If 12 cot^2 theta - 31 csc theta + 32 = 0,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation: $12 \cot^2 	heta - 31 \csc 	heta + 32 = 0$Using the identity $\cot^2 	heta = \csc^2 	heta - 1$,we substitute:$12(\csc^2

Vedclass · Accepted Answer

$\frac{4}{5}$ or $\frac{3}{4}$

Vedclass · Answer

(C) Given the equation: $12 \cot^2 	heta - 31 \csc 	heta + 32 = 0$Using the identity $\cot^2 	heta = \csc^2 	heta - 1$,we substitute:$12(\csc^2 	heta - 1) - 31 \csc 	heta + 32 = 0$Simplify the equation:$12 \csc^2 	heta - 12 - 31 \csc 	heta + 32 = 0$$12 \csc^2 	heta - 31 \csc 	heta + 20 = 0$Factor the quadratic equation in terms of $\csc 	heta$:$12 \csc^2 	heta - 16 \csc 	heta - 15 \csc 	heta + 20 = 0$$4 \csc 	heta (3 \csc 	heta - 4) - 5 (3 \csc 	heta - 4) = 0$$(4 \csc 	heta - 5)(3 \csc 	heta - 4) = 0$This gives two possible values for $\csc 	heta$:$\csc 	heta = \frac{5}{4}$ or $\csc 	heta = \frac{4}{3}$Since $\sin 	heta = \frac{1}{\csc 	heta}$,we have:$\sin 	heta = \frac{4}{5}$ or $\sin 	heta = \frac{3}{4}$.

If $12 \cot^2 \theta - 31 \csc \theta + 32 = 0$,then the value of $\sin \theta$ is

Similar Questions

If the general solution of the equation $\frac{\tan 3x - 1}{\tan 3x + 1} = \sqrt{3}$ is $x = \frac{n\pi}{p} + \frac{7\pi}{q}$ where $n, p, q \in \mathbb{Z}$,then $\frac{p}{q}$ is

If $2 \cos^{2} \theta + 3 \cos \theta = 2$,then the permissible value of $\cos \theta$ is

If $\tan (\cot x) = \cot (\tan x),$ then $\sin 2x =$

The general solution of $1+\sin ^{2} x=3 \sin x \cdot \cos x$,where $\tan x \neq \frac{1}{2}$,is

The number of values of $x$ with $0 \leq x \leq 2 \pi$ satisfying the equation $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module