Math, asked by VedicLord, 6 months ago

Prove that
 \frac{ \sec( \alpha )  -  \tan( \alpha ) }{ \cosec( \alpha )  +  \cot( \alpha ) }  =  \frac{ \cosec( \alpha )  -  \cot( \alpha ) }{ \cosec( \alpha ) +  \tan( \alpha )  }

Answers

Answered by Anonymous
2

In the denominator it should be sec instead of cosec.

Question:

 \frac{ \sec( \alpha )  -  \tan( \alpha ) }{ \csc( \alpha )  +  \cot( \alpha ) }  =  \frac{ \csc( \alpha ) -  \cot( \alpha )  }{ \sec( \alpha ) +  \tan( \alpha  )  }

Solution:

 =  \frac{ \sec( \alpha )  -  \tan( \alpha ) }{ \csc( \alpha )   + \cot( \alpha ) }  \times  \frac{ \csc( \alpha )  -  \cot( \alpha ) }{ \csc( \alpha )   -  \cot( \alpha ) }

 =  \frac{( \sec( \alpha  ) -  \tan( \alpha )  )( \csc( \alpha )  -  \cot( \alpha ) )}{ {csc}^{2}( \alpha ) -   {cot}^{2}  ( \alpha )}

 = ( \sec( \alpha )  -  \tan( \alpha ) )( \csc( \alpha )  -  \cot( \alpha ) )

 = ( \sec( \alpha )  -  \tan( \alpha ) )( \csc( \alpha )  -  \cot( \alpha ) ) \times  \frac{( \sec( \alpha )  +  \tan( \alpha ) ) }{ ( \sec( \alpha )   +  \tan( \alpha ) ) }

 =  \frac{( \csc( \alpha )  -  \cot( \alpha ) )( {sec}^{2}( \alpha ) -  {tan}^{2}  ( \alpha )}{( \ \sec( \alpha )  +  \tan( \alpha ) ) }

 =  \frac{( \csc( \alpha ) -  \cot( \alpha )  )}{( \sec( \alpha ) +  \tan( \alpha ))  }

Hope it helps you..!! ❤️

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