Question

21.
The minimum value of 9 tan^2x + 4 cot^2x is..,...........​

Answer 1

Answer:

min value is 12

Step-by-step explanation:

use AM is greater than or equal to gm

(9tan^2x+4cot^2x)/2 $\geq$ $\sqrt{(9tan^2x)(4cot^2x)}$

as,tanx=1/cotx

(9tan^2x+4cot^2x)  $\geq$  2*$\sqrt{9*4*1}$

(9tan^2x+4cot^2x)$\geq$2*6

(9tan^2x+4cot^2x) $\geq$ 12

hence min value is 12