The number of solutions of the given equation | vedclass

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

Question

(B) Given equation: $\sec 	heta + 	an 	heta = \sqrt{3}$ ... $(i)$We know that $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given equation: $\sec 	heta + 	an 	heta = \sqrt{3}$ ... $(i)$We know that $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Substituting $(i)$ into this,we get $\sec 	heta - 	an 	heta = \frac{1}{\sqrt{3}}$ ... $(ii)$Adding $(i)$ and $(ii)$ gives $2 \sec 	heta = \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$,so $\sec 	heta = \frac{2}{\sqrt{3}}$,which implies $\cos 	heta = \frac{\sqrt{3}}{2}$.Subtracting $(ii)$ from $(i)$ gives $2 	an 	heta = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$,so $	an 	heta = \frac{1}{\sqrt{3}}$.Since $\cos 	heta = \frac{\sqrt{3}}{2}$ and $	an 	heta = \frac{1}{\sqrt{3}}$,$	heta$ must be in the first quadrant.For $0 Thus,there is only $1$ solution.

The number of solutions of the given equation $\tan \theta + \sec \theta = \sqrt{3}$,where $0 < \theta < 2\pi$ is

Similar Questions

The general solution of the equation $\tan x + \tan 2x - \tan 3x = 0$ is

The number of solution$(s)$ of the equation $\sin 2\theta + \cos 2\theta = -\frac{1}{2}$ for $\theta \in \left( 0, \frac{\pi}{2} \right)$ is:

If $(2\cos x - 1)(3 + 2\cos x) = 0$ for $0 \le x \le 2\pi$,then $x = $

The number of solutions of the given equation $a \sin x + b \cos x = c$,where $|c| > \sqrt{a^2 + b^2}$,is

If $4\sin^2 \theta + 2(\sqrt{3} + 1)\cos \theta = 4 + \sqrt{3}$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module