Math, asked by blessedmwanyama02, 9 months ago

PROVE THE IDENTITIES A) sin A/1+cos A≡1-cos A/sin A B) cos A/1- tan A + sin A/1-cot A ≡sin A + cos A

Answers

Answered by saounksh
2

Step-by-step explanation:

A)

LHS =

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

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

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

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

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

= RHS

Hence Proved

B)

LHS =

 \frac{ \cos( \alpha ) }{1 -  \tan( \alpha ) }  +  \frac{ \sin( \alpha ) }{1 -  \cot( \alpha ) }

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

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

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

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

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

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

 =  \sin( \alpha )  +  \cos( \alpha )

= RHS

Hence Proved

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