SSC CGL 9 Aug 2015 Shift 2 Paper

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Question : 110 of 200

Marks: +1, -0

If

{cosα}/{sinβ}

= n and

{cosα}/{cosβ}

= m , then the value of

cos^2

β is

$m^2/{m^2+n^2}$
$1/{m^2+n^2}$
$n^2/{m^2+n^2}$
0

Solution:

{cosα}/{cosβ}

= n
and

{cosα}/{cosβ}

= m
⇒ cos α - n sin β and cos α = m cos β
∴

n^2 sin^2 β

=

m^2 cos^2 β

⇒

n^2(1-cos^2 β)

=

m^2 cos^2 β

⇒

n^2-n^2 cos^2 β

=

m^2 cos^2 β

⇒

m^2 cos^2 β + n^2 cos^2 β

=

n^2

⇒

cos^2 β(m^2+n^2)

=

n^2

⇒

cos^2 β

=

n^2/{m^2+n^2}

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