If , check whether or not.

Understand the Question Given , so let and , where is a positive number. Evaluate LHS using . For RHS (), use Pythagoras theorem to find Hypotenuse, then compute and . Step 1 · Find and Calculate Left Hand Side Given Evaluating LHS Step 2 · Find , , and Calculate Right Hand Side Consider right triangle with . Given Let and for some positive number . By Pythagoras theorem Now find and Evaluating RHS () By comparing equations (i) and (ii), we get Answer > Since , the statement is true. Common Mistakes Reciprocal Error: Incorrectly taking instead of taking the reciprocal . Trig Identity: Both and equal , so LHS and RHS are equal for all valid angles.

Introduction to Trigonometry | Exercise 8.1

Question 8

If $3 \cot A = 4$ , check whether $\dfrac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ or not.

Solution

Understand the Question

Given $3 \cot A = 4 \implies \cot A = \dfrac{4}{3} = \dfrac{\text{Adjacent}}{\text{Opposite}}$ , so let $\text{Adjacent} = 4k$ and $\text{Opposite} = 3k$ , where $k$ is a positive number.
Evaluate LHS using $\tan A = \dfrac{1}{\cot A} = \dfrac{3}{4}$ .
For RHS ( $\cos^2 A - \sin^2 A$ ), use Pythagoras theorem to find Hypotenuse, then compute $\sin A = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$ and $\cos A = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$ .

Step 1 · Find $\tan A$ and Calculate Left Hand Side

Given $3 \cot A = 4 \implies \cot A = \dfrac{4}{3}$

$\tan A = \dfrac{1}{\cot A} = \frac{1}{4/3} = \dfrac{3}{4}$

$\tan^2 A = \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}$

Evaluating LHS

\begin{aligned} \frac{1 - \tan^2 A}{1 + \tan^2 A} &= \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} \\[1.1em] &= \frac{\frac{7}{16}}{\frac{25}{16}} \\[1.1em] &= \frac{7}{25} \quad \dots (i) \end{aligned}

Step 2 · Find $\sin A$ , $\cos A$ , and Calculate Right Hand Side

Consider right triangle $\text{ABC}$ with $\angle \text{B} = 90^\circ$ .

Given $\cot A = \dfrac{\text{Adjacent side}}{\text{Opposite side}} = \dfrac{\text{AB}}{\text{BC}} = \dfrac{4}{3}$ Diagram 1

Let $\text{AB} = 4k$ and $\text{BC} = 3k$ for some positive number $k$ .

By Pythagoras theorem

\begin{aligned} \text{Hypotenuse}^2 &= \text{Opposite}^2 + \text{Adjacent}^2 \\ &= (3k)^2 + (4k)^2 \\ &= 9k^2 + 16k^2 \\ &= 25k^2 \\ \text{Hypotenuse} &= \sqrt{25k^{2}} = 5k \end{aligned}

Now find $\sin A$ and $\cos A$

\begin{aligned} \sin A &= \dfrac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{3k}{5k} = \frac{3}{5} \end{aligned} \\[0.6em] \begin{aligned} \cos A &= \dfrac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{4k}{5k} = \frac{4}{5} \end{aligned}

Evaluating RHS ( $\cos^2 A - \sin^2 A$ )

\begin{aligned} \cos^2 A &= \frac{16}{25} \\[0.6em] \sin^2 A &= \frac{9}{25} \end{aligned} \\[0.6em] \begin{aligned} \cos^2 A - \sin^2 A &= \frac{16}{25} - \frac{9}{25} \\[0.6em] &= \frac{16 - 9}{25} \\[0.6em] &= \frac{7}{25} \quad \dots (ii) \end{aligned}

By comparing equations (i) and (ii), we get $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$

Answer

Since $\text{LHS} = \text{RHS} = \dfrac{7}{25}$ , the statement is true.

$\dfrac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$

Common Mistakes

Reciprocal Error: Incorrectly taking $\tan A = \cot A = \dfrac{4}{3}$ instead of taking the reciprocal $\tan A = \dfrac{3}{4}$ .
Trig Identity: Both $\dfrac{1-\tan^2 A}{1+\tan^2 A}$ and $\cos^2 A - \sin^2 A$ equal $\cos 2A$ , so LHS and RHS are equal for all valid angles.

Question 8

More questions in Exercise 8.1