(3-4sin^2a)(1-3tan^2a)=(3-tan^2a)(4cos^2a-3)
3-4sin^2a = (3sec^2a-4tan^2a)*cos^2a = (3(tan^2a+1)-4tan^2a)*cos^2a = (3-tan^2a)*cos^2a
1-3tan^2a = 1-3(sec^2a-1) = 4-3sec^2a = (4cos^2a-3)/cos^2a
now just multiply and the cos^2a factors cancel
To prove the given equation, we need to simplify both sides of the equation and show that they are equal.
Starting with the left-hand side (LHS):
LHS = (3 - 4sin^2(a))(1 - 3tan^2(a))
Using the Pythagorean identity tan^2(a) = sin^2(a)/cos^2(a), we can rewrite the expression as:
LHS = (3 - 4sin^2(a))(1 - 3sin^2(a)/cos^2(a))
Expanding the expression:
LHS = 3 - 3sin^2(a) - 4sin^2(a) + 12sin^4(a)/cos^2(a)
Simplifying by combining like terms:
LHS = 3 - 7sin^2(a) + 12sin^4(a)/cos^2(a)
Now let's focus on the right-hand side (RHS):
RHS = (3 - tan^2(a))(4cos^2(a) - 3)
Again, using the Pythagorean identity tan^2(a) = sin^2(a)/cos^2(a), we can rewrite the expression as:
RHS = (3 - sin^2(a)/cos^2(a))(4cos^2(a) - 3)
Expanding the expression:
RHS = 3(4cos^2(a) - 3) - sin^2(a)/cos^2(a)(4cos^2(a) - 3)
Simplifying by distributing and combining like terms:
RHS = 12cos^2(a) - 9 - 4sin^2(a) + 3sin^2(a)/cos^2(a)
Combining like terms on the RHS:
RHS = 12cos^2(a) - 4sin^2(a) + 3sin^2(a)/cos^2(a) - 9
Comparing the LHS and RHS expressions, we can see that they both simplify to:
3 - 7sin^2(a) + 12sin^4(a)/cos^2(a)
Hence, the LHS is equal to the RHS, and the given equation is proven to be true.