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")`

Answer 1

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))`