The diagram shows part of the curve $y = \sqrt{4x + 5}$. The normal of the curve at $P$ meets the $x$-axis at $T$. The $x$-coordinate of $P$ is 5. Given that $PT$ is twice of $PQ$ and $RQ$ is parallel to the $x$-axis,

1. Find the $y$-coordinate of $P$.

When $x = 5$,
$y = \sqrt{4(5) + 5} = \sqrt{25} = 5$
So, $P$ is $(5, 5)$.

2. Find the gradient of the tangent at $P$.

$y = (4x + 5)^{\frac{1}{2}}$
$\frac{dy}{dx} = \frac{1}{2}(4x + 5)^{-\frac{1}{2}} \cdot 4 = \frac{2}{\sqrt{4x + 5}}$
At $x = 5$, gradient of tangent $m_t = \frac{2}{\sqrt{4(5) + 5}} = \frac{2}{5}$.

3. Find the equation of the normal at $P$.

Gradient of normal $m_n = -\frac{1}{m_t} = -\frac{5}{2}$.
Equation of the normal:
$y - 5 = -\frac{5}{2}(x - 5)$
$y = -\frac{5}{2}x + \frac{25}{2} + 5$
$y = -\frac{5}{2}x + \frac{35}{2}$

1. Find the derivative of the curve equation

$y = (4x + 5)^{\frac{1}{2}}$
$\frac{dy}{dx} = \frac{1}{2}(4x + 5)^{-\frac{1}{2}}(4)$
$\frac{dy}{dx} = \frac{2}{\sqrt{4x + 5}}$

2. Explain why the derivative cannot be zero

For a stationary point, $\frac{dy}{dx} = 0$.
However, the numerator of $\frac{2}{\sqrt{4x + 5}}$ is a non-zero constant (2), so $\frac{dy}{dx} \neq 0$ for any real value of $x$.
Therefore, the curve has no stationary points.

1. Find the coordinates of $P$, $T$, and $Q$ from earlier parts to establish the boundaries

From the curve equation $y = \sqrt{4x + 5}$, at $x = 5$, $y = \sqrt{25} = 5$. So $P$ is $(5, 5)$.
$\frac{dy}{dx} = \frac{1}{2}(4x+5)^{-1/2}(4) = \frac{2}{\sqrt{4x+5}}$.
At $x = 5$, gradient of tangent $= \frac{2}{5}$, so gradient of normal $= -\frac{5}{2}$.
Equation of normal: $y - 5 = -\frac{5}{2}(x - 5) \implies y = -2.5x + 17.5$.
At $T$, $y = 0 \implies 2.5x = 17.5 \implies x = 7$. So $T(7, 0)$.
Since $PT = 2PQ$ and $Q$ is above $P$, the horizontal distance from $Q$ to $P$ is half of that from $P$ to $T$.
$x_P - x_Q = \frac{1}{2}(x_T - x_P) \implies 5 - x_Q = \frac{1}{2}(7 - 5) = 1 \implies x_Q = 4$.
$y_Q = -2.5(4) + 17.5 = 7.5$. So $Q(4, 7.5)$.

2. Calculate the total area under the upper boundaries ($RQ$ and $QP$)

The upper boundary consists of the horizontal line $RQ$ from $x=0$ to $x=4$, and the line segment $QP$ from $x=4$ to $x=5$.
Area under $RQ$ (rectangle) $= 4 \times 7.5 = 30$.
Area under $QP$ (trapezium) $= \frac{1}{2}(7.5 + 5) \times (5 - 4) = 6.25$.
Total area under upper boundary $= 30 + 6.25 = 36.25 = \frac{145}{4}$.

3. Calculate the area under the curve and subtract it from the upper area to find the shaded region

Area under the curve from $x=0$ to $x=5$:
$\int_0^5 (4x + 5)^{\frac{1}{2}} \, dx = \left[ \frac{(4x+5)^{\frac{3}{2}}}{\frac{3}{2} \times 4} \right]_0^5$
$= \left[ \frac{1}{6}(4x+5)^{\frac{3}{2}} \right]_0^5$
$= \frac{1}{6}(25^{\frac{3}{2}} - 5^{\frac{3}{2}})$
$= \frac{1}{6}(125 - 5\sqrt{5}) = \frac{125}{6} - \frac{5\sqrt{5}}{6}$

Shaded Area $= \text{Area under lines} - \text{Area under curve}$
$= \frac{145}{4} - \left( \frac{125}{6} - \frac{5\sqrt{5}}{6} \right)$
$= \frac{435}{12} - \frac{250}{12} + \frac{5\sqrt{5}}{6}$
$= \frac{185}{12} + \frac{5\sqrt{5}}{6} \approx 17.3 \text{ units}^2$

1. Express the terms on the right-hand side using sine and cosine.

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

2. Combine the terms into a single fraction and expand the square.

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

3. Use the Pythagorean identity to express the denominator in terms of cosine.

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

4. Factorise the denominator as a difference of squares and simplify to obtain the left-hand side.

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

1. Express $\sin 230^\circ$ in terms of cosine using complementary angles or reference angles

$\sin 230^\circ = \sin(180^\circ + 50^\circ) = -\sin 50^\circ$
Since $\sin \theta = \cos(90^\circ - \theta)$, we have $\sin 50^\circ = \cos 40^\circ$.
Therefore, $\sin 230^\circ = -\cos 40^\circ$.

2. Set up the equation for $\cos 2x$ and find the basic angle

$\cos 2x = -\cos 40^\circ$
The basic angle for $2x$ is $40^\circ$. Since $\cos 2x$ is negative, $2x$ must lie in the second or third quadrant.

3. Solve for $x$ within the given domain

Given $0^\circ < x < 180^\circ$, the domain for $2x$ is $0^\circ < 2x < 360^\circ$.
$2x = 180^\circ - 40^\circ = 140^\circ$ or $2x = 180^\circ + 40^\circ = 220^\circ$
$x = 70^\circ \text{ or } 110^\circ$