Question

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Answer 1

$\huge \: \mathfrak{solution}$

To prove :

$\boxed{ \red{\bold{\sqrt{ \frac{1 + sin x}{1 - sin x} } = sec x \: + tanx}}}$

LHS:

$\rightarrow \: \sqrt{ \frac{1 + sinx}{1 - sinx} } \\ \\ \bold{ now \: rationalise \: the \: denominator} \\ \\ \rightarrow \: \sqrt{ \frac{1 + sinx}{1 - sinx} \times \frac{1 + sinx}{1 + sinx} } \\ \\ \rightarrow \: \sqrt{ \frac{(1 + {sin}x) ^{2} }{1 - {sin}^{2}x } } \\ \\ \bold{ as \: we \: know \: that }\ \\ \bold{ {sin}^{2} x + {cos}^{2} x = 1} \\ \bold{ {cos}^{2} x = 1 - {sin}^{2} x} \\ \\ \rightarrow \: \sqrt{ \frac{(1 + {sinx})^{2} }{ {cos}^{2} x} } \\ \\ \rightarrow \: \frac{1 + sinx}{cosx} \\ \\ \rightarrow \: \frac{1}{cosx} + \frac{sinx}{cosx} \\ \\ \rightarrow\: secx \: + \: tanx$

LHS=RHS

$\boxed{ \blue{ \huge\mathfrak{{proved}}}}$

Answer 2

$\sf \underline{ \green{ \fbox{\huge{ \: \pink{ \orange{Solution : \: \: }}}}}}$

LHS

$\hookrightarrow{ \mathtt{ \sqrt{ \frac{1 + \sin(A) }{1 - \sin(A) } }}}$

Rationalising the denominator , we obtain

$\hookrightarrow \mathtt{ \sqrt{ \frac{1 + \sin(A) }{1 - \sin(A) } \times \frac{1 + \sin(A) }{1 + \sin(A) }} } \\ \\ \hookrightarrow \mathtt{ \sqrt{ \frac{ {(1 + \sin(A))}^{2} }{ {(1)}^{2} - {\sin}^{2} (A) } }} \\ \\ \hookrightarrow \mathtt{ \sqrt{ \frac{ {(1 + \sin(A))}^{2} }{ {\cos}^{2} (A) } } \: \: \: \bigg \{ \because \: \: 1 - {\sin}^{2} (A) = {\cos}^{2} (A) \bigg\}} \\ \\ \hookrightarrow \mathtt{\frac{1 + \sin(A) }{\cos(A)} } \\ \\ \hookrightarrow \mathtt{ \frac{1}{\cos(A)} + \frac{\sin(A)}{\cos(A)} }\\ \\ \hookrightarrow \mathtt{ \sec(A) + \tan(A)\: \: \: \bigg \{ \because \: \: \frac{1}{ \cos(A) } = \sec(A) \: \: and \: \: \frac{ \sin(A)}{ \cos(A)} = \tan(A) \bigg \}}$

$\large {\purple{ \star}}$LHS = RHS

Hence proved