2. Algebraically prove the trigonometric equation $\frac{\cos \theta+\cot \theta}{1+\sin \theta}=\cot \theta$ is an identity. [3 Marks]
\begin{tabular}{|l|l|}
\hline Left Side: $\frac{\cos \theta+\cot \theta}{1+\sin \theta}$ & Right Side: $\cot \theta$ \\
\hline & \\
\hline & \\
\hline
\end{tabular}
- State the non-permissible values for the identity in the interval $0^{\circ} \leqslant \theta \leqslant \leqslant 360^{\circ}$, and then state the non-permissible values in general form. [2 Marks]

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.