prove that 1-tan 4 A by1+tan 4 A =cosA + sinA by cosA - sinA
To prove the given equation, we will start with the left-hand side (LHS) and manipulate it to simplify it into the right-hand side (RHS) of the equation. Here's how we can proceed:
Step 1: Expand the numerator of LHS using the formula for the difference of two squares:
1 - tan^2(4A) = 1 - sin^2(4A)/cos^2(4A)
Step 2: Apply the trigonometric identities for sine and cosine of double angles:
1 - sin^2(4A)/cos^2(4A) = cos^2(4A)/cos^2(4A) - sin^2(4A)/cos^2(4A) = (cos^2(4A) - sin^2(4A))/cos^2(4A)
Step 3: Apply the Pythagorean identity for cosine and sine:
(cos^2(4A) - sin^2(4A))/cos^2(4A) = cos(2*4A)/cos^2(4A)
Step 4: Simplify the expression:
cos(8A)/[2cos^2(2A)-1]
Step 5: Rewrite the denominator using the double-angle formula for cosine:
cos(8A)/(cos(4A))^2 - sin(4A)^2 = cos(8A)/(cos(4A))^2 - (1 - cos(8A))/2
Step 6: Obtain a common denominator:
cos(8A)/[(cos(4A))^2 - (1 - cos(8A))/2] = cos(8A)/[(2cos(4A))^2 - 2(1 - cos(8A))/2]
Step 7: Simplify the expression:
cos(8A)/[4cos(4A)^2 - 2 + 2cos(8A)] = cos(8A)/(4cos(4A)^2 + 2cos(8A) - 2)
Step 8: Apply the double-angle formula for cosine:
cos(8A)/(2cos(2*4A)^2 + 2cos(8A) - 2) = cos(8A)/(2cos(16A)^2 + 2cos(8A) - 2)
Step 9: Apply the triple-angle formula for cosine:
cos(8A)/(2[4cos(8A)^3 - 3cos(8A)]^2 + 2cos(8A) - 2) = cos(8A)/(2[16cos(8A)^6 - 48cos(8A)^4 + 36cos(8A)^2 - 3] + 2cos(8A) - 2)
Step 10: Simplify the expression:
cos(8A)/[32cos(8A)^6 - 96cos(8A)^4 + 72cos(8A)^2 - 6 + 2cos(8A) - 2]
Step 11: Combine like terms:
cos(8A)/[32cos(8A)^6 - 96cos(8A)^4 + 74cos(8A)^2 - 8]
Finally, we have the expression cos(8A)/[32cos(8A)^6 - 96cos(8A)^4 + 74cos(8A)^2 - 8], which is the simplified form of the LHS. Comparing it with the RHS, cosA + sinA/(cosA - sinA), we can see that they are not equivalent. Therefore, the given equation 1 - tan 4A / (1 + tan 4A) = cosA + sinA / (cosA - sinA) is not true.