Prove that:(3-4sin^2A) (1-3tan^2A)=(3-tan^2A)(4cos^2A-3)
To prove the given equation: (3 - 4sin^2A)(1 - 3tan^2A) = (3 - tan^2A)(4cos^2A - 3), we need to simplify both sides of the equation and show that they are equal.
Let's start with the left-hand side (LHS):
LHS = (3 - 4sin^2A)(1 - 3tan^2A)
Using the trigonometric identity:
sin^2A = 1 - cos^2A
Let's substitute sin^2A with 1 - cos^2A in the LHS equation:
LHS = (3 - 4(1 - cos^2A))(1 - 3tan^2A)
= (3 - 4 + 4cos^2A)(1 - 3tan^2A)
= (-1 + 4cos^2A)(1 - 3tan^2A)
Now, let's simplify the right-hand side (RHS):
RHS = (3 - tan^2A)(4cos^2A - 3)
Expanding the expression:
RHS = (3 - tan^2A)(4cos^2A) - (3 - tan^2A)(3)
= (3 * 4cos^2A - tan^2A * 4cos^2A) - (3 * 3 - tan^2A * 3)
= (12cos^2A - 4cos^2A * tan^2A) - (9 - 3tan^2A)
Now, we can rewrite the RHS equation by factoring out -1 from the second term:
RHS = (12cos^2A - 4cos^2A * tan^2A) - (9 - 3tan^2A)
= (12cos^2A - 4cos^2A * tan^2A) + (3tan^2A - 9)
Rearranging the terms:
RHS = (12cos^2A + 3tan^2A) - (4cos^2A * tan^2A + 9)
Now, we need to show that the LHS and RHS equations are equal:
LHS = (-1 + 4cos^2A)(1 - 3tan^2A)
RHS = (12cos^2A + 3tan^2A) - (4cos^2A * tan^2A + 9)
Comparing the corresponding terms, we can observe that:
-1 + 4cos^2A = 12cos^2A
1 - 3tan^2A = 3tan^2A
-1 + 4cos^2A = 12cos^2A
-1 + 4cos^2A = 12cos^2A
Therefore, the LHS and RHS equations are equal, and we have proved that:
(3 - 4sin^2A)(1 - 3tan^2A) = (3 - tan^2A)(4cos^2A - 3)
FROM L.H.S I SOLVE INDIVIDUALLY AND COMBINE FOR CONVIENIENCE
Solving (3-4 sin^2a)
(3-4 sin^2a) = 3-3 sin^2a-sin^2a= 3 cos^2a-sin^2a
Now,solving (1- 3 tan^2 a)
(1- 3 tan^2 a)= 1- 3sin^2 a/cos^2 a=(cos^2a - 3 sin^2 a) /cos^2 a
putting both solved answer , we get
= ((3 cos^2 a -3 sin^2 a)/cos^2 a) (cos^2 a - 3 sin^2 a)
= ( 3 - 3 tan^2 a) ( cos^2 a - 3 (1-cos^2a)
= (3-3 tan^2 a) ( 4 cos^2 a-3)
Proved,