Let x -0- -2 and log 24sin x 24 cos x -3-2- then the value of cosec2 x is equal toA- 6B- 3 - 8D- 9

The correct option is D 9(24 sin x)^3/2=24 cos x ⇒√(24)(sin x)^3/2=cos x ⇒ 24 sin^3 x = cos^2 x = 1 sin^2 x; Let sin x=t ⇒ 24t^3 + t^2 1=0 ⇒ (3t 1) D lt;0(8t ...

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Byju's Answer

Standard XII

Mathematics

Theorems for Continuity

Let x ϵ0, π/2...

Question

Let $x ϵ (0, \frac{π}{2})$ and $l o g_{24 s i n x} (24 c o s x) = \frac{3}{2}$ , then the value of $c o s e c^{2} x$ is equal to

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Solution

The correct option is D 9
$(24 s i n x)^{\frac{3}{2}} = 24 c o s x$
$\Rightarrow \sqrt{24} (s i n x)^{\frac{3}{2}} = c o s x$
$\Rightarrow 24 s i n^{3} x = c o s^{2} x = 1 - s i n^{2} x$ ; Let $s i n x = t \Rightarrow 24 t^{3} + t^{2} - 1 = 0$
$\Rightarrow (3 t - 1) (8 t^{2} + 3 t + 1) D < 0      = 0$
$t = \frac{1}{3}$
$\Rightarrow s i n x = \frac{1}{3}$
$\Rightarrow c o s e c^{2} x = 9$

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Let xϵ(0,π2) and log24 sin x(24 cos x)=32, then the value of cosec2 x is equal to

The correct option is D 9(24 sin x)32=24 cos x ⇒√24(sin x)32=cos x ⇒24 sin3x=cos2x=1−sin2x; Let sin x=t⇒24t3+t2−1=0 ⇒(3t−1)(8t2+3t+1)D<0=0 t=13 ⇒sin x=13 ⇒cosec2 x=9

Let $x ϵ (0, \frac{π}{2})$ and $l o g_{24 s i n x} (24 c o s x) = \frac{3}{2}$ , then the value of $c o s e c^{2} x$ is equal to