Given #cosx = -3/5# and #0<x<180#, what is #tan(x+45)#?
1 Answer
Apr 16, 2018
Explanation:
#"using the "color(blue)"trigonometric identities"#
#•color(white)(x)tanx=sinx/cosx#
#•color(white)(x)tan(x+y)=(tanx+tany)/(1-tanxtany)#
#•color(white)(x)sin^2x+cos^2x=1#
#rArrsinx=+-sqrt(1-cos^2x)#
#"cosx<0" and "0 < x < 180#
#rArr" x is in the second quadrant where "sinx>0#
#sinx=+sqrt(1-(-3/5)^2)#
#color(white)(sinx)=sqrt(1-9/25)=sqrt(16/25)=4/5#
#rArrtanx=4/5xx-5/3=-4/3#
#rArrtan(x+45)=(-4/3+1)/(1-4/3)=(-1/3)/(-1/3)=1#
