If cot theta + cot left( frac{pi }{4} + theta | vedclass

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

Question

(D) Given equation: $\cot 	heta + \cot \left( \frac{\pi }{4} + 	heta ight) = 2$ Using $\cot A + \cot B = \frac{\sin(A+B)}{\sin A \sin B}$,we get:

Vedclass · Accepted Answer

$n\pi \pm \frac{\pi }{6}$

Vedclass · Answer

(D) Given equation: $\cot 	heta + \cot \left( \frac{\pi }{4} + 	heta ight) = 2$ Using $\cot A + \cot B = \frac{\sin(A+B)}{\sin A \sin B}$,we get: $\frac{\sin(	heta + \frac{\pi}{4} + 	heta)}{\sin 	heta \sin(\frac{\pi}{4} + 	heta)} = 2$ $\frac{\sin(\frac{\pi}{4} + 2	heta)}{\sin 	heta \sin(\frac{\pi}{4} + 	heta)} = 2$ Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$: $\sin(\frac{\pi}{4} + 2	heta) = \cos(\frac{\pi}{4} + 	heta - 	heta) - \cos(\frac{\pi}{4} + 	heta + 	heta)$ $\sin(\frac{\pi}{4} + 2	heta) = \cos(\frac{\pi}{4}) - \cos(\frac{\pi}{4} + 2	heta)$ $\sin(\frac{\pi}{4} + 2	heta) + \cos(\frac{\pi}{4} + 2	heta) = \frac{1}{\sqrt{2}}$ Multiply by $\frac{1}{\sqrt{2}}$: $\frac{1}{\sqrt{2}} \sin(\frac{\pi}{4} + 2	heta) + \frac{1}{\sqrt{2}} \cos(\frac{\pi}{4} + 2	heta) = \frac{1}{2}$ $\sin(\frac{\pi}{4} + 2	heta + \frac{\pi}{4}) = \frac{1}{2}$ $\sin(2	heta + \frac{\pi}{2}) = \frac{1}{2}$ $\cos(2	heta) = \frac{1}{2}$ $2	heta = 2n\pi \pm \frac{\pi}{3}$ $	heta = n\pi \pm \frac{\pi}{6}$

If $\cot \theta + \cot \left( \frac{\pi }{4} + \theta \right) = 2$,then the general value of $\theta$ is

Similar Questions

The number of solutions of $\tan(5\pi \cos \theta) = \cot(5\pi \sin \theta)$ for $\theta$ in $(0, 2\pi)$ is:

If $\cos x = |\sin x|$,then the general solution is

The number of solutions of the equation $4 \sin^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0$ for $x \in [-2\pi, 2\pi]$ is:

The number of distinct solutions of the equation $\log _{\frac{1}{2}}|\sin x|=2-\log _{\frac{1}{2}}|\cos x|$ in the interval $[0,2 \pi]$ is

The number of solutions in $[0, 2\pi]$ of the equation $16^{\sin^2 x} + 16^{\cos^2 x} = 10$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module