Question 8

Right-Angled Triangle: For angle A with sides opposite, adjacent, hypotenuse:

$\sin A = \frac{\text{Opp}}{\text{Hyp}}, \quad \cos A = \frac{\text{Adj}}{\text{Hyp}}, \quad \tan A = \frac{\text{Opp}}{\text{Adj}}, \quad \cot A = \frac{\text{Adj}}{\text{Opp}}$

Reciprocal relationship: $\tan A = \dfrac{1}{\cot A}$ — tan and cot are always reciprocals of each other.

Pythagoras Theorem: $\text{Hyp}^2 = \text{Opp}^2 + \text{Adj}^2$ — used to find the third side once two sides are known.

What we're verifying: The equation $\dfrac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A$ asks whether the LHS (in terms of tan) always equals the RHS (in terms of sin and cos). We verify it by computing both sides numerically for the given value and checking they match.

Strategy: From $3\cot A = 4$, get $\cot A = \frac{4}{3}$, then $\tan A = \frac{3}{4}$. Take adjacent $= 4k$, opposite $= 3k$, find hypotenuse via Pythagoras, then compute LHS and RHS separately.

We need to check if the given identity is true.

Step 1 — Find $\tan A$ and calculate the Left Hand Side.

We are given the value of $3 \cot A$. Let's find $\cot A$.

$3 \cot A = 4$

$\cot A = \frac{4}{3}$

Now, we can find $\tan A$. We know that $\tan A$ is the reciprocal of $\cot A$.

$\tan A = \frac{1}{\cot A}$

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

$\tan A = \frac{3}{4}$

Let's calculate the Left Hand Side (LHS) of the equation. The LHS is $\frac{1 - \tan^2 A}{1 + \tan^2 A}$.

$\frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - (3/4)^2}{1 + (3/4)^2}$

$= \frac{1 - 9/16}{1 + 9/16}$

$= \frac{(16 - 9)/16}{(16 + 9)/16}$

$= \frac{7/16}{25/16}$

$\boxed{\frac{7}{25}}$

Step 2 — Find $\sin A$ and $\cos A$ and calculate the Right Hand Side.

Let's use a right-angled triangle to find $\sin A$ and $\cos A$. We know $\cot A = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{4}{3}$. Let the adjacent side be 4k and the opposite side be 3k. We can find the hypotenuse using the Pythagoras theorem.

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

$\text{Hypotenuse}^2 = (3k)^2 + (4k)^2$

$\text{Hypotenuse}^2 = 9k^2 + 16k^2$

$\text{Hypotenuse}^2 = 25k^2$

$\text{Hypotenuse} = \sqrt{25k^2}$

$\text{Hypotenuse} = 5k$

Now we can find $\sin A$ and $\cos A$. $\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}}$.

$\sin A = \frac{3k}{5k}$

$\sin A = \frac{3}{5}$

$\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$.

$\cos A = \frac{4k}{5k}$

$\cos A = \frac{4}{5}$

Let's calculate the Right Hand Side (RHS) of the equation. The RHS is $\cos^2 A - \sin^2 A$.

$\cos^2 A - \sin^2 A = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2$

$= \frac{16}{25} - \frac{9}{25}$

$= \frac{16 - 9}{25}$

$\boxed{\frac{7}{25}}$

Answer

We found that the Left Hand Side is $\frac{7}{25}$. We also found that the Right Hand Side is $\frac{7}{25}$. Since both sides are equal, the statement is true.

(i) The value of $\frac{1 - \tan^2 A}{1 + \tan^2 A}$ is $\frac{7}{25}$. (ii) The value of $\cos^2 A - \sin^2 A$ is $\frac{7}{25}$. (iii) Yes, $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$.