SSC CHSL 20 Dec 2015 Shift 2 Paper

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Question : 124 of 200

Marks: +1, -0

If sin

θ=a/b

, then the value of sec

θ-cosθ

is where

0°<θ<90°

$b^2/{a√ {b^2 −a^2}}$
$√ {b^2 +a^2}/√ {b^2 -a^2}$
$a /{b√ {b^2 +a^2}}$
$a /{b√ {b^2 -a^2}}$

Solution:

Given : sin

θ=a/b

Thus, using

cosθ = √{1 − sin^2 θ}

⇒

cosθ =√{ 1 − ( a /b ) ^2}

⇒

cosθ =√ { b^2 −a^2}/2

To find : sec θ − cos θ
=

1/{cosθ} −cosθ

=

{1−cos ^2θ}/{cosθ} ={sin^2 θ}/{ cosθ }

=

( a ^2 /b ^2 ) ÷ ( √ {b^2 -a^2}/b )

( a ^2 /b ^2 ) × (b/ √ {b^2 -a^2} )

=

a /{b√ {b^2 -a^2}}

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