The value of csc frac{pi}{18} - sqrt{3} sec f | vedclass

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

Question

(D) Let $E = \csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18} = \frac{1}{\sin \frac{\pi}{18}} - \frac{\sqrt{3}}{\cos \frac{\pi}{18}}$.$E = \frac{\co

Vedclass · Accepted Answer

natural number

Vedclass · Answer

(D) Let $E = \csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18} = \frac{1}{\sin \frac{\pi}{18}} - \frac{\sqrt{3}}{\cos \frac{\pi}{18}}$.$E = \frac{\cos \frac{\pi}{18} - \sqrt{3} \sin \frac{\pi}{18}}{\sin \frac{\pi}{18} \cos \frac{\pi}{18}}$.Multiply numerator and denominator by $2$:$E = \frac{2(\frac{1}{2} \cos \frac{\pi}{18} - \frac{\sqrt{3}}{2} \sin \frac{\pi}{18})}{\frac{1}{2} \sin \frac{\pi}{9}} = \frac{4(\sin \frac{\pi}{6} \cos \frac{\pi}{18} - \cos \frac{\pi}{6} \sin \frac{\pi}{18})}{\sin \frac{\pi}{9}}$.Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$E = \frac{4 \sin(\frac{\pi}{6} - \frac{\pi}{18})}{\sin \frac{\pi}{9}} = \frac{4 \sin(\frac{3\pi - \pi}{18})}{\sin \frac{\pi}{9}} = \frac{4 \sin \frac{2\pi}{18}}{\sin \frac{\pi}{9}} = \frac{4 \sin \frac{\pi}{9}}{\sin \frac{\pi}{9}} = 4$.Since $4$ is a natural number,the correct option is $D$.

The value of $\csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18}$ is a

Similar Questions

If $\tan \left(\frac{\pi}{9}\right), x, \tan \left(\frac{7 \pi}{18}\right)$ are in arithmetic progression and $\tan \left(\frac{\pi}{9}\right), y, \tan \left(\frac{5 \pi}{18}\right)$ are also in arithmetic progression,then $|x-2 y|$ is equal to:

Evaluate: $\sin 21^{\circ} \cos 9^{\circ}-\cos 84^{\circ} \cos 6^{\circ}$

$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = $

In a triangle $ABC$,if $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$,then $\sin (A + B)$ is equal to

The product $(1+\tan 1^{\circ})(1+\tan 2^{\circ})(1+\tan 3^{\circ}) \dots (1+\tan 45^{\circ})$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module