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Question
Find the value of tan(α + β), given that cot α = `1/2`, α ∈ `(pi, (3pi)/2)` and sec β = `- 5/3` β ∈ `(pi/2, pi)`
Solution
sec β = ` 5/3`
sec2β – tan2β = 1
sec2β – 1 = tan2β
tan2β = `(- 5/3)^2 - 1`
= `25/9 - 1`
tan2β = `(25 - 9)/9`
= `16/9`
tan β = `+- 4/3`
Given that β lies in the second quadant.
tan β is negative.
∴ tan β = `- 4/3`
tan(α + β) = `(tan alpha + tan beta)/(1 - tan alpha tan beta)`
= `(2 - 4/3)/(1 - 2 xx (- 4/3))`
= `((6 - 4)/3)/(1 + 8/3)`
= `((6 - 4)/3)/((3 + 8)/3)`
tan(α + β) = `2/11`
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