Prove that (3-4sin^2A) (1-3Tan^2A)= (3-tan^2A) (4cos^2A-3)

Answer 1

(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

Answer 2

(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)

Answer 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

Answer 4

These answers were very helpfull thankyou

Answer 5

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...

Answer 6

Bshsvbb
Hh sin
Cos tan
Jonny
Miya

Answer 7

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.

Answer 8

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