Question

prove that
If cot a =1/2 sec B = -5/3, where π < a < 3π\2 and
<π\2<b<π, find the value of
tan(a + b ). State the quadrant in which a + B terminates.

Answer 1

Given

cotA = 1/2 , where π<a<3π/2

secB = -5/3 , where π/2<b<π

To find

tan(a+b) = ?

Solution

As we know that ,

tan(a+b) = tan a + tan b/ 1- tan a . tanB

tan a ==>

we know that tan is reciprocal of cot, so

tan a = 1/cotA

tana = 1/(1/2)

tan a = 2

As angle a lies in 3rd quadrant here tan functions as positive , so

tan a = 2

tan b ==>

we know that,

1+tan²b = sec²b, so

tanB. = √ sec²b - 1

tanB = √ (-5/3)²-1

tanB = √ 25/9 -1

tanB = 4/3

As angle B lies in 2nd quadrant here tan functions as negative , so

tan B = -4/3

Now we have to find tan( a+b)

tan (a+b) = tan a+ tan b/ 1- tana . tanB.

tan ( a+b) = 2 +(-4/3)/ 1- 2(-4/3)

tan ( a+b) = (2/3 )/ 11/3

tan ( a+b ) = 2/11

Answer 2

Answer:

2/11

Step-by-step explanation:

cot a=1/2 so, tan a=2

sec b=-5/3

1+tan²b=sec²b

tan²b=sec²b-1

tanb=√sec²b-1

tanb=√(-5/3)²-1

tanb=√(25/9)-1

tanb=√16/9

tanb=-4/3.

tan(a+b)=tana+tanb/1-tana.tanb

tan(a+b)=(2-4/3)/1-2(-4/3)

2/3×3/11

2/11