Question

If $m^2+{m^{\prime }}^2+2m{m}^{\prime }\cos \theta =1$,
$n^2+{n^{\prime }}^2+2n{n}^{\prime }\cos \theta =1$,
and $mn+m^{\prime }{n}^{\prime }+(m{n}^{\prime }+m^{\prime }n)=\cos \theta =\theta$, then
prove that $m^2+n^2={\csc }^2\theta$.

Answer 1

The given relation can be written as
$(m^{\prime }+m\cos {\theta }^2)+m^2-m^2{\cos }^2\theta =1$
or $(m^{\prime }+m\cos {\theta }^2)=1-m^2{\sin }^2\theta$
Similarly ${(n^{\prime }+n\cos \theta )}^2=1-n^2{\sin }^2\theta$
Now $(m^{\prime }+m\cos \theta )=(n^{\prime }+n\cos \theta )$
$m^{\prime }{n}^{\prime }+(m{n}^{\prime }+m^{\prime }n)\cos \theta +mn{\cos }^2\theta$
$=-mn+mn{\cos }^2\theta$ by given relation
$=-mn(1-{\cos }^2\theta )=-mn{\sin }^2\theta$
Now squaring both sides, we get
or ${(m^{\prime }+m\cos \theta )}^2{(n^{\prime }+n\cos \theta )}^2=m^2{n}^2{\sin }^4\theta$
Hence substituting from $\left(1\right)$ and $\left(2\right)$ in $\left(3\right)$, we get
$(1-m^2{\sin }^2\theta )(1-n^2{\sin }^2\theta )=m^2{n}^2{\sin }^4\theta$
or $=\left(m^2{+n}^2\right){\sin }^2\theta =1$ i.e. $=m^2{+n}^2={\csc }^2\theta$