Question #27b54

1 Answer
maganbhai P.
Feb 27, 2018

5

Explanation:

We have, #tanx=2/3#.
#(sin(3pi+x)-sin(pi/2-x))/(cos((3pi)/2+x)+cos(pi-x))=(-sinx-cosx)/(sinx-cosx)=((-sinx/cosx-cosx/cosx)/(sinx/cosx-cosx/cosx))=(-tanx-1)/(tanx-1)=(-2/3-1)/(2/3-1)=(-2-3)/(2-3)=(-5)/(-1)=5#
Hint:
#*sin(3pi+x)=-sinx#,(Third Quadrant #rArrsin is -ve#)
#*sin(pi/2-x)=cosx,# (First Quadrant#rArrsin is +ve#)
#*cos((3pi)/2+x)=sinx,# (Fourth Quadrant#rArrcos is +ve#)
#*cos((pi-x)=-cosx,# (Second Quadrant#rArrcos is -ve#)

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