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Question
If 2 sinA = 1 = `sqrt(2)` cosB and `pi/2` < A < `pi`, `(3pi)/2` < B < `2pi`, then find the value of `(tan"A" + tan"B")/(cos"A" - cos"B")`
Solution
Given: 2 sin A = 1
∴ sin A = `1/2`
Now, sin2A + cos2A = 1
∴ cos2A = 1 - sin2A
= `1 - (1/2)^2`
= `1 - 1/4`
`= 3/4`
∴ cos A = `± sqrt(3)/2`
But`pi/2 <"A" < pi`, i.e., A lies in the second quadrant.
∴ cos A is negative.
∴ cos A = `-sqrt(3)/2`
∴ tan A = `sin"A"/cos"A"`
= `((1/2))/(-(sqrt(3))/2)`
= `-1/sqrt(3)`
Also, `sqrt(2)` cos B = 1
∴ cos B = `1/sqrt(2)`
We know that,
∴ sin2B = 1 – cos2B
= `1 - (1/sqrt(2))^2`
= `1 - 1/2 = 1/2`
∴ sin B = `± 1/sqrt(2)`
But `(3pi)/2 < "B" < 2pi`, i.e., B lies in the fourth quadrant.
∴ sin B is negative
∴ sin B = `-1/sqrt(2)`
∴ tan B = `sin"B"/cos"B" = ((-1/sqrt(2)))/((1/sqrt(2))` = – 1
∴ `(tan"A" + tan"B")/(cos"A" - cos"B") = (-1/sqrt(3) - 1)/(-sqrt3/2 - 1/sqrt(2))`
= `(((-1 - sqrt(3))/sqrt(3)))/(((-sqrt(3) - sqrt(2))/2)`
= `(-(sqrt(3) + 1))/(sqrt(3)) xx 2/(-(sqrt(3)+sqrt(2))`
=`(2(1 + sqrt(3)))/(sqrt(3)(sqrt(3) + sqrt(2))`
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