Question

If sin A = `(-5)/13, pi < "A" < (3pi)/2` and cos B = `3/5, (3pi)/2 < "B" < 2pi` find tan (A + B)

Answer 1

Given, sin A = `(-5)/13`

We know that, 

cos²A = 1 – sin²A = `1 - (-5/13)^2`

= `1 - 25/169`

= `144/169`

∴ cos A = `±12/13`

Since, `pi < "A" < (3pi)/2`

∴ ‘A’ lies in the 3^rd quadrant

∴ cos A < 0

∴ cos A = `(-12)/13`

Also, cos B = `3/5`

∴ sin²B = 1 – cos²B = `1 - (3/5)^2`

= `1 - 9/25`

= `16/25`

∴ sin B = `±4/5`

Since, `(3pi)/2 < "B" < 2pi`

∴ ‘B’ lies in the 4^th quadrant.

∴ sin B < 0

∴ sin B = `(-4)/5`

tan A = `sin"A"/cos"A" = ((-5/13))/((-12/13)) = 5/12`

tan B = `sin"B"/cos"B" = ((-4/5))/((3/5)) = -4/3`

tan (A + B) = `(tan"A" + tan"B")/(1 - tan"A" tan"B")`

= `(5/12 - 4/3)/(1 - (5/12)(-4/3)`

= `((-33/36))/((56/36))`

= `-33/56`