Question

How do you find the exact value `tan(x+y)` if `cscx=5/3, cosy=5/13`?

Answer 1

You have to use the identities relating cos(x) and sin(x) (i.e. cos(x) = sqrt( 1-sin^2(x) ), and the compound angle identity for tan(a+b) (i.e. tan(x+y) = (tan(x) + tan(y)) / (1 - tan(x)tan(y)).

Explanation:

(I apologize in advance for the long answer. I've shown all working steps but you could probably skim over these if you already know them.)

Firstly you need to find the values of sin(x) and cos(x), and sin(y) and cos(y).
Starting with cos(y) (since it's already in a form that's easy to work with)...

`cos(y) = 5/13`

Then using the identity `sin (y) = sqrt( 1 - cos^2(y))` so...
`sin(y) = sqrt( 1 - (5/13)^2)`
`= sqrt( 1 - (25/169))`
`= sqrt( 144/169)`
`sin(y) = 12/13`

Then we know that `csc(x) = 1 / sin(x)` so...
`csc(x) = 5/3 = 1/sin(x)`
`sin(x) = 3/5`

And again, the same identity as above, since it applies both ways...
`cos(x) = sqrt( 1 - (sin^2(x))`
`= sqrt( 1 - (3/5)^2)`
`= sqrt( 1 - (9/25))`
`= sqrt( 16/25)`
`cos(x) = 4/5`

Once we have all our sin and cos values we can use them to find their tan values...

`tan(x) = sin(x) / cos(x)`
`= (3/5) / (4/5)`
`= (3/5) * (5/4)`
`tan(x) = 3/4`

and

`tan(y) = sin(y) / cos(y)`
`= (12/13) / (5/13)`
`= (12/13) * (13/5)`
`tan(y) = 12/5`

We then use the tan values to find the value of tan(x+y), using the identity `tan(x+y) = (tan(x) + tan(y)) / (1 - tan(x)tan(y))`

`tan(x+y) = (tan(x) + tan(y)) / (1 - tan(x)tan(y)).`
`= ((3/4) + (12/5)) / ( 1 - (3/4)(12/5))`
`= ((15/20) + (48/20)) / ( 1 - (36/20))`
`= (63/20) / ( -16/20)`
`= (63/20) * (-20/16)`
`tan(x+y) = -63/16`

Answer 2

`tan (x+y) = -3 15/16`

Explanation:

`tan (x+y) = (tanx + tany)/(1-tanx*tany)`

`csc x = 5/3 :. sin x = 3/5 = p/h`; b= base , p= perpendicular , h=hypotenuse)

`h^2= b^2+p^2 ; p=3 , h =5 :. b^2 = h^2 - p^2= 5^2-3^2= 16 :. b = sqrt 16= 4 :. sinx = p/h = 3/5 :. tanx = p/b = 3/4`

`cos y = 5/13 = b/h`

`h^2= b^2+p^2 ; b=5 , h =13 :. p^2 = h^2 - b^2= 13^2-5^2= 144 :. p = sqrt 144= 12 :. siny = p/h = 12/13 :. tany = p/b = 12/5`

`tan (x+y) = (tanx + tany)/(1-tanx*tany) = (3/4+12/5)/(1-3/4*12/5` or

`tan (x+y) = ((15+48)/20)/ (1-36/20) = (63/20)/ (-16/20)` or

`tan (x+y) = (63/cancel20)*(-cancel20/16) = -63/16 =-3 15/16` [Ans]