sin 20^{circ}(4+sec 20^{circ})= | vedclass

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

Question

(A) दिया गया व्यंजक: $\sin 20^{\circ}(4+\sec 20^{\circ})$$= \sin 20^{\circ} \left(4 + \frac{1}{\cos 20^{\circ}}\right)$$= \frac{4 \sin 20^{\circ} \cos

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(A) दिया गया व्यंजक: $\sin 20^{\circ}(4+\sec 20^{\circ})$$= \sin 20^{\circ} \left(4 + \frac{1}{\cos 20^{\circ}}\right)$$= \frac{4 \sin 20^{\circ} \cos 20^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{2(2 \sin 20^{\circ} \cos 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{2 \sin 40^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{2 \sin(60^{\circ} - 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{2(\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{\sqrt{3} \cos 20^{\circ}}{\cos 20^{\circ}} = \sqrt{3}$

$\sin 20^{\circ}(4+\sec 20^{\circ})=$

Similar Questions

यदि $\sin \theta = \frac{1}{2} (x + \frac{1}{x})$ है,तो $\sin 3 \theta + \frac{1}{2} (x^3 + \frac{1}{x^3}) = $

$\cos^3 110^{\circ} + \cos^3 10^{\circ} + \cos^3 130^{\circ} = $

यदि $(m + 2)\sin \theta + (2m - 1)\cos \theta = 2m + 1$ है,तो निम्नलिखित में से कौन सा सत्य है?

यदि $\cos \theta + \cos 7\theta + \cos 3\theta + \cos 5\theta = 0$ है,तो $\theta$ का मान ज्ञात कीजिए:

यदि $x = 3 \sin \theta$,$y = 3 \cos \theta \cos \phi$,और $z = 3 \cos \theta \sin \phi$ है,तो $x^{2} + y^{2} + z^{2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module