1-sin^2x/sinx-cscx How do I solve this problem? I keep getting the wrong answer.

Answer:-sinxStep-by-step explanation:a trig identity that is crucial to solving this problem is: sin^2 + cos^2 = 1with knowing that, you can manipulate that and turn it into 1 - sin^2x = cos^xso 1-sin^2x/sinx - cscx becomes cos^2x/sinx - cscxit is also important to know that cscx is the same thing as 1/sinxknowing this information, cscx can be replaced with 1/sinx(cos^2x)/(sinx - 1/sinx)now sinx and 1/sinx do not have the same denominator, so we need to multiply top and bottom of sinx by sinx; it becomes....cos^2x---------------------(sin^2x - 1)/sinxnotice how in the denominator it has sin^2x-1 which is equal to -cos^2xso now it becomes:cos^2x---------------cos^2x/sinxbecause we have a fraction over a fraction, we need to flip itcos^2x          sinx---------- * ----------------1                  -  cos^2xbecause the cos^2x can cancel out, it becomes 1now the answer is -sinx