Find minimum value of y=((1−sin^2 x)/(2 tan^2 x))


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Question Number 192343 by cortano12 last updated on 15/May/23

$Find minimum value of$ $y = \frac{1 - \sin^{2} x}{2 \tan^{2} x}$
	Answered by manxsol last updated on 15/May/23
	$x≠kπ(π/2) y>0 2ysin^2 x=(1−sin^2 x)^2 sin^4 x−(2+2y)sin^2 x+1 Δ>0 (2+2y)^2 −4>0 y<−2 ∩ y>0 sol y>0−{f(x)/x=kπ/2} min y>0$
	$x \neq k π \frac{π}{2} y > 0$ $2 {y s i n}^{2} x = {(1 - {s i n}^{2} x)}^{2}$ ${s i n}^{4} x - (2 + 2 y) {s i n}^{2} x + 1$ $Δ > 0$ ${(2 + 2 y)}^{2} - 4 > 0$ $y < - 2 \cap y > 0$ $s o l y > 0 - {f (x) / x = k π / 2}$ $m i n y > 0$
	Commented by mehdee42 last updated on 15/May/23

	$o k$ $t h e r e i s n o t m i n i m u m$
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