frac{1}{sin 250^{circ}}+frac{sqrt{3}}{cos 290 | vedclass

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

Question

(B) दिया गया व्यंजक: $\frac{1}{\sin 250^{\circ}}+\frac{\sqrt{3}}{\cos 290^{\circ}}$$= \frac{1}{\sin(270^{\circ}-20^{\circ})} + \frac{\sqrt{3}}{\cos(27

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) दिया गया व्यंजक: $\frac{1}{\sin 250^{\circ}}+\frac{\sqrt{3}}{\cos 290^{\circ}}$$= \frac{1}{\sin(270^{\circ}-20^{\circ})} + \frac{\sqrt{3}}{\cos(270^{\circ}+20^{\circ})}$$= -\frac{1}{\cos 20^{\circ}} + \frac{\sqrt{3}}{\sin 20^{\circ}}$$= \frac{-\sin 20^{\circ} + \sqrt{3} \cos 20^{\circ}}{\cos 20^{\circ} \sin 20^{\circ}}$$= \frac{2 \left( -\frac{1}{2} \sin 20^{\circ} + \frac{\sqrt{3}}{2} \cos 20^{\circ} \right)}{\frac{1}{2} \sin 40^{\circ}}$$= \frac{2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$$= \frac{2 \sin(60^{\circ}-20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$$= \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$अतः,विकल्प $B$ सही है.

$\frac{1}{\sin 250^{\circ}}+\frac{\sqrt{3}}{\cos 290^{\circ}} = $

Similar Questions

यदि $\sec x + \tan x = 3, x \in (0, \frac{\pi}{2})$ है,तो $\sin x =$

$\sin^{2} 51^{\circ} + \sin^{2} 39^{\circ}$ का मान है

निम्नलिखित रेडियन माप के संगत डिग्री माप ज्ञात कीजिए ($\pi = \frac{22}{7}$ का प्रयोग करें):
$-4$ रेडियन

$\sin \left( \frac{\pi}{10} \right) \sin \left( \frac{3\pi}{10} \right) = $

$\sin 690^{\circ} \times \sec 240^{\circ} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module