Step 1 :First, we need to prove the identity. We start with the left side of the equation: $\frac{\cos \theta+\cot \theta}{1+\sin \theta}$
Step 2 :We know that $\cot \theta = \frac{\cos \theta}{\sin \theta}$, so we can substitute $\cot \theta$ with $\frac{\cos \theta}{\sin \theta}$ in the equation. So, the left side becomes $\frac{\cos \theta+\frac{\cos \theta}{\sin \theta}}{1+\sin \theta}$
Step 3 :We can simplify this to $\frac{\cos \theta \sin \theta + \cos \theta}{\sin \theta(1+\sin \theta)}$
Step 4 :We can factor out $\cos \theta$ from the numerator to get $\frac{\cos \theta(\sin \theta + 1)}{\sin \theta(1+\sin \theta)}$
Step 5 :We can see that $(\sin \theta + 1)$ cancels out in the numerator and the denominator, leaving us with $\frac{\cos \theta}{\sin \theta}$, which is equal to $\cot \theta$, the right side of the equation. So, the identity is proven.
Step 6 :Next, we need to find the non-permissible values for $\theta$ in the interval $0^\circ \leqslant \theta \leqslant 360^\circ$. The denominator of the left side of the equation is $1+\sin \theta$, so $\theta$ cannot be a value that makes $1+\sin \theta = 0$
Step 7 :We know that $-1 \leqslant \sin \theta \leqslant 1$, so $\sin \theta = -1$ is the only value that makes $1+\sin \theta = 0$. Therefore, $\theta = 270^\circ$ is a non-permissible value in the interval $0^\circ \leqslant \theta \leqslant 360^\circ$
Step 8 :In general form, $\theta = 180n + 270^\circ$ are non-permissible values, where $n$ is an integer.