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If m^2+m^('2)+2mn'cos theta =1 n^2+n^(...

If `m^2+m^('2)+2mn'cos theta =1`
`n^2+n^('2)+2"nn"' cos theta=1 and mn +m'n'+(mn'+m'n)cos theta=0` then show that `m^2+n^2=m^('2) +n^('2) =cosec^2 theta`.

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The correct Answer is:
Hence , we have , `m^2+n^2=m^('2)+n^('2) =cosec^2theta`
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