Trigonometric identity expressing $\sec \theta+\text{cosec } \theta$ in terms of sine and cosine > $\large{\text{cosec }\theta+\sec{\theta}=\dfrac{\sin\theta+\cos\theta}{\sin\theta\,\cos\theta}}$ I know that cosecant is the inverse of sine, and secant is the inverse of cosine. However, that does not equal the right hand side of the equation. I know nothing about what to do next.

Perhaps most algebraically natural is to go from right to left. We have $$\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\frac{\sin\theta}{\sin\theta\cos\theta}+\frac{\cos\theta}{\sin\theta\cos\theta}=\frac{1}{\cos\theta}+\frac{1}{\sin\theta}=\sec\theta+\csc\theta.$$

However, going from left to right is also in a certain sense natural. Express the left side in terms of sines and cosines. We have $$\csc\theta+\sec\theta=\frac{1}{\sin\theta}+\frac{1}{\cos\theta}.$$ Now bring the expression on the right to a common denominator $\sin\theta\cos\theta$.