find the exact value of tan(x+y) if cot x=6/5 and sec y=3/2

Question

is there any formulas I can use to solve this kind of question?

Answer 1

tan(A) = sin(A)/cos(A)

 

First let's deal with the x:

 

if cot x = 6/5, that means adjacent is 6 and opposite is 5.

 

From the Pythagorean formula, we know that:

 

opposite^2 + adjacent^2 = hypotenuse^2

 

So,

 

5^2 + 6^2 = hyp^2

 

hyp = sqrt(61)

 

sin = opp/hyp

cos = adj/hyp

 

So:

 

sin(x) = 5/sqrt(61)

cos(x) = 6/sqrt(61)

 

Now onto the y:

 

if sec y = 3/2, that means cos(y) = 2/3

 

if cos(y) = 2/3, that means that adjacent is 2 and hypotenuse is 3.

 

Again, from the Pythagorean formula, we know that:

 

opposite^2 + adjacent^2 = hypotenuse^2

 

So,

 

opp^2 + 2^2 = 3^2.

 

opp = sqrt(5)

 

this means that:

 

sin(y) = sqrt(5)/3

 

Now since you want to know tan(x+y), recognize that:

 

tan(x+y) = sin(x+y)/cos(x+y)

 

From trig you'll remember:

 

sin(x+y) = sin(x)cos(y) + sin(y)cos(x)

cos(x+y) = cos(x)cos(y) + sin(x)sin(y)

 

Gathering all of your information above into one place:

 

sin(x) = 5/sqrt(61)

cos(x) = 6/sqrt(61)

sin(y) = sqrt(5)/3

cos(y) = 2/3

 

tan(x+y) = (sin(x)cos(y) + sin(y)cos(x)) / (cos(x)cos(y) + sin(x)sin(y))

 

That's a lot to type in, but just plug and simplify.