Answer
$$\sin\theta+\cos\theta=\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}$$
The expression has been proved to be an identity by simplifying the right side.
Work Step by Step
$$\sin\theta+\cos\theta=\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}$$
The right side looks more complex, so we would deal with it first.
$$A=\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}$$
We would transform $\cot\theta$ and $\tan\theta$:
$$\cot\theta=\frac{\cos\theta}{\sin\theta}\hspace{2cm}\tan\theta=\frac{\sin\theta}{\cos\theta}$$
Therefore, $A$ would be $$A=\frac{\sin\theta}{1-\frac{\cos\theta}{\sin\theta}}+\frac{\cos\theta}{1-\frac{\sin\theta}{\cos\theta}}$$
$$A=\frac{\sin\theta}{\frac{\sin\theta-\cos\theta}{\sin\theta}}+\frac{\cos\theta}{\frac{\cos\theta-\sin\theta}{\cos\theta}}$$
$$A=\frac{\sin^2\theta}{\sin\theta-\cos\theta}+\frac{\cos^2\theta}{\cos\theta-\sin\theta}$$
$$A=\frac{\sin^2\theta}{\sin\theta-\cos\theta}-\frac{\cos^2\theta}{\sin\theta-\cos\theta}$$ (we change the sign of the numerator since the sign of the denominator has changed)
$$A=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta-\cos\theta}$$
$$A=\frac{(\sin\theta-\cos\theta)(\sin\theta+\cos\theta)}{\sin\theta-\cos\theta}$$ (for $a^2-b^2=(a-b)(a+b)$)
$$A=\sin\theta+\cos\theta$$
The left side is therefore equal with the right side. The expression has been proved to be an identity.