If $α$ , $β$ , $γ$ are the angles which a line makes with positive direction of the axes, then

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Solution

Given $α, β, γ$ are angles which a line makes with positive direction of axes
Let the line be $- - \to O P$ where $P (x, y, z)$
Projections of $¯ ¯¯¯¯¯¯ ¯ O P$ on axes are $x, y, z$ respectively
From trigonometry
$cos α = \frac{x}{∣ ∣ ¯ ¯¯¯¯¯¯ ¯ O P ∣ ∣} cos α = \frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} (1) S i m i l a r l y, cos β = \frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} (2) cos γ = \frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} (3)$
Squaring on both sides of $(1), (2), (3)$ and adding
${cos}^{2} α + {cos}^{2} β + {cos}^{2} γ = \frac{x^{2} + y^{2} + z^{2}}{x^{2} + y^{2} + z^{2}} {cos}^{2} α + {cos}^{2} β + {cos}^{2} γ = 1 (4)$
We know,
${cos}^{2} θ + {sin}^{2} θ = 1 {sin}^{2} θ = 1 - {cos}^{2} θ (5) F r o m (4), (5) 1 - {sin}^{2} α + 1 - {sin}^{2} β + 1 - {sin}^{2} γ = 1 {sin}^{2} α + {sin}^{2} β + {sin}^{2} γ = 2$
We know,
$cos 2 α = 2 {cos}^{2} α - 1 \frac{cos 2 α + 1}{2} = {cos}^{2} α (6) F r o m (4), (6) \frac{cos 2 α}{2} + \frac{1}{2} + \frac{cos 2 β}{2} + \frac{1}{2} + \frac{cos 2 γ}{2} + \frac{1}{2} = 1 cos 2 α + cos 2 β + cos 2 γ = - 1$

${cos}^{2} α + cos (β + γ) c o s (β - γ) = ? (7)$
Formula: $cos (β + γ) = cos β cos γ - sin β sin γ cos (β - γ) = cos β cos γ + sin β sin γ$
On multiplying $cos (β + γ) c o s (β - γ) = {cos}^{2} β {cos}^{2} γ - {sin}^{2} β {sin}^{2} γ$
$= {cos}^{2} β {cos}^{2} γ - (1 - {cos}^{2} β) (1 - {cos}^{2} γ) (∵ {cos}^{2} θ + {sin}^{2} θ = 1) = {cos}^{2} β {cos}^{2} γ - 1 - {cos}^{2} β {cos}^{2} γ + {cos}^{2} β + {cos}^{2} γ = - 1 + {cos}^{2} β + {cos}^{2} γ$
Adding ${cos}^{2} α$ on both sides
${cos}^{2} α + cos (β + γ) c o s (β - γ) = - 1 + {cos}^{2} β + {cos}^{2} γ + {cos}^{2} α {cos}^{2} α + cos (β + γ) c o s (β - γ) = - 1 + 1 f r o m (4) {cos}^{2} α + cos (β + γ) c o s (β - γ) = 0$

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