Question

prove that under root 1 minus sin theta upon 1 + sin theta is equals to sec theta + tan theta​

Answer 1

Correct Question :

To Prove $\sf{ \dfrac{ \sqrt{1 + \sin \theta } }{ \sqrt{1 - \sin \theta } } = \sec \theta + \tan \theta }$

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AnswEr :

To Prove :

$\sf{ \dfrac{ \sqrt{1 + \sin \theta } }{ \sqrt{1 - \sin \theta } } = \sec \theta + \tan \theta }$

Proof :

$\longrightarrow\sf{ \dfrac{ \sqrt{1+ \sin \theta } }{ \sqrt{1 - \sin \theta } }}$

• Rationalizing this

$\longrightarrow\sf{ \dfrac{ \sqrt{1 + \sin \theta } }{ \sqrt{1 - \sin \theta } } \times \dfrac{\sqrt{1 + \sin \theta } }{ \sqrt{1 + \sin \theta } }}$

$\longrightarrow\sf{ \dfrac{ (\sqrt{1+ \sin \theta }) ^{2} }{ \sqrt{1 - \sin^{2} \theta } }}$

$\longrightarrow\sf{ \dfrac{1 + \sin \theta}{ \sqrt{ \cos^{2} \theta } } }$

$\longrightarrow\sf{ \dfrac{1 + \sin \theta}{ \cos \theta }}$

$\longrightarrow\sf{ \dfrac{1}{\cos \theta} + \dfrac{\sin \theta}{ \cos \theta }}$

$\longrightarrow\sf{ \sec \theta + \tan \theta }$