tan left(-frac{23 pi}{3}right)-cot left(theta | vedclass

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

Question

(D) We have the expression: $	an \left(-\frac{23 \pi}{3}ight)-\cot \left(	heta-\frac{13 \pi}{3}ight)$ Since $	an(-x) = -	an(x)$,we have $	an

Vedclass · Accepted Answer

$\sqrt{3}+\cot \left(\frac{\pi}{3}-	hetaight)$

Vedclass · Answer

(D) We have the expression: $	an \left(-\frac{23 \pi}{3}ight)-\cot \left(	heta-\frac{13 \pi}{3}ight)$ Since $	an(-x) = -	an(x)$,we have $	an \left(-\frac{23 \pi}{3}ight) = -	an \left(\frac{23 \pi}{3}ight) = -	an \left(8 \pi - \frac{\pi}{3}ight) = -(-	an \frac{\pi}{3}) = \sqrt{3}$. For the second term,$\cot \left(	heta-\frac{13 \pi}{3}ight) = \cot \left(	heta - (4 \pi + \frac{\pi}{3})ight) = \cot \left(	heta - \frac{\pi}{3}ight) = -\cot \left(\frac{\pi}{3} - 	hetaight)$. Substituting these back into the expression: $\sqrt{3} - (-\cot \left(\frac{\pi}{3} - 	hetaight)) = \sqrt{3} + \cot \left(\frac{\pi}{3} - 	hetaight)$. Thus,the correct option is $D$.

$\tan \left(-\frac{23 \pi}{3}\right)-\cot \left(\theta-\frac{13 \pi}{3}\right) =$

Similar Questions

If $\theta$ lies in the second quadrant,then the value of $\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} + \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}}$ is:

Find the values of the other five trigonometric functions if $\sec x = \frac{13}{5}$ and $x$ lies in the fourth quadrant.

$(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)$ is equal to

Factorize the expression: $2 \cot^2 \theta - \cot \theta - 3$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module