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

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(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 $D$.

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

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