Recall the Pythagorean identities for tangent and cotangent.

We know that:

$$1+\tan^2\theta = \sec^2\theta$$ $$1+\cot^2\theta = \csc^2\theta$$

Substitute these into the LHS of the equation:

$$\text{LHS} = \frac{1}{\sec^2\theta} + \frac{1}{\csc^2\theta}$$

We know that:

$$\frac{1}{\sec\theta} = \cos\theta \implies \frac{1}{\sec^2\theta} = \cos^2\theta$$ $$\frac{1}{\csc\theta} = \sin\theta \implies \frac{1}{\csc^2\theta} = \sin^2\theta$$

Substitute these into the expression:

$$\text{LHS} = \cos^2\theta + \sin^2\theta$$

We know that:

$$\sin^2\theta + \cos^2\theta = 1$$

Therefore:

$$\text{LHS} = 1$$

Since the LHS simplifies to 1, which is equal to the RHS, the identity is proven.