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THE MINIMUM VALUE OF 9 tan2A + 4 cot2A is
(a) 13 (b) 9 (c) 6 (d) 12
Explain how...
9 tan sqer a + 4 c0t swer a
=9 sin ^4 A + 4 cos^4 A
------------------------------- = [3sn^2 A+ 2cos^2 A]^2 - 12 sin^2 A COS ^2 A (1+sin ^2 A)^2
sin^2 A COS ^2 A ---------------------------------------------------- = --------------- -12
sin^2 A COS ^2 A sin^2 A COS ^2 A
SO MINIMUM VALUE IS 12........
use AM >= GM
(9 tan2A + 4 cot2A)/2 >=sqrt ( 9tan2A * 4cot2A) =6
(9 tan2A + 4 cot2A) >= 12
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