If {\cosec} \\theta= \frac{1}{4 x}+x, then what isthe value of {\cosec} \\theta+\cot \\theta ?

Correct Answer is : 2x or 1(2x)

SSC CGL 16 Aug 2017 Shift 1 Paper

Question : 71 of 100

Marks: +1, -0

{\cosec} \θ= {1}/{4 x}+x,

then what isthe value of

{\cosec} \θ+\cot \θ ?

Solution:

{\cosec} \θ= {1}/{4 x}+x= {1+4 x^{2}}/{4 x}

{\cosec} \θ+\cot \θ= {\cosec} \θ+ {\cos \θ}/{\sin \θ }

={\cosec} \θ+{\cosec} \θ . \cos \θ

={\cosec} \θ[1+\cos \θ]

={\cosec} \θ[1+√{1-\sin ^{2} θ }]

={\cosec} \θ[1+√{1- {1}/{{\cosec} ^{2} \θ}}]

= [ {1+4 x^{2}}/{4 x} ]

[1+√ {1- {1}/{ ( {1+4 x^{2}}/{4 x} )^{2}}}]

=[ {1+4 x^{2}}/{4 x} ]{[1+√{1- {16 x^{2}}/{ (1+4 x^{2} )^{2}}}]}

=[ {1+4 x^{2}}/{4 x} ]

[1+√{ {1+16 x^{4}-8 x^{2}-16 x^{2}}/{ (1+4 x^{2} )^{2}}}]

[ {1+4 x^{2}}/{4 x} ]

[1+√{\ {1+16 x^{4}-8 x^{2}}/{ (1+4 x^{2} )^{2}}}]

= [ {1+4 x^{2}}/{4 x} ][1+√{{ (1-4 x^{2} )^{2}}/{ (1+4 x^{2} )^{2}}}]

= [ {1+4 x^{2}}/{4 x} ][1+ {{ (1-4 x^{2} ) }/{ (1+4 x^{2} ) }}]

= [ {1+4 x^{2}}/{4 x} ]

[ {1+4 x^{2}+1-4 x^{2}}/{1+4 x^{2}} ]

= {2}/{4 x}= {1}/{2 x}

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