SSC MTS Model Paper 5

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Question : 33 of 148

Marks: +1, -0

If

4 x = \sec θ

and

{ 4 } /{ x } = \tan θ ,

then 8

( x ^ { 2 } - { 1 } /{ x ^ { 2 } } )

is equals to-

$1/2$
$1/4$
$1/8$
$1/16$

Solution:

∵4x = sec θ

\;\;\;\;\;\;\;∴\x={\sec\θ}/4

and

4/\x

= tan θ

\;\;\;\;∴\x=4/{\tan\θ}

8 ( x ^ { 2 } - { 1 } /{ x ^ { 2 } } ) = 8 ( { \sec ^ { 2 } θ } /{ 16 } - {1/16}/{\tan^2\θ})

= 8 ( { \sec ^ { 2 } \θ } /{ 16 } - { \tan ^ { 2 } \θ }/ { 16 } )

=

8×{ 1 } { 16 } = { 1 } /{ 2 }

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