If frac{1-10 i cos theta}{1-10 sqrt{3} i sin | vedclass

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

Question

(A) To make the expression purely real,its imaginary part must be zero. Let $z = \frac{1-10 i \cos 	heta}{1-10 \sqrt{3} i \sin 	heta}$. Multiply the

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) To make the expression purely real,its imaginary part must be zero. Let $z = \frac{1-10 i \cos 	heta}{1-10 \sqrt{3} i \sin 	heta}$. Multiply the numerator and denominator by the conjugate of the denominator: $z = \frac{(1-10 i \cos 	heta)(1+10 \sqrt{3} i \sin 	heta)}{(1-10 \sqrt{3} i \sin 	heta)(1+10 \sqrt{3} i \sin 	heta)}$ $z = \frac{1 + 10 \sqrt{3} i \sin 	heta - 10 i \cos 	heta + 100 \sqrt{3} \sin 	heta \cos 	heta}{1 + 300 \sin^2 	heta}$ The imaginary part is $\frac{10 \sqrt{3} \sin 	heta - 10 \cos 	heta}{1 + 300 \sin^2 	heta}$. Setting the imaginary part to $0$: $10 \sqrt{3} \sin 	heta - 10 \cos 	heta = 0$ $10 \sqrt{3} \sin 	heta = 10 \cos 	heta$ $	an 	heta = \frac{10}{10 \sqrt{3}} = \frac{1}{\sqrt{3}}$ Since $	an 	heta = \frac{1}{\sqrt{3}}$,we have $	heta = \frac{\pi}{6}$.

If $\frac{1-10 i \cos \theta}{1-10 \sqrt{3} i \sin \theta}$ is purely real,then one of the values of $\theta$ is

Similar Questions

The modulus-amplitude form of $\frac{(1-i)^3(2-i)}{(2+i)(1+i)}$ is

If $\frac{(2-i) x+(1+i)}{2+i}+\frac{(1-2 i) y+(1-i)}{1+2 i}=1-2 i$,then $2 x+4 y=$

$(r, \theta)$ denotes $r(\cos \theta + i \sin \theta)$. If $x = (1, \alpha)$,$y = (1, \beta)$,$z = (1, \gamma)$ and $x + y + z = 0$,then $\sum \cos (2\alpha - \beta - \gamma) = $

The value of $\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3$ is

If $z=\cos 6^{\circ}+i \sin 6^{\circ}$,then $\sum_{n=1}^{20} \operatorname{Im}\left(z^{2 n-1}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module