1-sin^2x/sinx-cscx

how do i solve this problem? i keep getting the wrong answer.

-sinx

Step-by-step explanation:

a trig identity that is crucial to solving this problem is: sin^2 + cos^2 = 1

with knowing that, you can manipulate that and turn it into 1 - sin^2x = cos^x

so 1-sin^2x/sinx - cscx becomes cos^2x/sinx - cscx

it is also important to know that cscx is the same thing as 1/sinx

knowing 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)/sinx

notice how in the denominator it has sin^2x-1 which is equal to -cos^2x

so now it becomes:

cos^2x

-cos^2x/sinx

because we have a fraction over a fraction, we need to flip it

cos^2x          sinx

*

1                  -  cos^2x

because the cos^2x can cancel out, it becomes 1

now the answer is -sinx

answer: slight chance that it will happen

step-by-step explanation: