Question

please solve the above trigonometry question​

Answer 1

$To\:prove:$

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

$Proof$

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

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$Hope \:it \:helps$

Answer 2

Given,

LHS

$= \sqrt{ \frac{1 - \sin \alpha }{1 + \sin\alpha } }$

RHS

$= \sec\alpha - \tan\alpha$

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