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Question
If sin A = `(-5)/13, pi < "A" < (3pi)/2` and cos B = `3/5, (3pi)/2 < "B" < 2pi` find tan (A + B)
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Solution
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 3rd 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 4th 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`
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