Question

(Sec A+tan A) (1 - sin A) baraber h :
(a) sec A
(b) sin A
(c) cosec A
(d) cos A​

Answer 1

$\huge{\mathfrak{\blue{\underline{Answer}}}}$

$( \sec( a ) + \tan(a) )(1 - \sin(a) ) \\ ( \frac{1}{ \cos( \alpha ) } + \frac{sin( \alpha )}{ \cos( \alpha ) } )( 1 - \sin( \alpha ) ) \\ ( \frac{1 + \sin( \alpha ) }{ \cos( \alpha ) } )(1 - \sin( \alpha ) ) \\ = (\frac{1 - \sin {}^{2} ( \alpha ) } { \cos( \alpha ) } ) \\ = \frac{ \cos {}^{2} ( \alpha ) }{ \cos( \alpha ) }$

$\large{\implies{\purple{Cos(A)}}}$

Answer 2

Hi !

Solution :-. from LHS

=> (secA + tanA ) ( 1 - sinA)

=> (1/cosA + sinA)cosA) ( 1-sinA)

=> ( 1 + sinA)(1-sinA)/cosA

=> ( 1 - sin²A)/cosA

=> cos²A / cosA

=> cosA = RHS

Hence option (D) Are correct option

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Hope it helps you

@Rajkumar111