Help Me?!? plzz...?

Question

(sin^4x - cos^4x) / (sin^2x * cos^2x) = sec^2x - csc^2x
solve the left side to make it equal the right side, plzzzzzzzzzzz

Answer 1

LHS = (sin^4(x) - cos^4(x))/(sin^2(x)cos^2(x))

Factor as difference of squares

[(sin^2(x) - cos^2(x)) (sin^2(x) + cos^2(x))]/(sin^2(x)cos^2(x))

Note that sin^2(x) + cos^2(x) = 1, so we can effectively remove it.

(sin^2(x) - cos^2(x)) / (sin^2(x)cos^2(x))

Now, we split this up into two fractions with the same denominator.

(sin^2(x))/(sin^2(x)cos^2(x)) - (cos^2(x))/(sin^2(x)cos^2(x))

Note the terms that cancel in each fraction.

1/(cos^2(x)) - 1/(sin^2(x))

And now, by the definition of secant and cosecant,

sec^2(x) - csc^2(x) = RHS

Answer 2

Factor the numerator (it's the difference between two squares)

Write the right side in terms of fundamental trig functions:

sec = 1/cos
csc = 1/sin

this will help you see how to use the denominator of the left side.

It's really just a bunch of shuffling.

Answer 3

(sin^4x - cos^4x) / (sin^2(x) * cos^2(x) = (1/cos^2(x)) - (1/sin^2(x))

On the right side, multiply top and bottom of both fractions by sin^2(x) * cos^2(x).

Answer 4

min^1x - min^x2