BITSAT 2015 Paper

Let l, m and n be the direction cosines.
Then, l = cos θ, m = cos β, n = cos θ
we have l2+m2+n2 = 1
⇒ cos2 θ + cos2 β + cos2 θ = 1
⇒ 2cos2 θ + 1 – sin2 β = 1
⇒ 2cos2 θ – sin2 β = 0
⇒ 2cos2 θ – 3sin2 β = 0

[Since sin2β = 3sin2θ (given)]
⇒ tan2θ = 2/3
∴ cos2 θ =

1

1+tanθ

=

1

1+2∕3

=

3

5