$P$ is the length of the perpendicular from the origin upon the straight line joining the two points whose coordinates are
$(a cos α, a sin α)$ and $(a cos β, a sin β)$ .
If $P = a cos (\frac{α - β}{λ}),$ then $λ = ?$

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Solution

Equation of line joining $(a cos α, a sin α)$ and $(a cos β, a sin β)$ is
$y - a sin α = (\frac{a sin β - a sin α}{a cos β - a cos α}) (x - a cos α) y - a sin α = (\frac{sin β - sin α}{cos β - cos α}) (x - a cos α) y - a sin α = ⎛ ⎜ ⎜ ⎜ ⎝ \frac{2 cos \frac{β + α}{2} sin \frac{β - α}{2}}{- 2 sin \frac{β + α}{2} sin \frac{β - α}{2}} ⎞ ⎟ ⎟ ⎟ ⎠ (x - a cos α) y - a sin α = - ⎛ ⎜ ⎜ ⎜ ⎝ \frac{cos \frac{β + α}{2}}{sin \frac{β + α}{2}} ⎞ ⎟ ⎟ ⎟ ⎠ (x - a cos α) sin \frac{β + α}{2} y - a sin α sin \frac{β + α}{2} = - cos \frac{β + α}{2} x + a cos α cos \frac{β + α}{2} x cos \frac{β + α}{2} + y sin \frac{β + α}{2} = a (cos α cos \frac{β + α}{2} + sin α sin \frac{β + α}{2})$
$x cos \frac{β + α}{2} + y sin \frac{β + α}{2} = a (cos (α - \frac{β + α}{2}))$
$x cos \frac{β + α}{2} + y sin \frac{β + α}{2} = a cos \frac{α - β}{2}$
Let the length of perpendicular from origin be $p$
$p = \frac{∣ ∣ ∣ 0 (cos \frac{β + α}{2}) - 0 (sin \frac{β + α}{2}) - a cos \frac{α - β}{2} ∣ ∣ ∣}{\sqrt{{cos}^{2} \frac{β + α}{2} + {sin}^{2} \frac{β + α}{2}}} p = \frac{∣ ∣ ∣ - a cos \frac{α - β}{2} ∣ ∣ ∣}{1} \Rightarrow p = a cos \frac{α - β}{2}$

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P is the length of the perpendicular from the origin upon the straight line joining the two points whose coordinates are(acosα,asinα) and (acosβ,asinβ).If P=acos(α−βλ), then λ=?

$P$ is the length of the perpendicular from the origin upon the straight line joining the two points whose coordinates are
$(a cos α, a sin α)$ and $(a cos β, a sin β)$ .
If $P = a cos (\frac{α - β}{λ}),$ then $λ = ?$