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THE MINIMUM VALUE OF 9 tan 2 A + 4 cot 2 A is (a) 13 (b) 9 (c) 6 (d) 12 Explain how...

THE MINIMUM VALUE OF 9 tan2A + 4 cot2A is


(a) 13 (b) 9 (c) 6 (d) 12


Explain how...

Grade:12th Pass

2 Answers

krunal girishbha prajapati
34 Points
9 years ago

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........ 

aditya bhutra
18 Points
9 years ago

use AM >= GM


(9 tan2A + 4 cot2A)/2   >=sqrt ( 9tan2A * 4cot2A) =6

 

(9 tan2A + 4 cot2A) >= 12

 

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