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Question

Mathematics Question on Three Dimensional Geometry

If a straight line makes angles α,β,γ\alpha , \beta , \gamma with the coordinate axes, then 1tan2α1+tan2α+1sec2β2sin2γ=\frac{1-\tan^{2} \alpha }{1+tan^{2} \alpha} +\frac{1}{sec 2 \beta} -2\sin^{2} \gamma =

A

-1

B

1

C

-2

D

2

Answer

-2

Explanation

Solution

If a straight line makes angles α,β,γ\alpha, \beta, \gamma with the coordinate axes.
Then, cos2α+cos2β+cos2γ=1\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1
1+cos2α2+1+cos2β2+1sin2γ=1\Rightarrow \frac{1+\cos 2 \alpha}{2}+\frac{1+\cos 2 \beta}{2}+1-\sin ^{2} \gamma=1
1+cos2α+1+cos2β+22sin2γ=2\Rightarrow 1+\cos 2 \alpha+1+\cos 2 \beta+2-2 \sin ^{2} \gamma=2
cos2α+cos2β2sin2γ=2\Rightarrow \cos 2 \alpha+\cos 2 \beta-2 \sin ^{2} \gamma=-2
1tan2α1+tan2α+1sec2β2sin2γ=2\Rightarrow \frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}+\frac{1}{ sec\, 2 \beta}-2 \sin ^{2} \gamma=-2