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

\(\tan \left(-\frac{23 \pi}{3}\right)-\cot \left(\theta-\frac{13 \pi}{3}\right)=\)
  1. \(\sqrt{3}+\cot \theta\)
  2. \(\sqrt{3}-\tan \left(\frac{\pi}{6}+\theta\right)\)
  3. \(\sqrt{3}+\tan \theta\)
  4. \(\sqrt{3}+\cot \left(\frac{\pi}{3}-\theta\right)\)

Solution

\(\begin{aligned} & \tan \left(-\frac{23 \pi}{3}\right)-\cot \left(\theta-\frac{13 \pi}{3}\right) \\ & =-\tan \left(7 \pi+\frac{2 \pi}{3}\right)+\cot \left(4 \pi+\left(\frac{\pi}{3}-\theta\right)\right) \\ & =-\tan \frac{2 \pi}{3}+\cot \left(\frac{\pi}{3}-\theta\right) \\ & =-\tan \left(\pi-\frac{\pi}{3}\right)+\cot \left(\frac{\pi}{3}-\theta\right) \\ & =\tan \frac{\pi}{3}+\cot \left(\frac{\pi}{3}-\theta\right)=\sqrt{3}+\cot \left(\frac{\pi}{3}-\theta\right) \end{aligned}\) Hence, option (d) is correct.

Asked in: AP EAMCET 2020 (21 Sep Shift 1)

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