Trig (inverse functions)
posted by Dennis .
Problem:
tan[arccsc(5/3) + arctan(1/4)]
My work:
let arccsc(5/3)=X and let arctan(1/4)=Y
where pi/2<=X<=pi/2, X cannot be 0
and where pi/2<Y<pi/2
so that cscX=5/3 and tanY=1/4
The problem can now be written as
tan(X+Y)
=tanX + tanY/1tanXtanY which could also be written as
=sinXcosY + cosXsinY/cosXcosY  sinXsinY
I drew the respective triangles and came up with (for the X triangle) a=4, b=3, c=5. (for the Y triangle) a=1, b=4,c=sqrt17
so sinX=3/5 cosX=4/5
sinY=4/sqrt17 cosY=1/sqrt17
giving:
=[(3/5)(1/sqrt17) + (4/5)(4/sqrt17)]/[(4/5)(1/sqrt17)(3/5)(4/sqrt17)]
=[13/(5*sqrt17)]/[16/(5*sqrt17)]
This is not the correct answer. The book shows an answer of 8/19
Can someone please tell me where I went wrong. You don't have to give me the whole working, just where I started to go wrong.
Thank you so much. (hopefully all that typed stuff makes sense, it's hard without being able to use mathematical characters)

Trig (inverse functions) 
Dennis
I found my mistake, please disregard.
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