Prove that $\dfrac{\tan x}{1+\sec x}+\dfrac{1+\sec x}{\tan x}=2\csc x$. Hence, solve $\dfrac{\tan 2x}{1+\sec 2x}+\dfrac{1+\sec 2x}{\tan 2x}=3\sec x$ for $-\pi \le x \le \pi$. Use the result in part (a) to find the range of values of $k$ for which the equation $\dfrac{\tan x}{1+\sec x}+\dfrac{1+\sec x}{\tan x}=k$ has no solutions.

1. Combine the fractions on the left-hand side over a common denominator.

$\text{LHS} = \frac{\tan^2 x + (1+\sec x)^2}{\tan x(1+\sec x)}$

2. Expand the numerator.

$\text{LHS} = \frac{\tan^2 x + 1 + 2\sec x + \sec^2 x}{\tan x(1+\sec x)}$

3. Apply the Pythagorean identity $\tan^2 x + 1 = \sec^2 x$.

$\text{LHS} = \frac{\sec^2 x + 2\sec x + \sec^2 x}{\tan x(1+\sec x)} = \frac{2\sec^2 x + 2\sec x}{\tan x(1+\sec x)}$

4. Factorise the numerator and simplify the fraction.

$\text{LHS} = \frac{2\sec x(\sec x + 1)}{\tan x(1+\sec x)} = \frac{2\sec x}{\tan x}$

5. Convert to sine and cosine to reach the final form.

$\text{LHS} = \frac{2\left(\frac{1}{\cos x}\right)}{\frac{\sin x}{\cos x}} = \frac{2}{\sin x} = 2\csc x = \text{RHS}$

1. Substitute the identity from part (a) into the equation.

From part (a), we have $\frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}=2\csc x$.
So the equation becomes:
$2\csc x = k$
$\csc x = \frac{k}{2}$

2. Determine the range of values for which there are no solutions, considering the domain of the original expression.

We know that $\csc x = \frac{1}{\sin x}$. For the equation to have solutions, we generally need $|\csc x| \ge 1$, which means $\left|\frac{k}{2}\right| \ge 1 \implies |k| \ge 2$.
However, we must also consider the domain of the original expression. The terms $\tan x$ and $\sec x$ are undefined when $\cos x = 0$.
When $\cos x = 0$, $\sin x = \pm 1$, which corresponds to $\csc x = \pm 1$.
Therefore, $\csc x$ cannot be $1$ or $-1$ because the original equation would be undefined.
Thus, the valid range for $2\csc x$ is strictly $(-\infty, -2) \cup (2, \infty)$.
For the equation to have no solutions, $k$ must fall outside this valid range, including the undefined boundary values.

3. State the final range of $k$.

The equation has no solutions when:
$-2 \le k \le 2$