Question 3

We will use fundamental trigonometric identities to simplify each expression.

Step 1 — Simplify $9 \sec^2 A - 9 \tan^2 A$

Let's factor out the common term, which is 9.

$9 \sec^2 A - 9 \tan^2 A = 9 (\sec^2 A - \tan^2 A)$

Pythagorean Identity: $\sec^2 A - \tan^2 A = 1$ (derived from dividing $\sin^2 A + \cos^2 A = 1$ by $\cos^2 A$).

We know a very important identity. The identity states that $\sec^2 A - \tan^2 A$ is always equal to 1.

$= 9 (1)$

$\boxed{9}$

Step 2 — Simplify $(1 + \tan \theta + \sec \theta) (1 + \cot \theta - \text{cosec } \theta)$

Let's rewrite all terms using $\sin \theta$ and $\cos \theta$. This helps us combine them.

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

Now, we find a common denominator for each bracket.

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

Difference of Squares: $(x+y)(x-y) = x^2 - y^2$. Here $x = (\cos\theta + \sin\theta)$ and $y = 1$.

Let's group $(\cos \theta + \sin \theta)$ together. This looks like $(x + y)(x - y)$ form. Here, $x = (\cos \theta + \sin \theta)$ and $y = 1$.

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

Now, we expand the numerator. Remember $(a+b)^2 = a^2 + b^2 + 2ab$.

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

Pythagorean Identity: $\sin^2\theta + \cos^2\theta = 1$.

We know that $\cos^2 \theta + \sin^2 \theta$ is equal to 1.

$\frac{1 + 2 \sin \theta \cos \theta - 1}{\cos \theta \sin \theta}$

Let's simplify the numerator.

$\frac{2 \sin \theta \cos \theta}{\cos \theta \sin \theta}$

We can cancel out $\sin \theta \cos \theta$ from top and bottom.

$\boxed{2}$

Step 3 — Simplify $(\sec A + \tan A) (1 - \sin A)$

Let's convert $\sec A$ and $\tan A$ into $\sin A$ and $\cos A$.

$(\frac{1}{\cos A} + \frac{\sin A}{\cos A}) (1 - \sin A)$

Now, we combine the terms in the first bracket.

$(\frac{1 + \sin A}{\cos A}) (1 - \sin A)$

Let's multiply the numerators. This is in the form $(a+b)(a-b) = a^2 - b^2$.

$\frac{1^2 - \sin^2 A}{\cos A}$

Pythagorean Identity: $\sin^2 A + \cos^2 A = 1 \Rightarrow 1 - \sin^2 A = \cos^2 A$.

We know that $1 - \sin^2 A$ is equal to $\cos^2 A$.

$\frac{\cos^2 A}{\cos A}$

We can cancel one $\cos A$ from the numerator and denominator.

$\boxed{\cos A}$

Step 4 — Simplify $\frac{1 + \tan^2 A}{1 + \cot^2 A}$

Pythagorean Identities: $1 + \tan^2 A = \sec^2 A$ (divide $\sin^2+\cos^2=1$ by $\cos^2$) and $1 + \cot^2 A = \csc^2 A$ (divide by $\sin^2$).

We will use two important identities here. We know that $1 + \tan^2 A = \sec^2 A$. Also, we know that $1 + \cot^2 A = \text{cosec}^2 A$.

$\frac{\sec^2 A}{\text{cosec}^2 A}$

Now, let's rewrite $\sec^2 A$ and $\text{cosec}^2 A$ using $\sin A$ and $\cos A$. $\sec^2 A = \frac{1}{\cos^2 A}$ and $\text{cosec}^2 A = \frac{1}{\sin^2 A}$.

$\frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}}$

To divide fractions, we multiply by the reciprocal.

$\frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1}$

$\frac{\sin^2 A}{\cos^2 A}$

We know that $\frac{\sin A}{\cos A}$ is equal to $\tan A$.

$\boxed{\tan^2 A}$

Answer

(i) The correct option is (B). (ii) The correct option is (C). (iii) The correct option is (D). (iv) The correct option is (D).