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  • If cotx =1/2 , secy=-5/3 , where pi<x<3pi/2 and pi/2<y<pi Find the value of tan(x+y).

If cotx =1/2 , secy=-5/3 , where pi

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Solution: we have ,

\ cotx=1/2Rightarrow tanx=2

Since y lies in second quadrant . Therefore ,tany is negative .Consequently,

\ 1+tan^2y=sec^2yRightarrow tany=-sqrtsec^2y-1=-sqrt25/9 -1=-4/3

Hence , tan(x+y)=tanx+tany/1-tanxcdot tany=(2-4/3)/1-2 imes -4/3=2/11

Posted by

Deependra Verma

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