Step-by-step explanation:
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that lines makes an angle α, β, γ with x - axis, y - axis and z - axis respectively.
So, By definition of direction cosines,
[tex]\rm :\longmapsto\:l = cos \alpha [/tex]
[tex]\rm :\longmapsto\:m = cos \beta [/tex]
[tex]\rm :\longmapsto\:n = cos \gamma [/tex]
So,
[tex]\rm :\longmapsto\: {l}^{2} + {m}^{2} + {n}^{2} = 1[/tex]
[tex]\rm :\longmapsto\: {cos}^{2} \alpha + {cos}^{2} \beta + {cos}^{2} \gamma = 1[/tex]
On multiply by 2 on both sides we get
[tex]\rm :\longmapsto\: 2{cos}^{2} \alpha + 2{cos}^{2} \beta + 2 {cos}^{2} \gamma = 2[/tex]
can be further rewritten as
[tex]\rm :\longmapsto\: 2{cos}^{2} \alpha - 1 + 1 + 2{cos}^{2} \beta - 1 + 1 + 2 {cos}^{2} \gamma - 1 + 1 = 2[/tex]
[tex]\rm :\longmapsto\: (2{cos}^{2} \alpha - 1)+ (2{cos}^{2} \beta - 1)+ (2 {cos}^{2} \gamma - 1) + 3= 2[/tex]
[tex]\rm :\longmapsto\:cos2 \alpha + cos2 \beta + cos2 \gamma + 3= 2[/tex]
[tex] \red{ \bigg\{ \sf \: \because \: cos2x = {2cos}^{2}x - 1 \bigg\}}[/tex]
[tex]\rm :\longmapsto\:cos2 \alpha + cos2 \beta + cos2 \gamma= 2 - 3[/tex]
[tex]\rm :\longmapsto\:cos2 \alpha + cos2 \beta + cos2 \gamma= - 1[/tex]
Hence,
[tex]\bf\implies \:\boxed{\tt{ \: cos2 \alpha + cos2 \beta + cos2 \gamma = - 1 \: }}[/tex]
So, option (d) is correct.
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
MORE TO KNOW
Direction cosines of a line segment is defined as the cosines of the angle which a line makes with the positive direction of respective axis.
The scalar components of unit vector always give direction cosines.
The scalar components of a vector gives direction ratios.
