Find the value of  \(\frac{sec45°\;+\;cot45°}{sin30°\;+\;tan45°}\) .

0 1 \(\frac{2√2 + 2 }{3}\) \(\frac{2√2 + 2 }{5}\)

\(\frac{2√2 + 2 }{3}\)

⇒ \(\frac{sec45°\;+\;cot45°}{sin30°\;+\;tan45°}\) = \(\frac{√2 + 1 }{ \frac{1}{2} + 1 }\) =  \(\frac{2√2 + 2 }{3}\)