Question

If ${m^2} + {m'^2} + 2mm'\cos \theta = 1$, ${n^2} + {n'^2} + 2nn'\cos \theta = 1$ and $mn + m'n' + (mn' + m'n)\cos \theta = 0$, then prove that ${m^2} + {n^2} = {\csc ^2}\theta$.

Answer 1

Hint: In the above question you were asked to prove that ${m^2} + {n^2} = {\csc ^2}\theta$ and you are given with some conditions. As you can see that the proving part is in terms of the square so you will have to square the given terms and use them to solve. Also, $\csc \theta$ is the reciprocal of $\sin \theta$. So let us see how we can solve this problem.

Complete Step by Step Solution:
In the given problem we have to prove ${m^2} + {n^2} = {\csc ^2}\theta$.
The given relation ${m^2} + {m'^2} + 2mm'\cos \theta = 1$ can be rewritten as
On adding and subtracting ${m^2}co{s^2}\theta$ in the above expression we get,
 $\Rightarrow {m^2} + {m'^2} + 2mm'cos\theta + {m^2}co{s^2}\theta - {m^2}co{s^2}\theta = 1$
 $\Rightarrow {(m\prime + mcos\theta )^2} + {m^2} - {m^2}co{s^2}\theta = 1\;$
On taking ${m^2}$ as common we get
 $\Rightarrow {(m' + mcos\theta )^2} + {m^2}(1 - co{s^2}\theta ) = 1\;$
We know that, ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and $1 - {\cos ^2}\theta = {\sin ^2}\theta$ , therefore
 $\Rightarrow {(m' + mcos\theta )^2} = 1 - {m^2}si{n^2}\theta$ --(i)
Similarly, from ${n^2} + {n'^2} + 2nn'\cos \theta = 1$ we get,
 $\Rightarrow {(n' + ncos\theta )^2} = 1 - {n^2}si{n^2}\theta$ --(ii)
 $(m' + m\cos \theta ) + (n' + n\cos \theta ) = m'n' + (mn' + nm')\cos \theta + mn{\cos ^2}\theta$
From the given relation: $m'n' + (mn' + nm')\cos \theta = - nm$
 $\therefore (m' + mcos\theta )(n' + ncos\theta ) = - mn + mn{\cos ^2}\theta = - mn\sin {}^2\theta$
On squaring both sides of the above equation we get,
 $\Rightarrow (m' + mcos\theta ){}^2(n' + ncos\theta ){}^2 = m{}^2n{}^2si{n^4}\theta$
Using (i) and (ii) we get,
 $\Rightarrow (1 - m{}^2\sin {}^2\theta )(1 - n{}^2\sin {}^2\theta ) = m{}^2n{}^2si{n^4}\theta$
 $\Rightarrow m{}^2\sin {}^2\theta + n{}^2\sin {}^2\theta = 1$
On dividing both the sides with $\sin {}^2\theta$ we get,
$\Rightarrow {m^2} + {n^2} = \dfrac{1}{{{{\sin }^2}\theta }}$
We know that $\csc \theta$ is the reciprocal of $\sin \theta$
 $\Rightarrow {m^2} + {n^2} = {\csc ^2}\theta$

Hence, it is proved that ${m^2} + {n^2} = {\csc ^2}\theta$.

Note:
In the above solution we are given with three equations. First, we solved these three equations and then we substituted those values to prove ${m^2} + {n^2} = {\csc ^2}\theta$ . Also, we used basic formulas like ${\sin ^2}\theta = 1 - {\cos ^2}\theta$ and $\dfrac{1}{{{{\sin }^2}\theta }} = {\csc ^2}\theta$.