Question

#HelpASAP

Prove : $\rm {\sqrt{ \dfrac{1 + sin \: \theta}{1 - sin \: \theta} } = sec \: \theta + tan \: \theta}$​

Answer 1

Correct Question :-

Prove : $\tt {\sqrt{ \dfrac{1 + sin \: \theta}{1 - sin \: \theta} } = sec \: \theta + tan \: \theta}$

Correct Answer :-

Solving $ \large \mathcal{LHS}$ —

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

• Rationalising...

$\leadsto \sf { = \sqrt{ \dfrac{1 + sin \: \theta}{1 - sin \: \theta} \times \dfrac{1 + sin \: \theta}{1 + sin \: \theta} }}$

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

• Identity : sin²∅ + cos²∅ = 1

$\leadsto { \sf {=\sqrt{ \dfrac{{(1 + sin \: \theta)}^{2} }{{cos}^{2} \theta} }} }$

$\leadsto \sf {= \sqrt{{ \bigg( \dfrac{1 + sin \: \theta}{cos \: \theta} \bigg)}^{2} }}$

$\leadsto \sf{ =\dfrac{1}{cos \: \theta} + \dfrac{sin \: \theta}{cos \: \theta} }$

$\leadsto \sf {=sec \: \theta + tan \: \theta }$

— Results $ \large \mathcal{RHS}$

∴ Hence, Proved!!!

@MrCyber