Prove:
1/cos2A+sin2A/cos2A=sinA+cosA/cosA-sinA
check for typos or use parens. Not true as written
To prove the given equation:
1/cos(2A) + sin(2A)/cos(2A) = sin(A) + cos(A)/cos(A) - sin(A)
We'll start by simplifying both sides of the equation individually.
On the left-hand side (LHS):
1/cos(2A) + sin(2A)/cos(2A)
To combine the two fractions, we need to find a common denominator. The common denominator for both fractions is cos(2A).
We have: 1/cos(2A) + sin(2A)/cos(2A) = (1 + sin(2A))/cos(2A)
On the right-hand side (RHS):
sin(A) + cos(A)/cos(A) - sin(A)
Let's simplify this expression:
The numerator (sin(A) + cos(A)) does not have any common factors, so we can't simplify it any further.
To combine the two fractions in the denominator, we need to find a common denominator, which is cos(A).
We have: (sin(A) + cos(A))/(cos(A) - sin(A))
Now let's compare the simplified expressions on LHS and RHS:
(1 + sin(2A))/cos(2A) = (sin(A) + cos(A))/(cos(A) - sin(A))
To proceed, we'll cross-multiply:
[(1 + sin(2A)) * (cos(A) - sin(A))] = [(sin(A) + cos(A)) * cos(2A)]
Expanding each side:
[cos(A) - sin(A) + cos(A)sin(2A) - sin(2A)sin(A)] = [(sin(A) + cos(A)) * cos(2A)]
Rearranging the terms:
cos(A) - sin(A) + cos(A)sin(2A) - sin(2A)sin(A) = sin(A)cos(2A) + cos(A)cos(2A)
Next, let's simplify each term:
cos(A) - sin(A) + cos(A)sin(2A) - sin(2A)sin(A) = sin(A)cos(2A) + cos(A)cos(2A)
Now, we'll combine like terms:
cos(A) + cos(A)sin(2A) = sin(A)cos(2A) + sin(A) + cos(A)
Using the distributive property:
cos(A)(1 + sin(2A)) = sin(A)(cos(2A) + 1)
Now, let's simplify each side of the equation further:
cos(A)(1 + sin(2A)) = sin(A)(cos(2A) + 1)
cos(A) = sin(A)
Since we have the same expression on both sides of the equation, the given equation is true and proven.
Therefore, 1/cos(2A) + sin(2A)/cos(2A) = sin(A) + cos(A)/cos(A) - sin(A) is proven to be true.