The general value of theta satisfying the equ | vedclass

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

Question

(B) Given the equation: $	an 	heta + 	an \left( \frac{\pi}{2} - 	heta ight) = 2$ Since $	an \left( \frac{\pi}{2} - 	heta ight) = \cot 	heta

Vedclass · Accepted Answer

$n\pi + \frac{\pi}{4}$

Vedclass · Answer

(B) Given the equation: $	an 	heta + 	an \left( \frac{\pi}{2} - 	heta ight) = 2$ Since $	an \left( \frac{\pi}{2} - 	heta ight) = \cot 	heta$,the equation becomes: $	an 	heta + \cot 	heta = 2$ $	an 	heta + \frac{1}{	an 	heta} = 2$ Multiplying by $	an 	heta$: $	an^2 	heta - 2 	an 	heta + 1 = 0$ $(	an 	heta - 1)^2 = 0$ $	an 	heta = 1 = 	an \frac{\pi}{4}$ The general solution for $	an 	heta = 	an \alpha$ is $	heta = n\pi + \alpha$. Therefore,$	heta = n\pi + \frac{\pi}{4}$.

The general value of $\theta$ satisfying the equation $\tan \theta + \tan \left( \frac{\pi}{2} - \theta \right) = 2$ is:

Similar Questions

Find the general solution of $4 \cos 2x - 4 \sqrt{3} \sin 2x + \cos 3x - \sqrt{3} \sin 3x + \cos x - \sqrt{3} \sin x = 0$.

If $\sin^2 \theta = \frac{1}{4},$ then the most general value of $\theta$ is

If $\cot (\alpha + \beta ) = 0,$ then $\sin (\alpha + 2\beta ) = $

The number of solutions of the equation $\sin A - 5 \sin 2A + \sin 3A = \cos A - 5 \cos 2A + \cos 3A$ in the interval $(0, \pi)$ is:

The number of values of $x$ in the interval $\left(\frac{\pi}{4}, \frac{7 \pi}{4}\right)$ for which $14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 21 - 4 \cos^{2} x$ holds,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module