If α, β and γ are angle of inclinations of a line with x, y and z axes respectively, then the value of 2(cos 2α + cos 2β + cos 2γ) is

When a line is drawn in 3D space, it makes angles with all three axes:

• $\alpha$α = angle with x-axis
• $\beta$β = angle with y-axis
• $\gamma$γ = angle with z-axis

The direction cosines satisfy:

$$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$cos2α+cos2β+cos2γ=1

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Using the double angle formula $\cos 2\theta = 2\cos^2\theta - 1$cos2θ=2cos2θ−1:

$$\cos 2\alpha = 2\cos^2\alpha - 1$$cos2α=2cos2α−1

$$\cos 2\beta = 2\cos^2\beta - 1$$cos2β=2cos2β−1

$$\cos 2\gamma = 2\cos^2\gamma - 1$$cos2γ=2cos2γ−1

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Adding these expressions:

$$\cos 2\alpha + \cos 2\beta + \cos 2\gamma$$cos2α+cos2β+cos2γ

$$= (2\cos^2\alpha - 1) + (2\cos^2\beta - 1) + (2\cos^2\gamma - 1)$$=(2cos2α−1)+(2cos2β−1)+(2cos2γ−1)

$$= 2\cos^2\alpha + 2\cos^2\beta + 2\cos^2\gamma - 3$$=2cos2α+2cos2β+2cos2γ−3

$$= 2(\cos^2\alpha + \cos^2\beta + \cos^2\gamma) - 3$$=2(cos2α+cos2β+cos2γ)−3

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Substituting $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$cos2α+cos2β+cos2γ=1:

$$= 2(1) - 3$$=2(1)−3

$$= -1$$=−1

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Therefore:

$$2(\cos 2\alpha + \cos 2\beta + \cos 2\gamma)$$2(cos2α+cos2β+cos2γ)

$$= 2 \times (-1)$$=2×(−1)

$$= -2$$=−2

The answer is $-2$−2.