Question

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Answer 1

Answer:

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Answer 2

$\sqrt{ \dfrac{1 \: + \: sin \: A}{1 \: - \: sin \: A} }$ = sec A + tan A

_________ [GIVEN EQUATION]

Taking L.H.S.

$\sqrt{ \dfrac{1 \: + \: sin \: A}{1 \: - \: sin \: A} }$

Rationalize it

$\implies$ $\sqrt{ \dfrac{1 \: + \: sin \: A}{1 \: - \: sin \: A} } \: \times \: \sqrt{ \dfrac{1 \: + \: sin \: A}{1 \: + \: sin \: A} }$

(a + b) (a - b) = a² - b²

$\implies$ $\sqrt{ \dfrac{ {(1 \: + \: sin \: A)}^{2} }{ {1}^{2} \: - \: {sin}^{2}A } }$

$\implies$ $\sqrt{ \dfrac{ {(1 \: + \: sin \: A)}^{2} }{{cos}^{2}A } }$

$\implies$ $\bigg( {\sqrt{ \dfrac{ {1 \: + \: sin \: A} }{cos \: A }}\bigg)}^{2}$

$\implies$ $\dfrac{1 \: + \: sin \: A }{cos \: A }$

• sec A = $\dfrac{1}{cos\:A}$

• tan A = $\dfrac{sin\:A}{cos\:A}$

$\implies$ $\dfrac{1}{cos \: A }$ + $\dfrac{sin \: A }{cos \: A }$

$\implies$ sec A + tan A

L.H.S. = R.H.S.

______ [HENCE PROVED]

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