Question

( sec A + tan A) ( 1 - sin A ) = ?

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Answer 1

Answer:

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Answer 2

❍ GIVEN :-

( sec A + tan A) ( 1 - sin A )

⇝AS WE KNOW THAT :-

$= (\tt\frac{1}{cos \: A} + \frac{sin \: A}{cos \: A} )( \tt1 - sin \: A)$

$\tt \: = ( \frac{1 + sin \:A}{ cos\: A} )( \tt1 - sin \: A)$

$= \tt \frac{(1 + sin \: A)\:( \tt1 - sin \: A) }{ \tt\cos \: A }$

$\tt = \frac{(1 - {sin}^{2} A)}{cos \: A}$

$= \tt \frac{ {cos}^{2}A }{cos \: A} = \tt \: {cos \: A}$

$\tt \therefore( sec \: A + tan \: A) ( 1 - sin \: A ) = cos \: A$