sqrt{3} csc 20^circ - sec 20^circ = | vedclass

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

Question

(C) $\sqrt{3} \csc 20^\circ - \sec 20^\circ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}$ $= \frac{\sqrt{3} \cos 20^\circ - \sin 20^\cir

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) $\sqrt{3} \csc 20^\circ - \sec 20^\circ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}$ $= \frac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}$ અંશ અને છેદને $2$ વડે ગુણતા: $= \frac{2 \left( \frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ \right)}{\frac{1}{2} (2 \sin 20^\circ \cos 20^\circ)}$ $= \frac{2 (\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ)}{\frac{1}{2} \sin 40^\circ}$ $= \frac{2 \sin(60^\circ - 20^\circ)}{\frac{1}{2} \sin 40^\circ} = \frac{2 \sin 40^\circ}{\frac{1}{2} \sin 40^\circ} = 4$.

$\sqrt{3} \csc 20^\circ - \sec 20^\circ = $

Similar Questions

$\sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8} = $

જો $7 \sin^{2} \theta + 3 \cos^{2} \theta = 4$ હોય,તો $\tan \theta$ ની કિંમત શોધો (જ્યાં $\theta$ લઘુકોણ છે).

જો $\pi < \alpha < \frac{3\pi}{2}$ હોય,તો $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

જો $\csc \theta = \frac{p + q}{p - q}$ હોય,તો $\cot \left( \frac{\pi}{4} + \frac{\theta}{2} \right) = $

$\sin x \cos x$ ની મહત્તમ અને ન્યૂનતમ કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module