Question

Simplify the expression.

(cos^2x + sin^2x)/(cot^2x - csc^2x)

Answer 1

can I get a fan and medal if I get it correct?

Answer 2

Hint: Pythagorean identities.

Answer 3

the ans is 1

Answer 4

they both are identities

Answer 5

(cos^2x+sin^2x) / (csc^2x-cot^2x )=
-(1/1)=
-1

Answer 6

-1 is correct

Answer 7

Thank you.

Answer 8

the numerator = 1
for the denominator use the identittyy

csc^x = 1 + cot^2x

Answer 9

yup -1 is correct i thought it was cosec^2x - cot^2x

Answer 10

\(\color{red}{\sin^2 x + \cos^2 x = 1}\) Eq. 1

Divide both sides of Eq. 1 by \(\sin^2 x\):

\(1 + \cot^2x = \csc^2 x\)

\(\color{green}{\cot^2 x - \csc^2 x = -1}\) Eq. 2

Eq. 1 and Eq. 2 above are the identities you need.

\(\dfrac{\color{red}{\cos^2 x + \sin^2 x}}{\color{green}{\cot^2 x - \csc^2 x}} = \dfrac{\color{red}{1}}{\color{green}{-1}} = -1\)

Answer 11

(cos^2x + sin^2x)/(cot^2x - csc^2x)

Fundamental identity shows: cos^2x+sin^2x=1
Plug in for (cos^2x+sin^2x)

=1/(cot^2x-csc^2x)

Fundamental identity shows: 1+cot^2x=csc^2x
-csc^2x -csc^2x
1+cot^2x-csc^2=0
-1 -1
cot^2x-csc^2x=-1
Plug in for (cot^2x-csc^2x)

=(1/-1)
=-1