Question

if cosO = 2x/1+x² find the value of sinO and cotO
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Answer 1

Step-by-step explanation:

cosO=2x/1+x^2=base/hypotenuse

Therefore,( altitudes) ^2+(base)^2=(hypotenuse)^2

( Altitude) ^2=(hypo)^2-(base)^2

=(1+x^2)^2-(2x)^2

=1+2x^2+x^4-4x^2

=1-2x^2+x^4

=(1-x^2)^2

Altitude=1-x^2

sinO=alt/hypo=(1-x^2)/(1+x^2)

cotO=base/alt=2x/1-x^2

Answer 2

$\large \dag$ Question :-

$\rm If \: cos\theta = \dfrac{2x}{1 + {x}^{2} } \: then \: find : -$

$\rm sin \theta \: and \: cot \theta$

$\large \dag$ Answer :-

• $\purple{ {{{\bf sin \theta = \dfrac{1 - {x}^{2} }{1+x^2}} }}}$

• $\purple{ {{{\bf cot \theta = \dfrac{1 - {x}^{2} }{2x}} }}}$

$\large \dag$ Step by step Explanation :-

We have,

$\rm cos \theta = \frac{2x}{1 + {x}^{2}}$

$:\longmapsto \rm cos {}^{2} \theta = \bigg( \frac{2x}{ {1 + x}^{2} } \bigg)^{2}$

$:\longmapsto \rm cos {}^{2} \theta = \frac{ {4x}^{2} }{1 + {x}^{4} + {2x}^{2} }$

Also we know that,

$\\ \green{ \underline \red{\rm{sin}^{2} \theta = 1 - {cos}^{2} \theta}}$

$:\longmapsto \rm {sin}^{2} \theta = 1 - \frac{ {4x}^{2} }{1 + {x}^{4} + {2x}^{2} }$

$:\longmapsto \rm {sin}^{2} \theta = \frac{1 + {x}^{4} + {2x}^{2} - {4x}^{2} }{1 + {x}^{4} + {2x}^{2} }$

$:\longmapsto \rm {sin}^{2} \theta = \frac{1 + {x}^{4} - {2x}^{2} }{1 + {x}^{4} + {2x}^{2} }$

$:\longmapsto \rm {sin}^{2} \theta = \frac{(1 - {x {}^{2} )}^{2} }{ {(1 + {x}^{2} )}^{2} }$

$:\longmapsto \rm {sin}^{2} \theta = \bigg( \frac{1 - {x}^{2} }{1 + {x}^{2} } \bigg)^{2}$

$:\longmapsto \rm {sin}\theta = \sqrt{\bigg( \frac{1 - {x}^{2} }{1 + {x}^{2} } \bigg)^{2} }$

$\purple{ \large :\longmapsto \underline {\boxed{{\bf {sin}\theta = \frac{1 - {x}^{2} }{1 + {x}^{2} } } }}}$

Also we know that,

$\\ \rm cot \theta = \frac{sin \theta}{cos \theta}$

$:\longmapsto \rm cot \theta = \dfrac{ \dfrac{1 - {x}^{2} }{ \cancel{1 + {x}^{2}} } }{ \dfrac{2x}{ \cancel{1 + {x}^{2}} } }$

$\purple{ \large :\longmapsto \underline {\boxed{{\bf cot \theta = \frac{1 - {x}^{2} }{2x}} }}}$