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 $2$:$E = \frac{2 \left( \frac{1}{2} \cos \frac{\pi}{18} - \frac{\sqrt{3}}{2} \sin \frac{\pi}{18} \right)}{\frac{1}{2} \sin \frac{2\pi}{18}} = \frac{2 \left( \sin \frac{\pi}{6} \cos \frac{\pi}{18} - \cos \frac{\pi}{6} \sin \frac{\pi}{18} \right)}{\frac{1}{2} \sin \frac{\pi}{9}}$.Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$E = \frac{2 \sin(\frac{\pi}{6} - \frac{\pi}{18})}{\frac{1}{2} \sin \frac{\pi}{9}} = \frac{4 \sin(\frac{3\pi - \pi}{18})}{\sin \frac{\pi}{9}} = \frac{4 \sin \frac{2\pi}{18}}{\sin \frac{\pi}{9}} = \frac{4 \sin \frac{\pi}{9}}{\sin \frac{\pi}{9}} = 4$.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 $x \cos \theta = y \cos \left( \theta + \frac{2\pi}{3} \right) = z \cos \left( \theta + \frac{4\pi}{3} \right)$,then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is equal to

The value of $\cot \frac{\pi}{20} \cot \frac{3 \pi}{20} \cot \frac{5 \pi}{20} \cot \frac{7 \pi}{20} \cot \frac{9 \pi}{20}$ is

If $\sec \theta = x + \frac{1}{4x}$ where $0^{\circ} < \theta < 90^{\circ}$,then $\sec \theta + \tan \theta$ is equal to:

If $\left| a\sin^2 \theta + b\sin \theta \cos \theta + c\cos^2 \theta - \frac{1}{2}(a + c) \right| \le \frac{1}{2}k,$ then $k^2$ is equal to

If $\sin A, \cos A$ and $\tan A$ are in $G.P.$,then $\cos^3 A + \cos^2 A$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module