SSC CGL 19 Oct 2014 Shift 2 Paper

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Question : 131 of 200

Marks: +1, -0

If sin 21° =

x/y

, then sec 21° - sin 69° is equal to

$x^2/{y√{y^2-x^2}}$
$y^2/{x√{y^2-x^2}}$
$x^2/{y√{x^2-y^2}}$
$y^2/{x√{x^2-y^2}}$

Solution:

sin 21° =

x/y

cos 21° =

√{1-sin^2 21°}

=

√{1-x^2/y^2}

=

√{y^2-x^2}/y

∴ sec 21° =

y/√{y^2-x^2}

∴ sec 21° – sin 69°
= sec 21° – sin (90° – 21°)
= sec 21° – cos 21°
=

y/√{y^2-x^2}

-

√{y^2-x^2}/y

=

{y^2-(y^2-x^2)}/{y√{y^2-x^2}}

=

x^2/{y√{y^2-x^2}}

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