The number of distinct solutions of sec theta | vedclass

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

Question

$\sec 	heta+	an 	heta=\sqrt{3}$

$\Rightarrow \quad 1+\sin 	heta=\sqrt{3} \cos 	heta(\cos 	heta 
eq 0)$

$\Rightarrow \quad \sqrt{3} \cos \

Vedclass · Accepted Answer

None of these

Vedclass · Answer

$\sec 	heta+	an 	heta=\sqrt{3}$

$\Rightarrow \quad 1+\sin 	heta=\sqrt{3} \cos 	heta(\cos 	heta 
eq 0)$

$\Rightarrow \quad \sqrt{3} \cos 	heta-\sin 	heta=1$

$\Rightarrow \quad \frac{\sqrt{3}}{2} \cos 	heta-\frac{1}{2} \sin 	heta=\frac{1}{2}$

$\Rightarrow \quad \cos 	heta \cos \frac{\pi}{6}-\sin 	heta \sin \frac{\pi}{6}=\cos \frac{\pi}{3}$

$\Rightarrow \quad \cos \left(	heta+\frac{\pi}{6}ight)=\cos \frac{\pi}{3}$

$\Rightarrow \quad 	heta+\frac{\pi}{6}=2 n \pi \pm \frac{\pi}{3}$             $\left( {n \in Z} ight)$

$\Rightarrow \quad 	heta=2 n \pi-\frac{\pi}{6} \pm \frac{\pi}{3}$              $\left( {n \in Z} ight)$

$=2 n \pi+\frac{\pi}{6} 	ext { or } 2 n \pi-\frac{\pi}{2}$

But we reject the values $	heta=2 \mathrm{n} \pi-\frac{\pi}{2}$

$\because \quad \cos 	heta=0$ for this values of $	heta$

$	herefore $ $	heta=2 \mathrm{n} \pi+\frac{\pi}{6}, \mathrm{n} \in \mathrm{I}$

Thus, $	heta=\frac{\pi}{6}$ is the only value of $	heta \in[0,2, \pi]$

The number of distinct solutions of $\sec \theta \,\, + \,\,\tan \theta \, = \,\sqrt 3 \,,\,0\,\, \leqslant \,\,\theta \,\, \leqslant \,\,2\pi$

Similar Questions

The number of solutions of the equation $\cos \left(x+\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x, x \in[-3 \pi$ $3 \pi]$ is

If $\sec x\cos 5x + 1 = 0$, where $0 < x < 2\pi $, then $x =$

Let $S=\left[-\pi, \frac{\pi}{2}\right)-\left\{-\frac{\pi}{2},-\frac{\pi}{4},-\frac{3 \pi}{4}, \frac{\pi}{4}\right\}$. Then the number of elements in the set $=\{\theta \in S : \tan \theta(1+\sqrt{5} \tan (2 \theta))=\sqrt{5}-\tan (2 \theta)\}$ is $...$

If $A, B, C, D$ are the angles of a cyclic quadrilateral taken in order, then
$cos(180^o + A) + cos(180^o -B) + cos(180^o -C) -sin(90^o -D)=$

Find the principal and general solutions of the equation $\cot x=-\sqrt{3}$