Math, asked by myshindedevidas341, 1 year ago

Sec /
1+cos 0
-= 2 coseco
sin 0
Q.9
sino
Prove that
1+cos
Com the following​

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Answers

Answered by prathmeshmalewar007
2

Proof:

LHS

  = \frac{ \sin( \alpha ) }{(1 +  \cos( \alpha ) )}  +  \frac{(1 +  \cos( \alpha ) }{ \sin( \alpha ) }

  = \frac{ { \sin( \alpha ) }^{2} +  {(1 +  \cos( \alpha ) )}^{2}  }{ \sin( \alpha ) (1 +  \cos( \alpha ) )}

  = \frac{1 -  { \cos( \alpha ) }^{2} + 1 +  { \cos( \alpha ) }^{2} + 2 \cos( \alpha )    }{ \sin( \alpha ) (1 +  \cos( \alpha )  )}

 =  \frac{2 + 2 \cos( \alpha ) }{ \sin( \alpha )(1 +  \cos( \alpha ))  }

  = \frac{2(1 +  \cos( \alpha )) }{ \sin( \alpha )(1 +  \cos( \alpha ))  }

  = \frac{2}{ \sin( \alpha ) }  = 2 \cosec( \alpha )

Hope it helps

feel free to mark as brainlliest.....

Answered by duragpalsingh
2

Hey there!

Taking L.H.S:

sin Ф / (1 + cos Ф) + (1 + cos Ф) / sin Ф

= [ sin²Ф + ( 1 + cos Ф)² ]  /  ( 1 + cosФ) . sin Ф

= [ sin² Ф + 1² + cos²Ф + 2 cosФ ] / (1 + cos Ф) . sin Ф

= [sin²Ф + cos²Ф + 1 + 2 cosФ]  /  (1 + cos Ф) . sin Ф

{ ∵ sin² Ф + cos² Ф = 1}

= [ 1 + 1  + 2 cosФ ] / ( 1 + cosФ) . sinФ

= [2 + 2 cosФ] / ( 1 + cosФ) .sinФ

= [ 2 ( 1 + cosФ)] / ( 1+ cosФ). sinФ

= 2 * 1 / sin Ф

= 2 cosec Ф         { ∵ 1 / sinФ = cosec Ф}

= R.H.S

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