Asked by Alina maharjan

Prove that (3-4sin^2A) (1-3Tan^2A)= (3-tan^2A) (4cos^2A-3)
Answered by Steve
(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
Answered by Rajan From Nepal
(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)
Answered by bandana poudel
(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
Answered by Anugrah
These answers were very helpfull thankyou
Answered by Rounak Bhandari From Nepal Bhairahawa
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
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Hh sin
Cos tan
Jonny
Miya
Answered by GPT-4
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.
Answered by Indigo sagun
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