How do you find the exact value $tan(x+y)$ if $cotx=6/5,secy=3/2$ ?

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Answer 1 · 2017-07-19T13:53:51

If $y$ is in $Q1$ , $tan(x+y)=(5+3sqrt5)/(6-5sqrt5)$ and
if $y$ is in $Q4$ , $tan(x+y)=(5-3sqrt5)/(6+5sqrt5)$

As $cotx=6/5$ , $tanx=1/cotx=1/(6/5)=5/6$

and as $secy=3/2$ , $tany=+-sqrt((3/2)^2-1)=+-sqrt(9/4-1)=+-sqrt(5/4)=+-sqrt5/2$

Observe that $tany=sqrt5/2$ , when $y$ is in $Q1$ and $tany=-sqrt5/2$ , when $y$ is in $Q4$ .

Hence there canj be two values of $tan(x+y)$

If $y$ is in $Q1$ , $tan(x+y)=(tanx+tany)/(1-tanxtany)=(5/6+sqrt5/2)/(1-5/6xxsqrt5/2)=(5+3sqrt5)/(6-5sqrt5)$

If $y$ is in $Q4$ , $tan(x+y)=(tanx+tany)/(1-tanxtany)=(5/6-sqrt5/2)/(1+5/6xxsqrt5/2)=(5-3sqrt5)/(6+5sqrt5)$

How do you find the exact value tan(x+y)tan(x+y) if cotx=6/5,secy=3/2cotx=6/5,secy=3/2?