Question

(Sec A + Tan A)(1 - Sin A) =

Answer 1

Answer:

cos A

Step-by-step explanation:

(Sec A + Tan A)(1 - Sin A)

= sec A + tan A - tan A - sin^2 A / cos A

= 1 / cos A - sin^2 A / cos A = (1 - sin^2 A) / cos A = cos^2 A / cos A = cos A

Answer 2

$\sf{\star{\boxed{\large{\mathfrak{\pink{ Solution}}}}}}$

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$\sf : \implies\: {\bold{ ( sec A + Tan A) ( 1 - Sin A) }}$

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• $\sf{\red{\boxed{\bold{Sec A = \dfrac{1}{cosA}}}}}$

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$\sf : \implies\: {\bold{ ( \dfrac{1}{cos A} + Tan A) ( 1 - Sin A) }}$

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• $\sf{\red{\boxed{\bold{Tan A = \dfrac{SinA}{Cos A}}}}}$

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$\sf : \implies\: {\bold{ \bigg( \dfrac{1}{cosA} + \dfrac{SinA}{CosA} \bigg) ( 1 - Sin A) }}$

⠀⠀⠀⠀

$\sf : \implies\: {\bold{ \bigg( \dfrac{1 + SinA}{CosA} \bigg) ( 1 - Sin A) }}$

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$\sf : \implies\: {\bold{ \dfrac{ ( 1 + SinA)( 1 - SinA)}{Cos A}}}$

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• $\sf{\red{\boxed{\bold{( a + b )( a - b ) = a^2 - b^2}}}}$

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$\sf : \implies\: {\bold{ \dfrac{1^2 - Sin^2A}{CosA} }}$

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$\sf : \implies\: {\bold{ \dfrac{1 - Sin^2A}{ CosA}}}$

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• $\sf{\red{\boxed{\bold{1 - Sin^2A = Cos^2A}}}}$

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$\sf : \implies\: {\bold{ \dfrac {Cos^2A}{CosA} }}$

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$\sf : \implies\: {\bold{ \dfrac{CosA \times \cancel { CosA}}{\cancel{CosA}} }}$

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$\sf : \implies\: {\bold{ CosA}}$

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$\sf{\star{\boxed{\mathfrak{\orange{ CosA }}}}}$