Given that f(θ) = (cos θ / (1-tan θ)) + (sin θ / (1-cot θ)). (a) Prove that f(θ) ≡ cos θ + sin θ. (b) Express f(θ) in the form Rsin(θ + λ), where R > 0 and λ an acute angle. (c) Find the general solution of the equation f(θ) = 1.

$$f(\theta) = \cos \theta + \sin \theta$$

Thus, $f(\theta) \equiv \cos \theta + \sin \theta$ is proven.

We compare this to the expansion of $R\sin(\theta + \lambda) = R\sin\theta\cos\lambda + R\cos\theta\sin\lambda$.
By comparing coefficients:
$R\cos\lambda = 1$ (coefficient of $\sin\theta$)
$R\sin\lambda = 1$ (coefficient of $\cos\theta$)
To find $R$, square and add the two equations:

$$(R\cos\lambda)^2 + (R\sin\lambda)^2 = 1^2 + 1^2$$ $$R^2(\cos^2\lambda + \sin^2\lambda) = 2$$ $$R^2(1) = 2 \implies R = \sqrt{2} \quad (\text{since } R > 0)$$

$$\frac{R\sin\lambda}{R\cos\lambda} = \frac{1}{1}$$ $$\tan\lambda = 1$$

Since $\lambda$ is an acute angle, $\lambda = \arctan(1) = \frac{\pi}{4}$ radians or $45^\circ$.

This is $x = \frac{\pi}{4}$.
The general solution for $\sin X = \sin \alpha$ is $X = k\pi + (-1)^k \alpha$, where $k$ is an integer.
Let $X = \theta + \frac{\pi}{4}$ and $\alpha = \frac{\pi}{4}$.

$$\theta + \frac{\pi}{4} = k\pi + (-1)^k \frac{\pi}{4}$$

$$\theta = k\pi + (-1)^k \frac{\pi}{4} - \frac{\pi}{4}$$

We consider two cases for $k$:
Case 1: $k$ is an even integer (e.g., $k=2n$ for some integer $n$).

$$\theta = 2n\pi + (-1)^{2n} \frac{\pi}{4} - \frac{\pi}{4}$$ $$\theta = 2n\pi + \frac{\pi}{4} - \frac{\pi}{4}$$ $$\theta = 2n\pi$$

Case 2: $k$ is an odd integer (e.g., $k=2n+1$ for some integer $n$).

$$\theta = (2n+1)\pi + (-1)^{2n+1} \frac{\pi}{4} - \frac{\pi}{4}$$ $$\theta = (2n+1)\pi - \frac{\pi}{4} - \frac{\pi}{4}$$ $$\theta = (2n+1)\pi - \frac{2\pi}{4}$$ $$\theta = (2n+1)\pi - \frac{\pi}{2}$$ $$\theta = 2n\pi + \pi - \frac{\pi}{2}$$ $$\theta = 2n\pi + \frac{\pi}{2}$$

The general solution is:

$$\boxed{\theta = 2n\pi \quad \text{or} \quad \theta = 2n\pi + \frac{\pi}{2}, \quad \text{where } n \in \mathbb{Z}}$$