1/(sin^2x-cos^2x) = 1/[(sinx-cosx)(sinx+cosx)]
so, using a common denominator,
1/(sin^2x-cos^2x) - 2/(cosx-sinx)
= 1/[(sinx-cosx)(sinx+cosx)] + 2(sinx+cosx)/[(sinx-cosx)(sinx+cosx)]
= (1+2sinx+2cosx)/(sin^2x-cos^2x)
I see several typos in your posting. If I have interpreted things wrong, feel free to fix it, using extra parentheses if needed...
How do you simplify:
(1/(sin^2x-cos^2x))-(2/cosx-sinx)?
I tried factoring and creating a LCD of (sinx+cosx)(sinx cosx) (cosx-sinx), but cannot come up with the right answer.
The answer is (1+2sinx+2cosx)/sin^2x+cos^2x,but I don't know how the book arrived at that answer. I'd appreciate any help. Please explain your answer. Thank you.
2 answers
Factoring looks promising:
(1/(sin^2x-cos^2x))-(2/cosx-sinx)
= 1/( (sinx+cosx)(sinx-cosx) ) + 2/(sinx-cosx)
so (sinx+cosx)(sinx-cosx) is the LCD
= 1/( (sinx+cosx)(sinx-cosx) ) + 2(sinx+cosx)/( (sinx+cosx)(sinx-cosx) )
= ( 1 + 2sinx + 2cosx)/(sin^2 x - cos^2 x)
You have a typo in the answer you posted. Your denominator of the answer would be just 1.
(1/(sin^2x-cos^2x))-(2/cosx-sinx)
= 1/( (sinx+cosx)(sinx-cosx) ) + 2/(sinx-cosx)
so (sinx+cosx)(sinx-cosx) is the LCD
= 1/( (sinx+cosx)(sinx-cosx) ) + 2(sinx+cosx)/( (sinx+cosx)(sinx-cosx) )
= ( 1 + 2sinx + 2cosx)/(sin^2 x - cos^2 x)
You have a typo in the answer you posted. Your denominator of the answer would be just 1.
1 answer