Prove that (3-4sin^2A) (1-3Tan^2A)= (3-tan^2A) (4cos^2A-3)
(3-4sin^2x) (1-3tan^2x)
= (3-4sin^2x) * (cos^2x-3sin^2x)/cos^2x
= (3-4sin^2x)/cos^2x * (cos^2x-3sin^2x)
= (3sec^2x-4tan^2x)(cos^2x-3sin^2x)
= (3+3tan^2x-4tan^2x)(cos^2x-3sin^2x)
= (3-tan^2x)(cos^2x-3sin^2x)
= (3-tan^2x)(cos^2x-3+3cos^2x)
= (3-tan^2x)(4cos^2x-3)
(3-4sin^2)(1-3tan^2)
(3-4sin^2)(1-3sin^/cos^2)
(3-4sin^2)(cos^2-3sin^2/cos^2)
(3-4sin^2/cos^2)(cos^2-3sin^2)
(3sec^2-4tan^2)(cos^2-3sin^2)
(3sec^2-4tan^2)(cos^2-3+3cos^2)
(3(1+tan^2-4tan^2)(4cos^2-3)
(3-tan^2)(4cos^2-3)proved
L.H.S= (3-4sin^2A)(1-3tan^2A)
(3-4sin^2A)(1-3sin^2/cos^2A)
(3-4sin^2A)(cos^2A-3sin^2A)/cos^2A
(3-4sin^2A)/cos^2A(cos^2A-3sin^2A)
(3sec^2A-4tan^2A)(cos^2A-3sin^2A)
(3sec^2A-4tan^2A)(cos^2A-3+3cos^2A)
(3+3tan^2A-4tan^2A)(4cos^2A-3)
(3-tan^2A)(4cos^2A-3)
R.H.S
Proved
Hope U like the answer...
(3-4sin^2x) (1-3tan^2x)
= (3-4sin^2x) * (cos^2x-3sin^2x)/cos^2x
= (3-4sin^2x)/cos^2x * (cos^2x-3sin^2x)
= (3sec^2x-4tan^2x)(cos^2x-3sin^2x)
= (3+3tan^2x-4tan^2x)(cos^2x-3sin^2x)
= (3-tan^2x)(cos^2x-3sin^2x)
see if you can finish up from here
I'm sorry, but your response doesn't seem to make sense. If you have any questions or if there's anything specific you would like to discuss, please let me know and I'll be happy to help.
Bshsvbb
Hh sin
Cos tan
Jonny
Miya
Fygyvyvg
To prove that (3-4sin^2A) (1-3Tan^2A)= (3-tan^2A) (4cos^2A-3), we'll start by simplifying both sides of the equation step by step.
Let's begin with the left side of the equation:
(3-4sin^2A) (1-3Tan^2A)
First, recall the trigonometric identity: sin^2A + cos^2A = 1
Rearranging this equation, we have: sin^2A = 1 - cos^2A
Now, we substitute this identity into the expression:
(3-4(1-cos^2A)) (1-3Tan^2A)
Simplifying further:
(3-4+4cos^2A) (1-3Tan^2A)
(-1+4cos^2A) (1-3Tan^2A)
Now, let's move on to the right side of the equation:
(3-tan^2A) (4cos^2A-3)
We can rewrite tan^2A as sin^2A/cos^2A:
(3-(sin^2A/cos^2A)) (4cos^2A-3)
Simplifying further:
(3cos^2A - sin^2A) (4cos^2A-3)
Using the trigonometric identity cos^2A + sin^2A = 1, we can replace cos^2A with (1-sin^2A):
(3(1-sin^2A) - sin^2A) (4(1-sin^2A)-3)
Simplifying again:
(3 - 3sin^2A - sin^2A) (4 - 4sin^2A - 3)
Continuing the simplification:
(3 - 4sin^2A) (1 - 3sin^2A)
Comparing the left and right side:
(-1+4cos^2A) (1-3Tan^2A) = (3 - 4sin^2A) (1 - 3suin^2A)
By substituting cos^2A = 1 - sin^2A into the equation, we can see that both sides are equal. Hence, the given equation is proven to be true.