Question
Question: If \({m^2} + {m'^2} + 2mm'\cos \theta = 1\), \({n^2} + {n'^2} + 2nn'\cos \theta = 1\) and \(mn + m'n...
If m2+m′2+2mm′cosθ=1, n2+n′2+2nn′cosθ=1 and mn+m′n′+(mn′+m′n)cosθ=0, then prove that m2+n2=csc2θ.
Solution
In the above question you were asked to prove that m2+n2=csc2θ 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 is the reciprocal of sinθ. 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.
The given relation m2+m′2+2mm′cosθ=1 can be rewritten as
On adding and subtracting m2cos2θ in the above expression we get,
⇒m2+m′2+2mm′cosθ+m2cos2θ−m2cos2θ=1
⇒(m′+mcosθ)2+m2−m2cos2θ=1
On taking m2 as common we get
⇒(m′+mcosθ)2+m2(1−cos2θ)=1
We know that, sin2θ+cos2θ=1 and 1−cos2θ=sin2θ , therefore
⇒(m′+mcosθ)2=1−m2sin2θ --(i)
Similarly, from n2+n′2+2nn′cosθ=1 we get,
⇒(n′+ncosθ)2=1−n2sin2θ --(ii)
(m′+mcosθ)+(n′+ncosθ)=m′n′+(mn′+nm′)cosθ+mncos2θ
From the given relation: m′n′+(mn′+nm′)cosθ=−nm
∴(m′+mcosθ)(n′+ncosθ)=−mn+mncos2θ=−mnsin2θ
On squaring both sides of the above equation we get,
⇒(m′+mcosθ)2(n′+ncosθ)2=m2n2sin4θ
Using (i) and (ii) we get,
⇒(1−m2sin2θ)(1−n2sin2θ)=m2n2sin4θ
⇒m2sin2θ+n2sin2θ=1
On dividing both the sides with sin2θ we get,
⇒m2+n2=sin2θ1
We know that cscθ is the reciprocal of sinθ
⇒m2+n2=csc2θ
Hence, it is proved that m2+n2=csc2θ.
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 . Also, we used basic formulas like sin2θ=1−cos2θ and sin2θ1=csc2θ.