Question

(secA + tanA)(1 - sinA) = cosA
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Answer 1

Answer:

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Answer 2

To prove:-

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

Solution:-

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

$\dashrightarrow \Bigg (\dfrac{1}{\cos \ A} + \dfrac{\sin A}{\cos A} \Bigg )(1-\sin A)= \cos A \qquad ...\Bigg [ \because\quad\sec A = \dfrac{1}{\cos A } \ \ \rm and \ \ tan \ A = \dfrac{sin \ A}{cos \ A} \Bigg ]$

$\dashrightarrow \Bigg (\dfrac{1+\sin A}{\cos A} \Bigg )(1-\sin A)= \cos A$

Multiply both the sides of the equation by cos A

$\dashrightarrow (\cos A) \Bigg (\dfrac{1+\sin A}{\cos A} \Bigg )(1-\sin A)= (\cos A)(\cos A)$

$\dashrightarrow (1+\sin A )(1-\sin A)={\cos }^2 A$

$\dashrightarrow 1-{\sin}^2 A={\cos }^2 A$

Since, cos² A = 1 - sin² A

Therefore,

$\dashrightarrow {\cos}^2 A={\cos }^2 A$

Hence, Proved.