx/cosx)

Answer 1

$\dfrac{\sec x}{\sin x}-\dfrac{\sinx }{\cos x}=\dfrac{\sec x\cos x}{\sin x\cos x}-\dfrac{\sin^2x}{\sin x\cos x}=\dfrac{1-\sin^2x}{\sin x\cos x}=\dfrac{\cos^2x}{\sin x\cos x}=\dfrac{\cos x}{\sin x}=\cot x$