Asked by Rishabh
Prove that:(3-4sin^2A) (1-3tan^2A)=(3-tan^2A)(4cos^2A-3)
Answers
Answered by
SAGAR BASNET
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,
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,
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