SOLUTION: find the eact value is 0<x<pie/2 and 0<y<pie/2
tan(x+y) if cscx=5/3 and cosy=5/13
I am aware that I have to use the sum idenity for the tangent function but I have no clue how
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Trigonometry-basics
-> SOLUTION: find the eact value is 0<x<pie/2 and 0<y<pie/2
tan(x+y) if cscx=5/3 and cosy=5/13
I am aware that I have to use the sum idenity for the tangent function but I have no clue how
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Question 34256: find the eact value is 0
tan(x+y) if cscx=5/3 and cosy=5/13
I am aware that I have to use the sum idenity for the tangent function but I have no clue how to change csc and cos into a tangent idenity if that is even what I am supposed to do. I have tried a few times but I get different wrong answers each time. Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! tan(x+y)=(TAN(X)+TAN(Y))/(1-TAN(X)*TAN(Y))
if cscx=5/3=HYPOTENUSE/OPPOSITE SIDE....SO USING PYTHOGARUS
ADJ.SIDE^2+OPP.SIDE^2=HYP.^2.....ADJ.^2+3^2=5^2...ADJ.^2=25-9=16
ADJ.SIDE=4....TAN(X)=OPP.SIDE/ADJ.SIDE=3/4
and cosy=5/13 =ADJ.SIDE/HYP.
5^2+OPP.SIDE^2=13^2
OPP.SIDE^2=169-25=144
OPP.SIDE=12
TAN(Y)=OPP.SIDE/ADJ.SIDE=12/5
TAN(X+Y)=(3/4 + 12/5 )/(1 - (3/4)(12/5))
=(3*5+12*4)/(4*5-3*12)=(15+48)/(20-36)=-63/16
