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)`

Answer 1

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`