JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 19)
If cot$$\alpha$$ = 1 and sec$$\beta$$ = $$ - {5 \over 3}$$, where $$\pi < \alpha < {{3\pi } \over 2}$$ and $${\pi \over 2} < \beta < \pi $$, then the value of $$\tan (\alpha + \beta )$$ and the quadrant in which $$\alpha$$ + $$\beta$$ lies, respectively are :
$$ - {1 \over 7}$$ and IVth quadrant
7 and Ist quadrant
$$-$$7 and IVth quadrant
$$ {1 \over 7}$$ and Ist quadrant
Explanation
$\because \cot \alpha=1, \quad \alpha \in\left(\pi, \frac{3 \pi}{2}\right)$
then $\tan \alpha=1$
and $\sec \beta=-\frac{5}{3}, \quad \beta \in\left(\frac{\pi}{2}, \pi\right)$
then $\tan \beta=-\frac{4}{3}$
$\therefore \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}$
$$ \begin{aligned} &=\frac{1-\frac{4}{3}}{1+\frac{4}{3}} \\\\ &=-\frac{1}{7} \end{aligned} $$
$$ \alpha+\beta \in\left(\frac{3 \pi}{2}, 2 \pi\right) \text { i.e. fourth quadrant } $$
then $\tan \alpha=1$
and $\sec \beta=-\frac{5}{3}, \quad \beta \in\left(\frac{\pi}{2}, \pi\right)$
then $\tan \beta=-\frac{4}{3}$
$\therefore \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}$
$$ \begin{aligned} &=\frac{1-\frac{4}{3}}{1+\frac{4}{3}} \\\\ &=-\frac{1}{7} \end{aligned} $$
$$ \alpha+\beta \in\left(\frac{3 \pi}{2}, 2 \pi\right) \text { i.e. fourth quadrant } $$
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