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

1 Answer
Shwetank Mauria
Jul 19, 2017

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)#

Explanation:

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)#

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