If , evaluate : (i) (ii)

Understand the Question Given , so let and , where is a positive number. To evaluate the given expressions, we need and , where Therefore, our first step is to use the Pythagoras theorem to find the length of the hypotenuse. Part (i) Evaluate Step 1 · Find Sides of the Triangle Consider right triangle with and . Given Let and for some positive number . By Pythagoras theorem in Now, calculate and Step 2 · Evaluate the First Expression Using Substitute and Answer > (i) Part (ii) Evaluate Step 1 · Evaluate the Second Expression Since Answer > (ii) Common Mistakes Swapping Sides: , so and . Swapping these inverts and . Identity Shortcut: Note that . Both parts yield the exact same value.

Introduction to Trigonometry | Exercise 8.1

Question 7

If $\cot \theta = \dfrac{7}{8}$ , evaluate :

(i) $\dfrac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$

(ii) $\cot^2 \theta$

Solution

Understand the Question

Given $\cot \theta = \dfrac{7}{8} = \dfrac{\text{Adjacent}}{\text{Opposite}}$ , so let $\text{Adjacent} = 7k$ and $\text{Opposite} = 8k$ , where $k$ is a positive number.
To evaluate the given expressions, we need $\sin \theta$ $sin θ$ and $\cos \theta$ $cos θ$ , where
- $\sin \theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$
- $\cos \theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$
Therefore, our first step is to use the Pythagoras theorem to find the length of the hypotenuse.

(i) Evaluate $\dfrac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$

Step 1 · Find Sides of the Triangle

Consider right triangle $\text{ABC}$ with $\angle \text{B} = 90^\circ$ and $\angle \text{C} = \theta$ . Diagram 1

Given $\cot \theta = \dfrac{\text{Adjacent side}}{\text{Opposite side}} = \dfrac{\text{BC}}{\text{AB}} = \dfrac{7}{8}$

Let $\text{BC} = 7k$ and $\text{AB} = 8k$ for some positive number $k$ .

By Pythagoras theorem in $\Delta \text{ABC}$

\begin{aligned} \text{AC}^2 &= \text{AB}^2 + \text{BC}^2 \\ &= (8k)^2 + (7k)^2 \\ &= 64k^2 + 49k^2 \\ &= 113k^2 \\ \text{AC} &= \sqrt{113k^2} = k\sqrt{113} \end{aligned}

Now, calculate $\sin \theta$ and $\cos \theta$

\begin{aligned} \sin \theta &= \dfrac{\text{Opposite side}}{\text{Hypotenuse}} = \dfrac{\text{AB}}{\text{AC}} = \dfrac{8k}{k\sqrt{113}} = \dfrac{8}{\sqrt{113}} \end{aligned} \\[0.6em] \begin{aligned} \cos \theta &= \dfrac{\text{Adjacent side}}{\text{Hypotenuse}} = \dfrac{\text{BC}}{\text{AC}} = \dfrac{7k}{k\sqrt{113}} = \dfrac{7}{\sqrt{113}} \end{aligned}

Step 2 · Evaluate the First Expression

Using $(a+b)(a-b) = a^2 - b^2$

\begin{aligned} \frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)} &= \frac{1^2 - \sin^2 \theta}{1^2 - \cos^2 \theta} \\[0.6em] &= \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} \end{aligned}

Substitute $\sin^2\theta = \dfrac{64}{113}$ and $\cos^2\theta = \dfrac{49}{113}$

\begin{aligned} \frac{1 - \sin^2\theta}{1 - \cos^2\theta} &= \frac{1 - \frac{64}{113}}{1 - \frac{49}{113}} \\[1.1em] &= \frac{\frac{49}{113}}{\frac{64}{113}} \\[1.1em] &= \frac{49}{64} \end{aligned}

Answer

(i) $\dfrac{49}{64}$

(ii) Evaluate $\cot^2 \theta$

Step 1 · Evaluate the Second Expression

Since $\cot \theta = \dfrac{7}{8}$

\begin{aligned} \cot^2 \theta &= \left(\frac{7}{8}\right)^2 \\[0.6em] &= \frac{49}{64} \end{aligned}

Answer

(ii) $\dfrac{49}{64}$

Common Mistakes

Swapping Sides: $\cot \theta = \dfrac{\text{Adjacent}}{\text{Opposite}}$ , so $\text{BC} = 7k$ and $\text{AB} = 8k$ . Swapping these inverts $\sin \theta$ and $\cos \theta$ .
Identity Shortcut: Note that $\dfrac{1-\sin^2\theta}{1-\cos^2\theta} = \dfrac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta$ . Both parts yield the exact same value.

Question 7

More questions in Exercise 8.1