If `m^2 + m'^2 + 2mm' costheta = 1` and `n^2+n'^2+2n n'costheta=1`, `(mn + m' n' + (m n' + m' n) costheta = 0` prove that `m^2 + n^2 = cosec^2theta`

Explanation

The given relation can be written as

`(m′+mcostheta^2)+m^2−m^2cos^2theta=1 `

or `(m′+mcostheta^2)=1−m^2sin^2theta`

Similarly `(n′+ncostheta)^2=1−n^2sin^2theta`

Now `(m′+mcostheta)xx(n′+ncostheta)`

=`m′n′+(mn′+m′n)costheta+mncos^2theta`

=`−mn+mncos^2theta` [by given relation ]

=`−mn(1−cos^2theta)=−mnsin^2theta`

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