Question

If 5 tan A =√2 π<A<3π/2 and sec B = √3/2<B<2π Then find cosec A - tan B​

Answer 1

Given that 5 tan A =√2

therefore tan A =√2/5

cot A = 1/tan A

cot A = 1/√2/5 = 5/√2

we have 1+cot^2A = cosec^2 A

1 +(5/√2)^2 = cosec^2 A

1+25/2 = cosec^A

cosec A =+- √27/2

but π<A<3π/2

therefore A lies in 3rd quadrant

cosec A = - √27/2

and sec B = √11

we have,

1+ tan^2B=sec^2B

tan^2B = sec^2 B-1 = (√11)^2-1

=11-1

tan^2B = 10

tan B = +- √10

But 3π/2<B<2π

B lies in 4th quadrant

therefore tanB =-√10

therefore cosecA - cosec B

=-√27/2 -(-√10)

=-√27/2+√10