Sinx+cosecx=2 then find sin^nx-cos^nx?

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Answer 1 · 2018-04-23T11:26:49

#sin^n(x)-cos^n(x)=1#

Here,

#sinx+cscx=2#

#sinx+1/sinx=2#

#sin^2x+1=2sinx#

#sin^2x-2sinx+1=0#

#(sinx-1)^2=0#

#sinx-1=0#

#sinx=1#

#and cscx=1/sinx=1/1=1#

#i.e.sinx=1and cscx=1#

#color(violet)(=>sin^n(x)-csc^n(x)=(1)^n-(1)^n=0...to(extra)#

#cos^2x=1-sin^2x=1-(1)^2=0#

#=>cosx=0#

So,

#sin^n(x)-cos^n(x)=(1)^n-(0)^n=1-0=1#