Question 786289
if sin(theta)=5/6, theta in quadrant 2, find the exact value of tan (theta+pi/4)
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use x for theta
cosx=√(1-sin^2x)=√(1-25/36)=√(11/36)=√11/6
tan x=sinx/-cosx=-5/&#8730;11(In quadrant 2, sin>0, cos<0, and tan<0)
use tan addition formula:
tan(x+&#960;/4)=(tanx+tan(&#960;/4))/(1-tanx*tan(&#960;/4))
tan(&#960;/4)=1
tan(x+(&#960;/4))=(-5/&#8730;11)+1/(1-(-5/&#8730;11*1)
tan(x+(&#960;/4))=(1-5/&#8730;11)/(1+5/&#8730;11)