Question

1 + sin theta / 1 - sin theta = ( sec theta + tan theta )^2​

Answer 1

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• Prove that, 1 + sin theta / 1 - sin theta = ( sec theta + tan theta )^2

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$\sf \: \: \: \: \: \bull\: \: \: \: \: \frac{1 + \sin \theta}{1 - \sin \theta}$

$\implies \sf \frac{ \big(1 + \sin \theta\big)\big(1 + \sin \theta\big)}{\big(1 - \sin \theta\big)\big(1 + \sin \theta\big)}$

$\implies \sf \frac{ {\big(1 + \sin \theta\big)}^{2} }{\big(1 - { \sin}^{2} \theta\big) }$

$\implies \sf\frac{1 + 2 \sin \theta + { \sin}^{2} \theta }{ { \cos}^{2} \theta}$

$\implies \sf { \sec}^{2} \theta + 2 \tan \theta \sec \theta + { \tan}^{2} \theta$

$\implies { \underline{ \boxed{{\bf{\bigg( \sec \theta + { \tan} \theta \bigg) }^{2} }}}}\: \: \: \: \: \: \: \: \: \: \: hence\: \: proved.$

Answer 2

Answer:

rationalise lhs we get, (1+sin) ²/1²-sin²

(1+sin)²/cos²

(1+sin/cos) ²

(1/cos+sin/cos) ²

(sec+tan) ²