Question

If $m^2+{m^{\prime }}^2+2m{m}^{\prime }\cos x=1$, $n^2+{n^{\prime }}^2+2n{n}^{\prime }\cos x=1$ and $mn+m^{\prime }{n}^{\prime }+(m{n}^{\prime }+m^{\prime }n)\cos x=0$, prove that
(i) $m^2+n^2=co{\sec }^{2}x$
(ii) ${m^{\prime }}^2+{n^{\prime }}^2=co{\sec }^{2}x$

Answer 1

Consider the given equation,

$m^2+m{{}^{\prime }}^2+2m{m}^{\prime }\mathrm{cosx}=1......\left(1\right)$

$n^2+n{{}^{\prime }}^2+2n{n}^{\prime }\mathrm{cosx}=1......\left(2\right)$

Part (1).

Adding both side $m^2{\cos }^2\theta$ in equation (1) and Adding both side $n^2{\cos }^2\theta$ in equation (2) and we get,

$m^2+m{{}^{\prime }}^2+2m{m}^{\prime }\mathrm{cosx}+m^2{\cos }^{2}x=1+m^2{\cos }^{2}x$

$n^2+n{{}^{\prime }}^2+2n{n}^{\prime }\mathrm{cosx}+n^2{\cos }^{2}x=1+n^2{\cos }^{2}x$

${(m^{\prime }+m\mathrm{cosx})}^2=1-m^2{\sin }^{2}x......\left(3\right)$

${(n^{\prime }+n\mathrm{cosx})}^2=1-n^2{\sin }^{2}x......\left(4\right)$

Multiplying equation (3) and (4) to and we get,

${(m^{\prime }+m\mathrm{cosx})}^2{(n^{\prime }+n\mathrm{cosx})}^2={(m^{\prime }{n}^{\prime }+m^{\prime }n\mathrm{cosx}+m{n}^{\prime }\mathrm{cosx}+mn{\cos }^{2}x)}^2$

$=(m^{\prime }{n}^{\prime }+(m^{\prime }n+m{n}^{\prime })\mathrm{cosx}+mn{\cos }^{2}x).......\left(5\right)$

Given that,

$mn+m\prime n\prime +(mn\prime +m\prime n)cosx=0$

Then,

$m\prime n\prime +(mn\prime +m\prime n)cosx=-mn$

Put the value of equation (5) and we get,

${(m^{\prime }+m\mathrm{cosx})}^2{(n^{\prime }+n\mathrm{cosx})}^2={(m^{\prime }{n}^{\prime }+(m^{\prime }n+m{n}^{\prime })\mathrm{cosx}+mn{\cos }^{2}x)}^2$

$={(-mn+mn{\cos }^{2}x)}^2$

$={[-mn(1-{\cos }^{2}x)]}^2$

$={[-mn{\sin }^{2}x]}^2$

${(m^{\prime }+m\mathrm{cosx})}^2{(n^{\prime }+n\mathrm{cosx})}^2=m^2{n}^2{\sin }^{4}x......\left(6\right)$

Put the value of LHS by equation (3) and (4) to in equation (6), we get

$(1-m^2{\sin }^{2}x)(1-n^2{\sin }^{2}x)=m^2{n}^2{\sin }^{4}x$

$\Rightarrow 1-m^2{\sin }^{2}x-n^2{\sin }^{2}x+m^2{n}^2{\sin }^{4}x=m^2{n}^2{\sin }^{4}x$

$\Rightarrow 1-m^2{\sin }^{2}x-n^2{\sin }^{2}x=0$

$\Rightarrow m^2{\sin }^{2}x+n^2{\sin }^{2}x=1$

$\Rightarrow {\sin }^{2}x(m^2+n^2)=1$

$\Rightarrow m^2+n^2=\frac{1}{{\sin }^{2}x}$

$\Rightarrow m^2+n^2={\csc }^{2}x$

Hence proved

Part (2).

Similarly, we can proved it.

Hence, this is the answer.