Question

✓ 1-sinø/✓1+sinø = secø-tanø

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Answer 1

$\sf \rightarrow \quad \sqrt{ \dfrac{1 - sin \: \theta }{1 + sin \: \theta} } = sec \: \theta - tan \: \theta \\ \\ \sf \quad \: solving \: L.H.S \\ \\ \sf \rightarrow \quad \sqrt{ \frac{1 - sin \: \theta}{1 + sin \: \theta} } \times \sqrt{ \frac{1 - sin \: \theta}{1 - sin \: \theta} } \\ \\ \sf \rightarrow \quad \sqrt{ \frac{ {(1 - sin \: \theta)}^{2} }{(1 + sin \: \theta)(1 - sin \: \theta)} } \\ \\ \sf \rightarrow \quad \frac{1 - sin \: \theta}{ \sqrt{1 - {sin}^{2} \: \theta } } \qquad \bigg( \because \: (a + b)(a - b) = {a}^{2} - {b}^{2} \bigg) \\ \\ \sf \rightarrow \quad \frac{1 - sin \: \theta}{cos \: \theta} \qquad \bigg( \: \because \: {sin}^{2} \: \theta + {cos }^{2} \: \theta = 1 \bigg) \\ \\ \sf \rightarrow \quad \frac{1}{cos \: \theta} - \frac{sin \: \theta}{cos \: \theta} \\ \\ \sf \rightarrow \quad sec \: \theta - tan \: \theta \: \qquad \bigg( \because \: \frac{1}{cos \: \theta }= sec \: \theta \: \quad \: and \: \quad \frac{sin \: \theta}{cos \: \theta} = tan \: \theta \bigg) \\ \\ \sf \qquad L.H.S = R.H.S \\ \sf \bold{Hence, \: Proved}$