Math, asked by sonusharma45, 4 months ago

yeh dono me se koi bhi solve kar do plzzz ​

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Answered by ranveersingh28851
0

Answer:

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Answered by Anonymous
32

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First One -

 \rm \longrightarrow \:  \sqrt{ \dfrac{1 +  \cos( \theta) }{1 -  \cos( \theta) } }  +  \sqrt{ \dfrac{1 -  \cos( \theta) }{1 +  \cos( \theta) } }

  • Rationalize the denominator ...

 \implies \rm \:  \sqrt{ \frac{(1 +  \cos( \theta) )(1 -  \cos( \theta) )}{(1 -  \cos( \theta) )(1 +  \cos( \theta)) } }  +  \sqrt{ \frac{(1 -  \cos( \theta))(1 -  \cos( \theta))  }{(1 +   \cos( \theta))(1 -  \cos( \theta))  } }

 \implies \rm \:  \dfrac{1 +  \cos( \theta) }{ \sqrt{1 -  { \cos }^{2}  \theta} }  +  \dfrac{1 -  \cos( \theta) }{ \sqrt{1 -  { \cos }^{2}  \theta} }

 \implies \rm \:  \dfrac{1 +  \cos( \theta) }{ \sqrt{ { \sin}^{2} \theta } }  +  \dfrac{1 -  \cos( \theta) }{ \sqrt{ { \sin }^{2} \theta } }

 \implies \rm \:  \dfrac{1 +  \cos( \theta) }{ \sin( \theta) }  +  \dfrac{1 -  \cos( \theta) }{  \sin( \theta) }

 \implies \rm \:  \dfrac{1 +  \cos( \theta)  + 1 -  \cos( \theta) }{ \sin( \theta) }

 \implies \rm \:  \dfrac{2}{ \sin( \theta) }

 \implies \rm \: 2 \sin( \theta)

 \therefore\:  \rm \: hence \: proved

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Formula Used -

  \mapsto \rm \:  (x + y)(x - y) =  {x}^{2}  -  {y}^{2}

 \mapsto \rm \: 1 -  {cos}^{2}  \theta \:  =  \:  {sin}^{2}  \theta

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★ Second One -

 \rm \: answer \: in \: attachment

Formula Used -

 \mapsto \rm \:   {x}^{3}  +  {y}^{3}  = (x + y)( {x}^{2}  +  {y}^{2}  - xy)

 \mapsto \rm \:  {x}^{3}  -  {y}^{3} =  (x - y)( {x}^{2}  +  {y}^{2}  + xy)

 \mapsto \rm \:  {cos}^{2}  \theta +  {sin}^{2}  \theta = 1

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