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]

$sin^2x=m^2sin^2y$

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

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

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

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

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

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

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