Prove the identity cos theta - sin theta / cos theta + sin theta = sec 2 theta - tan 2 theta.
omitting all the θ's for simplicity,
(cos-sin)/(cos+sin)
multiply top and bottom by (cos-sin)
(cos-sin)^2 / (cos^2 - sin^2)
(cos^2 - 2sin*cos + sin^2)/cos2θ
(1 - 2sin*cos)/cos2θ
(1-sin2θ)/cos2θ
sec2θ - tan2θ
To prove the identity cos(theta) - sin(theta) / cos(theta) + sin(theta) = sec^2(theta) - tan^2(theta), we start with the left-hand side (LHS) of the equation and simplify it using trigonometric identities.
LHS: cos(theta) - sin(theta) / cos(theta) + sin(theta)
To simplify the LHS, we multiply the numerator and denominator by the conjugate of the denominator, which is cos(theta) - sin(theta):
LHS = (cos(theta) - sin(theta)) * (cos(theta) - sin(theta)) / (cos(theta) + sin(theta)) * (cos(theta) - sin(theta))
Using the identity (a - b)(a + b) = a^2 - b^2, we simplify the numerator:
LHS = (cos^2(theta) - 2sin(theta)cos(theta) + sin^2(theta)) / (cos^2(theta) - sin^2(theta))
Next, we can replace sin^2(theta) using the Pythagorean identity sin^2(theta) = 1 - cos^2(theta):
LHS = (cos^2(theta) - 2sin(theta)cos(theta) + (1 - cos^2(theta))) / (cos^2(theta) - (1 - cos^2(theta)))
Simplifying further, we combine like terms:
LHS = (2cos^2(theta) - 2sin(theta)cos(theta) + 1) / (2cos^2(theta) - 1)
Now, let's work on simplifying the right-hand side (RHS) of the equation.
RHS: sec^2(theta) - tan^2(theta)
Using the identity sec^2(theta) = 1/cos^2(theta) and tan^2(theta) = sin^2(theta) / cos^2(theta), we can substitute these values into the RHS:
RHS = 1/cos^2(theta) - sin^2(theta) / cos^2(theta)
Combining the two fractions:
RHS = (1 - sin^2(theta)) / cos^2(theta)
Again, using the Pythagorean identity sin^2(theta) = 1 - cos^2(theta):
RHS = cos^2(theta) / cos^2(theta)
Simplifying further, we get:
RHS = 1
Since the LHS and RHS are equal to each other, we have proven the identity:
cos(theta) - sin(theta) / cos(theta) + sin(theta) = sec^2(theta) - tan^2(theta)