Question #db20d

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Answer 1 · 2017-04-14T14:08:41

#cosx = pm sqrt((m^2-1)/(n^2-1))#

We have equivalently

#{(sinx/sqrt(1-sin^2x)=n siny/sqrt(1-sin^2y)),(siny=1/msinx):}#

or

#sinx/sqrt(1-sin^2x)=(n/m)sin x/sqrt(1-1/m^2sin^2x)#

or

#sqrt(1-1/m^2sin^2x)=(n/m)sqrt(1-sin^2x)#

and squaring both sides

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

solving for #sin^2x# we have

#sin^2x=(n^2-m^2)/(n^2-1)=1-cos^2x#

and finally

#cos^2x=(m^2-1)/(n^2-1)#

#cosx = pm sqrt((m^2-1)/(n^2-1))#

Answer 2 · 2017-04-15T06:14:55

Given

#tanx=ntany........[1]#

#sinx=nsiny........[2]#

Dividing [2] by [1]

#cosx=m/ncosy........[3]#

#=>cosy=n/mcosx#

Squaring [2]

#sin^2x=m^2sin^2y#

#=>1-cos^2x=m^2(1-cos^2y)#

#=>1-cos^2x=m^2(1-n^2/m^2cos^2x)#

#=>1-cos^2x=m^2-n^2cos^2x#

#=>n^2cos^2x-cos^2x=m^2-1#

#=>(n^2-1)cos^2x=m^2-1#

#=>cos^2x=(m^2-1)/(n^2-1)#

#=>cosx=pmsqrt((m^2-1)/(n^2-1))#