If \cos \\theta= {x^{2}-y^{2}}/{x^{2}+y^{2}} then the value of \cot\\theta is equal to [ If 0 ≤ θ ≤ 90°]

SSC CGL Tier II 12 Jan 2017 Paper 1

Question : 90 of 100

Marks: +1, -0

\cos \θ= {x^{2}-y^{2}}/{x^{2}+y^{2}}

then the value of

\cot\θ

is equal to [ If 0 ≤ θ ≤ 90°]

Solution:

(4)

\cos \θ= {x^{2}-y^{2}}/{x^{2}+y^{2}}

∴ \sin \θ=√{1-\cos ^{2} \θ}

=√{1-\ { (x^{2}-y^{2} )^{2}}/{ (x^{2}+y^{2} )^{2}}}

=√{ { (x^{2}+y^{2} )^{2}- (x^{2}-y^{2} )^{2}}/{ (x^{2}+y^{2} )^{2}}}

=\√{\ {4 x^{2} y^{2}}/{ (x^{2}+y^{2} )^{2}}}= {2 x y}/{x^{2}+y^{2}}

∴\cot \θ= {\cos \θ}/{\sin \θ}

= {{x^{2}-y^{2}}/{x^{2}+y^{2}} ÷{{2 x y}/{x^{2}+y^{2}}

= {x^{2}-y^{2}}/{2 x y}

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