If $\cot \alpha=\frac{1}{2}$ and $\sec \beta=-\frac{5}{3}$ where a $\in\left(\pi, \frac{3 \pi}{2}\right)$…

If $\cot \alpha=\frac{1}{2}$ and $\sec \beta=-\frac{5}{3}$ where a $\in\left(\pi, \frac{3 \pi}{2}\right)$ and $\beta \in\left(\frac{\pi}{2}, \pi\right)$, then $\tan (\alpha+\beta)$ has the value
  1. $\frac{3}{11}$
  2. $\frac{22}{9}$
  3. $\frac{9}{11}$
  4. $\frac{2}{11}$

Solution

$\cot \alpha=\frac{1}{2} \Rightarrow \tan \alpha=2$ $\sec \beta=\frac{-5}{3} \Rightarrow \tan \beta=-\frac{4}{3}$ Now, $\tan (\alpha+\beta)=\frac{\tan \theta+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=\frac{2+\left(\frac{-4}{3}\right)}{1-2 \times\left(\frac{-4}{3}\right)}=\frac{\frac{6-4}{3}}{\frac{3+8}{3}}=\frac{2}{11}$

Asked in: MHT CET 2022 (05 Aug Shift 1)

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