Prove:
1/cos2A+sin2A/cos2A=sinA+cosA/cosA-sinA
1/cos^2A -sin^2A + 2cosAsinA/cos^2A -sin^2A = 1+2cosAsinA/cos^2A -sin^2A
I used the trig identies to expand 1/cos2A and sin2A/cos2A and added the together because they have the same denominator.
How I got from 1 to cos^2+sin^2 is by trig identity cos^2+sin^2=1.
1+2cosAsinA/cos^2A -sin^2A ==> cos^2+sin^2+2cosAsinA/cos^2A -sin^2A (factorise this equation both the numerator and denominator which will give you the following: )
(cosA+sinA)(cosA+sinA)/(cosA+sinA)(cosA-sinA) (Cancel out the common factors)
cosA+sinA/cosA-sinA
Hope this makes sense!
To prove the equation
1/cos^2(A) + sin^2(A)/cos^2(A) = sin(A)/cos(A) + cos(A)/cos(A) - sin(A),
we will work on each side of the equation separately and simplify them:
Left side of the equation:
1/cos^2(A) + sin^2(A)/cos^2(A)
To combine the fractions on the left side, we need to find a common denominator. The common denominator for the two fractions is cos^2(A). Rewriting the fractions, we have:
1/cos^2(A) + sin^2(A)/cos^2(A) = (1 + sin^2(A))/cos^2(A).
Using the trigonometric identity sin^2(A) + cos^2(A) = 1, we can simplify the numerator further:
(1 + sin^2(A))/cos^2(A) = cos^2(A)/cos^2(A).
Cancelling out the common factor, we obtain:
cos^2(A)/cos^2(A) = 1.
Right side of the equation:
sin(A)/cos(A) + cos(A)/cos(A) - sin(A)
Combining the fractions on the right side, we have:
sin(A)/cos(A) + cos(A)/cos(A) - sin(A) = (sin(A) + cos(A) - sin(A))/cos(A).
Simplifying the numerator, we obtain:
(sin(A) + cos(A) - sin(A))/cos(A) = cos(A)/cos(A).
Cancelling out the common factor:
cos(A)/cos(A) = 1.
Therefore, the left side of the equation equals the right side of the equation:
1/cos^2(A) + sin^2(A)/cos^2(A) = sin(A)/cos(A) + cos(A)/cos(A) - sin(A).
Therefore, the given equation is proven.