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Question: If \({m^2} + {m'^2} + 2mm'\cos \theta = 1\), \({n^2} + {n'^2} + 2nn'\cos \theta = 1\) and \(mn + m'n...

If m2+m2+2mmcosθ=1{m^2} + {m'^2} + 2mm'\cos \theta = 1, n2+n2+2nncosθ=1{n^2} + {n'^2} + 2nn'\cos \theta = 1 and mn+mn+(mn+mn)cosθ=0mn + m'n' + (mn' + m'n)\cos \theta = 0, then prove that m2+n2=csc2θ{m^2} + {n^2} = {\csc ^2}\theta.

Explanation

Solution

In the above question you were asked to prove that m2+n2=csc2θ{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, cscθ\csc \theta is the reciprocal of sinθ\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 m2+n2=csc2θ{m^2} + {n^2} = {\csc ^2}\theta.
The given relation m2+m2+2mmcosθ=1{m^2} + {m'^2} + 2mm'\cos \theta = 1 can be rewritten as
On adding and subtracting m2cos2θ{m^2}co{s^2}\theta in the above expression we get,
m2+m2+2mmcosθ+m2cos2θm2cos2θ=1\Rightarrow {m^2} + {m'^2} + 2mm'cos\theta + {m^2}co{s^2}\theta - {m^2}co{s^2}\theta = 1
(m+mcosθ)2+m2m2cos2θ=1  \Rightarrow {(m\prime + mcos\theta )^2} + {m^2} - {m^2}co{s^2}\theta = 1\;
On taking m2{m^2} as common we get
(m+mcosθ)2+m2(1cos2θ)=1  \Rightarrow {(m' + mcos\theta )^2} + {m^2}(1 - co{s^2}\theta ) = 1\;
We know that, sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 and 1cos2θ=sin2θ1 - {\cos ^2}\theta = {\sin ^2}\theta , therefore
(m+mcosθ)2=1m2sin2θ\Rightarrow {(m' + mcos\theta )^2} = 1 - {m^2}si{n^2}\theta --(i)
Similarly, from n2+n2+2nncosθ=1{n^2} + {n'^2} + 2nn'\cos \theta = 1 we get,
(n+ncosθ)2=1n2sin2θ\Rightarrow {(n' + ncos\theta )^2} = 1 - {n^2}si{n^2}\theta --(ii)
(m+mcosθ)+(n+ncosθ)=mn+(mn+nm)cosθ+mncos2θ(m' + m\cos \theta ) + (n' + n\cos \theta ) = m'n' + (mn' + nm')\cos \theta + mn{\cos ^2}\theta
From the given relation: mn+(mn+nm)cosθ=nmm'n' + (mn' + nm')\cos \theta = - nm
(m+mcosθ)(n+ncosθ)=mn+mncos2θ=mnsin2θ\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,
(m+mcosθ)2(n+ncosθ)2=m2n2sin4θ\Rightarrow (m' + mcos\theta ){}^2(n' + ncos\theta ){}^2 = m{}^2n{}^2si{n^4}\theta
Using (i) and (ii) we get,
(1m2sin2θ)(1n2sin2θ)=m2n2sin4θ\Rightarrow (1 - m{}^2\sin {}^2\theta )(1 - n{}^2\sin {}^2\theta ) = m{}^2n{}^2si{n^4}\theta
m2sin2θ+n2sin2θ=1\Rightarrow m{}^2\sin {}^2\theta + n{}^2\sin {}^2\theta = 1
On dividing both the sides with sin2θ\sin {}^2\theta we get,
m2+n2=1sin2θ\Rightarrow {m^2} + {n^2} = \dfrac{1}{{{{\sin }^2}\theta }}
We know that cscθ\csc \theta is the reciprocal of sinθ\sin \theta
m2+n2=csc2θ\Rightarrow {m^2} + {n^2} = {\csc ^2}\theta

Hence, it is proved that m2+n2=csc2θ{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 m2+n2=csc2θ{m^2} + {n^2} = {\csc ^2}\theta . Also, we used basic formulas like sin2θ=1cos2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta and 1sin2θ=csc2θ\dfrac{1}{{{{\sin }^2}\theta }} = {\csc ^2}\theta.