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For-0-lt-x-lt-pi-6-all-the-values-of-tan-2-3x-cos-2-x-4tan-3x-sin-2x-16sin-2-x-lie-in-the-interval-a-0-121-36-b-1-121-9-c-1-0-d-None-of-these-




Question Number 95520 by PRITHWISH SEN 2 last updated on 25/May/20
For 0<x<(π/(6 )), all the values of  tan^2 (3x)cos^2 (x)−4tan (3x)sin (2x)+16sin^2 (x)  lie in the interval  (a). (0,((121)/(36))) (b).(1,((121)/9)) (c). (−1,0) (d). None of these.
$$\mathrm{For}\:\mathrm{0}<\mathrm{x}<\frac{\pi}{\mathrm{6}\:},\:\mathrm{all}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{3x}\right)\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{4tan}\:\left(\mathrm{3x}\right)\mathrm{sin}\:\left(\mathrm{2x}\right)+\mathrm{16sin}\:^{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$\mathrm{lie}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\left(\mathrm{a}\right).\:\left(\mathrm{0},\frac{\mathrm{121}}{\mathrm{36}}\right)\:\left(\mathrm{b}\right).\left(\mathrm{1},\frac{\mathrm{121}}{\mathrm{9}}\right)\:\left(\mathrm{c}\right).\:\left(−\mathrm{1},\mathrm{0}\right)\:\left(\mathrm{d}\right).\:\mathrm{None}\:\mathrm{of}\:\mathrm{these}. \\ $$

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