Question

If sin x = `15/17` and cos y = `12/13, 0 < x < pi/2, 0 < y < pi/2`, find the value of tan(x + y)

Answer 1

Given sin x = `15/17, 0 < x < pi/2`

We have cos²x + sin²x = 1

∴ cos²x = 1 – sin²x

= `1 - (15/17)^2`

= `1 - 225/289`

cos²x = `(289 - 225)/289 = 64/289`

cos x = `+-  sqrt(64/289)`

= `+-  8/17`

Given that `0 < x < pi/2`, that is x lies in the first quadrant

∴ cos x is positive.

cos x = `8/17`

Also given cos y = `12/13, 0 < x < pi/2`

We have cos²y + sin²y = 1

sin²y = 1 – cos²y

sin²y = `1 - (12/13)^2 = 1 - 14/169`

sin²y = `(169 - 144)/169 = 25/169`

sin y = `+-  sqrt(25/169) = +-  5/13`

Since `0 < y < pi/2, y lies in the first quadrant sin y is positive.

∴ sin y = `5/13`

sin x = `15/17`

sin y = `5/13`

cos x = `8/17`

cos y = `12/13`

tan(x + y)

tan x = `sinx/cosx = (15/17)/(8/17) = 15/8`

tan y = `siny/cosy = (5/13)/(12/13) = 5/12`

tan(x + y) = `(tanx + tany)/(1 - tanx tany)`

= `(15/8 + 5/12)/(1 - 15/8*5/12)`

= `((180 + 40)/96)/((96 - 75)/96)`

tan(x + y) = `220/21`