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1-sin^2x/sinx-cscx

How do I solve this problem? I keep getting the wrong answer.

1-sin^2x/sinx-cscx How do I solve this problem? I keep getting the wrong answer.-example-1
User Leanora
by
7.6k points

1 Answer

1 vote

Answer:

-sinx

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

User Conal Tuohy
by
8.6k points