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Mathematics Question on Trigonometric Functions

Find the value of tan(α+β)tan(\alpha + \beta), given that cotα=12,α∈(π,3π2)cot\alpha=\frac{1}{2}, \alpha\in\left(\pi, \frac{3\pi}{2}\right) and secβ=−53,β∈(π2,π)sec\beta=\frac{-5}{3}, \beta\in\left(\frac{\pi}{2}, \pi\right).

A

1/111/11

B

2/112/11

C

5/115/11

D

3/113/11

Answer

2/112/11

Explanation

Solution

Given, cotα=12cot\alpha=\frac{1}{2}, ⇒tanα=2\Rightarrow tan\alpha=2 and secβ=−53sec\beta=\frac{-5}{3} Then, tanβ=sec2 β−1tan\beta=\sqrt{sec^{2}\,\beta-1} ⇒tanβ=±43\Rightarrow tan\beta=\pm\frac{4}{3} But, tanβ=−43tan\beta=\frac{-4}{3} (∵ tanβ(\because\, tan\beta is -veve in IIII quadrant)) ∴tan(α+β)=tan α+tan β1−tan α⋅tan β\therefore tan\left(\alpha+\beta\right)=\frac{tan\,\alpha+tan\,\beta}{1-tan\,\alpha\cdot tan\,\beta} =2+(−43)1−(2)(−43)=211=\frac{2+\left(-\frac{4}{3}\right)}{1-\left(2\right)\left(\frac{-4}{3}\right)}=\frac{2}{11}