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

The value of $\sin \theta + \cos \theta$ will be greatest when

The value of $\cos \frac{\pi}{10} \cos \frac{2\pi}{10} \cos \frac{4\pi}{10} \cos \frac{8\pi}{10} \cos \frac{16\pi}{10}$ is:

The value of $\frac{1}{\sqrt{2}} \sin \frac{\pi}{6} \cos \frac{\pi}{4} - \cot \frac{\pi}{3} \sec \frac{\pi}{6} + \frac{5 \tan \frac{\pi}{4}}{12 \sin \frac{\pi}{2}}$ is equal to

If $\cos P = \frac{1}{7}$ and $\cos Q = \frac{13}{14}$,where $P$ and $Q$ both are acute angles,then the value of $P - Q$ is....$^o$

If $A > 0, B > 0$ and $A + B = \frac{\pi}{6}$,then the minimum value of $\tan A + \tan B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module