Hello, Lavlesh!

I haven't solved it yet, but I have some observations . . .

Given:

. . [1] . m² + m’² + 2mm’cosθ .= .1

. . [2] . n² + n’² + 2nn’cosθ .= .1

. . [3] .mn + m’n’ + (mn’ + m’n)cosθ .= .0

Prove: .m² + n² .= .csc²θ

Add [1], [2], and 2 times [3]:

. . . .[1] . m² + m’² + 2mm’cosθ .= .1

. . . .[2] . n² + n’² + 2nn’cosθ .= .1

. . 2x[3] .2mn + 2m’n’ + 2(mn’ + m’n)cosθ .= .0

Their sum is:

. . m² + 2mn + n² + m'² +2m'n' + n'² + 2(mm' + mn' + m'n + nn')cosθ .= .2

which simplifies to:

. . (m + n)² + (m' + n')² + 2(m + n)(m' + n')cosθ .= .2

Did that help? . . . Not a bit!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The given equations look suspiciously like the Law of Cosines.

I modified them:

. . [1] . m² + m’² - 2mm’cos(180°-θ) .= .1

. . [2] . n² + n’² - 2nn’cos(180°-θ) .= .1

. . [3] .mn + m’n’ - (mn’ + m’n)cos(180°-θ) .= .0

Adding [1], [2], and 2 times [3],

. . we get: .(m + n)² + (m' + n')² - 2(m + n)(m' + n')cos(180°-θ) .= .2

So I have a triangle . . . Code:

*
* *
* * 2
m + n * *
* *
θ * *
- - - * * * * * * * * * * *
m' + n'

But this too failed to inspire me.