The number of solutions of the equation csc t | vedclass

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

Question

(D) Given equation: $\csc 	heta - \cot 	heta = 1$Using $\csc 	heta = \frac{1}{\sin 	heta}$ and $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$,we

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given equation: $\csc 	heta - \cot 	heta = 1$Using $\csc 	heta = \frac{1}{\sin 	heta}$ and $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$,we get:$\frac{1 - \cos 	heta}{\sin 	heta} = 1$$\Rightarrow 1 - \cos 	heta = \sin 	heta$Using half-angle formulas $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$:$2 \sin^2 \frac{	heta}{2} = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$$\Rightarrow 2 \sin \frac{	heta}{2} (\sin \frac{	heta}{2} - \cos \frac{	heta}{2}) = 0$Case $1$: $\sin \frac{	heta}{2} = 0$ $\Rightarrow \frac{	heta}{2} = 0, \pi$ $\Rightarrow 	heta = 0, 2\pi$. However,at $	heta = 0$ and $	heta = 2\pi$,$\csc 	heta$ and $\cot 	heta$ are undefined. So,these are not solutions.Case $2$: $\sin \frac{	heta}{2} - \cos \frac{	heta}{2} = 0 \Rightarrow 	an \frac{	heta}{2} = 1$$\Rightarrow \frac{	heta}{2} = \frac{\pi}{4}$ $\Rightarrow 	heta = \frac{\pi}{2}$.Checking $	heta = \frac{\pi}{2}$: $\csc(\frac{\pi}{2}) - \cot(\frac{\pi}{2}) = 1 - 0 = 1$. This is a valid solution.Thus,there is only $1$ solution.

The number of solutions of the equation $\csc \theta - \cot \theta = 1$ in the interval $[0, 2\pi]$ is:

Similar Questions

The number of real roots of the equation $\sin^{2020} x - \cos^{2020} x + 2019 = 2020$ in the interval $\left(-\frac{3\pi}{2}, \frac{5\pi}{2}\right)$ is:

The solution set of the equation $\sin^2 \theta - \cos \theta = \frac{1}{4}$ in the interval $[0, 2\pi]$ is

If $\cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 0$,where $0 < \theta < 180^\circ$,then $\theta = $

The general solution of $\cos 4\theta = \cos 3\theta$ is $\theta =.....$

If $n$ is any integer,then the general solution of the equation $\cos x - \sin x = \frac{1}{\sqrt{2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module