Question

plzz ...prove that..​

Answer 1

$lhs = \frac{1 + \sec(x) }{ \sec(x) } \\ \: \: \: \: \: \: \: = \frac{1 + \frac{1}{ \cos(x) } }{ \frac{1}{ \cos(x) } } = \frac{ \frac{ \cos(x) + 1}{ \cos(x) } }{ \frac{1}{ \cos(x) } } \\ \: \: \: \: \: \: \: = \cos(x) + 1$

$rhs = \frac{ \sin {}^{2} (x) }{1 - \cos(x) } = \frac{1 - \cos {}^{2} (x) }{1 - \cos(x) } \\ \: \: \: \: \: \: \: \: = \frac{(1 + \cos(x))(1 - \cos(x)) }{1 - \cos(x) } \\ \: \: \: \: \: \: \: \: = 1 + \cos(x)$

Identities used are :

sin²x = 1-cos²x

a²-b² = (a+b)(a-b)

Therefore LHS = RHS

hence proved

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