Question

√1+cos theta/√1-cos theta = ​

Answer 1

$\sf \Rightarrow \quad \dfrac{ \sqrt{1 + cos \: \theta} }{ \sqrt{1 - cos \: \theta} } \\ \\ \sf \Rightarrow \quad \sqrt{ \frac{1 + cos \: \theta}{1 - cos \: \theta} } \times \sqrt{ \frac{1 + cos \: \theta}{1 + cos \: \theta} } \\ \\ \sf \Rightarrow \quad \sqrt{ \frac{ {(1 + cos \: \theta)}^{2} }{(1 + cos \: \theta)(1 - cos \: \theta)} } \\ \\ \sf \Rightarrow \quad \sqrt{ \frac{ {(1 + cos \: \theta)}^{2} }{1 - {cos}^{2} \: \theta } } \qquad \bigg\{ \because (a + b)(a - b) = {a}^{2} - {b}^{2} \bigg\} \\ \\ \sf \Rightarrow \quad \frac{ \sqrt{ {(1 + cos \: \theta)}^{2} } }{ \sqrt{ {sin}^{2} \: \theta } } \qquad \bigg( \because {sin}^{2} \: \theta + {cos}^{2} \: \theta = 1 \bigg) \\ \\ \sf \Rightarrow \quad \frac{1 + cos \: \theta}{sin \: \theta} \\ \\ \sf \Rightarrow \quad \frac{1}{sin \: \theta} + \frac{cos \: \theta}{sin \: \theta} \\ \\ \sf \Rightarrow \quad \boxed{ \sf \red{cosec \: \theta + cot \: \theta }}\qquad \bigg( \because \frac{1}{sin \: \theta} = cosec \: \theta \quad , \frac{cos \: \theta}{sin \: \theta} = cot \: \theta \bigg) \\ \\ \\ \underline{ \sf Some \: Formulae} \\ \\ \sf \star \quad \frac{1}{cos \: \theta} = sec \: \theta \\ \\ \star \quad \sf \frac{sin \: \theta}{cos \: \theta} = tan \: \theta \\ \\ \star \sf \quad 1 + {tan}^{2} \: \theta = {sec}^{2} \: \theta \\ \\ \star \sf \quad 1 + {cot}^{2} \: \theta = {cosec}^{2} \: \theta$

Answer 2

$\huge{\boxed{\underline{\bold{Solution:)}}}}$