Given,

\(\frac{sin 30° + tan 45° - cosec 60°}{sec 30° + cos 60° + cot 45°}\)

We know that,

sin 30° = \(\frac{1}{2}\)

tan 45° = 1

cosec 60° = 2/√3

sec 30° = 2/√3

cos 60° = \(\frac{1}{2}\)

cot 45° = 1

Substitute the values in the given problem, we get

\(\frac{sin 30° + tan 45° - cosec 60°}{sec 30° + cos 60° + cot 45°}\) 

= \(\frac{\frac{1}{2} + 1 - \frac{2}{\sqrt3}}{\frac{2}{\sqrt3}+\frac{1}{2}+1}\)

= \(\frac{\frac{\sqrt2+2\sqrt3-4}{2\sqrt3}}{\frac{4+\sqrt3+2\sqrt3}{2\sqrt3}}\)

Now, cancel the term \(2\sqrt3\), in numerator and denominator,

we get,

= \(\frac{\sqrt3 +2\sqrt3-4}{4+\sqrt3+2\sqrt3}\)

= \(\frac{3\sqrt3-4}{3\sqrt3+4}\)

Now, rationalize the terms

= \(\frac{3\sqrt3-4}{3\sqrt3+4}\times\)\(\frac{3\sqrt3-4}{3\sqrt3-4}\)

= \(\frac{27-12\sqrt3-12\sqrt3+16}{27-12\sqrt3+12\sqrt3+16}\)

= \(\frac{27-24\sqrt3+16}{11}\)

= \(\frac{43-24\sqrt3}{11}\)

\(\frac{sin 30° + tan 45° - cosec 60°}{sec 30° + cos 60° + cot 45°}\)

= \(\frac{43-24\sqrt3}{11}\)