Question 7

Right-Angled Triangle: For any acute angle $\theta$:

$\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}}, \quad \cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}, \quad \cot\theta = \frac{\text{Adjacent}}{\text{Opposite}}$

Pythagoras Theorem: $\text{Hypotenuse}^2 = \text{Opposite}^2 + \text{Adjacent}^2$

Useful identity: $(1 + \sin\theta)(1 - \sin\theta) = 1 - \sin^2\theta$ and $(1 + \cos\theta)(1 - \cos\theta) = 1 - \cos^2\theta$. These come from the difference of squares formula $(a+b)(a-b) = a^2 - b^2$.

Since $\cot\theta = \frac{7}{8}$, we take adjacent $= 7k$, opposite $= 8k$, find hypotenuse via Pythagoras, then compute all required values.

We will use a right-angled triangle to find the values of $\sin \theta$ and $\cos \theta$.

Step 1 — Find sides of the triangle

Let's draw a right-angled triangle ABC. Let angle B be $\mathbf{90^\circ}$. Let angle C be $\theta$. We are given $\cot \theta = \frac{7}{8}$. We know $\cot \theta = \frac{\text{Adjacent side}}{\text{Opposite side}}$. So, $\text{BC}$ is the adjacent side. $\text{AB}$ is the opposite side. Let $\text{BC} = \mathbf{7k}$ and $\text{AB} = \mathbf{8k}$ for some positive number $\mathbf{k}$. Now, we find the hypotenuse $\text{AC}$. We use the Pythagoras theorem.

$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$

$= (8k)^2 + (7k)^2$

$= 64k^2 + 49k^2$

$= 113k^2$

$\text{AC} = \sqrt{113k^2}$

$\boxed{\text{AC} = k\sqrt{113}}$

Now, we find $\sin \theta$ and $\cos \theta$. We know $\sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}}$.

$\sin \theta = \frac{\text{AB}}{\text{AC}}$

$= \frac{8k}{k\sqrt{113}}$

$\boxed{\sin \theta = \frac{8}{\sqrt{113}}}$

We know $\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$.

$\cos \theta = \frac{\text{BC}}{\text{AC}}$

$= \frac{7k}{k\sqrt{113}}$

$\boxed{\cos \theta = \frac{7}{\sqrt{113}}}$

Step 2 — Evaluate the first expression

Let's evaluate the first expression. The expression is $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$. We use the identity $(a+b)(a-b) = a^2 - b^2$. So, the numerator becomes $1^2 - \sin^2 \theta$. The denominator becomes $1^2 - \cos^2 \theta$.

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

Now, we substitute the values of $\sin \theta$ and $\cos \theta$.

$= \frac{1 - \left(\frac{8}{\sqrt{113}}\right)^2}{1 - \left(\frac{7}{\sqrt{113}}\right)^2}$

$= \frac{1 - \frac{64}{113}}{1 - \frac{49}{113}}$

$= \frac{\frac{113 - 64}{113}}{\frac{113 - 49}{113}}$

$= \frac{\frac{49}{113}}{\frac{64}{113}}$

$= \frac{49}{113} \times \frac{113}{64}$

$\boxed{\frac{49}{64}}$

Step 3 — Evaluate the second expression

Let's evaluate the second expression. The expression is $\cot^2 \theta$. We are given that $\cot \theta = \frac{7}{8}$.

$\cot^2 \theta = \left(\frac{7}{8}\right)^2$

$\boxed{\frac{49}{64}}$

Answer

(i) $\frac{49}{64}$ (ii) $\frac{49}{64}$