Question

answer this question ​

Answer 1

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Answer 2

Answer:

$\sqrt{ \frac{1 + sin \: a }{1 - sin \: a} } = \sec \: a + \tan \: a$

$\sqrt{ \frac{1 + sin \: a }{1 - sin \: a } } \times \sqrt{ \frac{1 + sin \: a }{1 + sin \: a } } = \sec \: a + \tan \: a$

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

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

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

$\frac{1 + sin \: a }{ cos \: a } = \sec \: a + \tan \: a$

$\frac{1}{ cos \: a} + \frac{ sin \: a }{ cos \: a } = \sec \: + \tan \: a$

$sec\:a+tan\:a=sec\:a+tan \: a$

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

hence proved