How do you find a numerical value of one trigonometric function of x given #(1+cosx)/sinx+sinx/(1+cosx)=4#?

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Answer 1 · 2016-11-12T10:00:20

#:.x=pi/6 +2pin or x=(5pi)/6 +2pin#

#(1+cosx)/sinx + sinx/(1+cosx)=4#

#((1+cosx)(1+cosx) + sinxsinx)/((1+cosx)sinx)=4#

#(1+2cosx+cos^2x+ sin^2x)/((1+cosx)sinx)=4#

#(1+2cosx+1)/((1+cosx)sinx)=4#

#(2+2cosx)/((1+cosx)sinx)=4#

#(2(1+cosx))/((1+cosx)sinx)=4#

#(2cancel(1+cosx))/(cancel(1+cosx)sinx)=4#

#2/sinx=4#

#2=4sinx#

#2/4=sinx#

#1/2=sinx#

#sin^-1(1/2)=x#

#:.x=pi/6 +2pin or x=(5pi)/6 +2pin#