Question

If tan α = `1/7`, sin β = `1/sqrt10`. Prove that α + 2β = `pi/4`  where 0 < α < `pi/2` and 0 < β < `pi/2`.

Answer 1

Given that tan α = `1/7`

We wish to find tan(α + 2β)

AB² = AC² - BC²

AB² = 10 - 1

AB² = 9

AB = 3

sin β = `1/sqrt10 = "Opposite side"/"Hypotenuse"`

tan β = `"Opposite side"/"Adjacent side" = 1/3`   (Here β is an acute angle)

Now tan 2β = `(2 tan beta)/(1 - tan^2 beta)`

`= (2 (1/3))/(1 - (1/3)^2) = (2/3)/(8/9)`

`= 2/3 xx 9/8 = 3/4`

Consider tan (α + 2β) = `(tan alpha + tan 2 beta)/(1 - tan alpha  tan 2beta)`

`= (1/7 + 3/4)/(1 - 1/7 xx 3/4)`

`= ((1 xx 4 + 3xx7)/28)/(1 - 3/28)`

`= (25/28)/(25/28)` = 1

tan (α + 2β) = `tan pi/4  (because tan pi/4 =1)` 

∴ α + 2β = `pi/4`