Show that $\tan x + \frac{\cos x}{1 + \sin x} = \sec x$. Explain, without the use of calculator, why the equation $\tan x + \frac{\cos x}{1 + \sin x} = -\frac{1}{3}$ has no solution.

1. Express $\tan x$ in terms of sine and cosine, and combine the terms into a single fraction.

$\text{LHS} = \frac{\sin x}{\cos x} + \frac{\cos x}{1 + \sin x}$
$= \frac{\sin x(1 + \sin x) + \cos^2 x}{\cos x(1 + \sin x)}$

2. Expand the numerator and apply the Pythagorean identity.

$= \frac{\sin x + \sin^2 x + \cos^2 x}{\cos x(1 + \sin x)}$
$= \frac{\sin x + 1}{\cos x(1 + \sin x)}$

3. Simplify the fraction to obtain the right-hand side.

$= \frac{1}{\cos x}$
$= \sec x = \text{RHS}$

1. Use the result from part (i) to simplify the equation

From (i), $\tan x + \frac{\cos x}{1 + \sin x} = \sec x$.
So the equation becomes:
$\sec x = -\frac{1}{3}$

2. Express in terms of cosine and explain why there is no solution

$\frac{1}{\cos x} = -\frac{1}{3}$
$\cos x = -3$
Since the range of $\cos x$ is $-1 \leqslant \cos x \leqslant 1$ for all real $x$, $\cos x$ cannot be $-3$. Therefore, the equation has no solution.