Choose the correct option. Justify your choice. (i) (A) (B) (C) (D) (ii) (A) (B) (C) (D) (iii) (A) (B) (C) (D) (iv) (A) (B) (C) (D)

Understand the Question We simplify each trigonometric expression using standard Pythagorean and reciprocal identities: Converting trigonometric functions into terms of and is a reliable strategy when identities are not immediately obvious. Part (i) (A) (B) (C) (D) Step 1 · Factor and Apply Pythagorean Identity Factor out and use the identity : Answer > (i) (B) Part (ii) (A) (B) (C) (D) Step 1 · Convert to Sine and Cosine Rewrite all terms in terms of and : Step 2 · Simplify Using Algebraic and Trigonometric Identities Using difference of squares where and : Answer > (ii) (C) Part (iii) (A) (B) (C) (D) Step 1 · Convert to Sine and Cosine and Simplify Express and in terms of and : Answer > (iii) (D) Part (iv) (A) (B) (C) (D) Step 1 · Apply Pythagorean Identities and Simplify Using and : Answer > (iv) (D) Common Mistakes Sign Error in Identities: Misremembering as instead of . Remember . Algebraic Expansion in Part (ii): Failing to group as a single term when applying the difference of squares identity . Reciprocal Division in Part (iv): Incorrectly inverting fractions when simplifying .

Introduction to Trigonometry | Exercise 8.3

Question 3

Choose the correct option. Justify your choice.

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

(A) $1$ (B) $9$ (C) $8$ (D) $0$

(ii) $(1 + \tan \theta + \sec \theta) (1 + \cot \theta - \csc \theta) =$

(A) $0$ (B) $1$ (C) $2$ (D) $-1$

(iii) $(\sec A + \tan A) (1 - \sin A) =$

(A) $\sec A$ (B) $\sin A$ (C) $\csc A$ (D) $\cos A$

(iv) $\dfrac{1 + \tan^2 A}{1 + \cot^2 A} =$

(A) $\sec^2 A$ (B) $-1$ (C) $\cot^2 A$ (D) $\tan^2 A$

Solution

Understand the Question

We simplify each trigonometric expression using standard Pythagorean and reciprocal identities:
- $\sin^2 A + \cos^2 A = 1 \implies 1 - \sin^2 A = \cos^2 A$
- $1 + \tan^2 A = \sec^2 A \implies \sec^2 A - \tan^2 A = 1$
- $1 + \cot^2 A = \csc^2 A$
Converting trigonometric functions into terms of $\sin$ and $\cos$ is a reliable strategy when identities are not immediately obvious.

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

(A) $1$ (B) $9$ (C) $8$ (D) $0$ **

Step 1 · Factor and Apply Pythagorean Identity

Factor out $9$ and use the identity $\sec^2 A - \tan^2 A = 1$ :

\begin{aligned} 9 \sec^2 A - 9 \tan^2 A &= 9 (\sec^2 A - \tan^2 A) \\[0.6em] &= 9(1) \\[0.6em] &= 9 \end{aligned}

Answer

(i) (B) $9$

**(ii) $(1 + \tan \theta + \sec \theta) (1 + \cot \theta - \csc \theta) =$

(A) $0$ (B) $1$ (C) $2$ (D) $-1$ **

Step 1 · Convert to Sine and Cosine

Rewrite all terms in terms of $\sin \theta$ and $\cos \theta$ :

\begin{aligned} (1 + \tan \theta + \sec \theta) (1 + \cot \theta - \csc \theta) &= \left(1 + \dfrac{\sin \theta}{\cos \theta} + \dfrac{1}{\cos \theta}\right) \left(1 + \dfrac{\cos \theta}{\sin \theta} - \dfrac{1}{\sin \theta}\right) \\[0.6em] &= \left(\dfrac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \left(\dfrac{\sin \theta + \cos \theta - 1}{\sin \theta}\right) \end{aligned}

Step 2 · Simplify Using Algebraic and Trigonometric Identities

Using difference of squares $(a+b)(a-b) = a^2 - b^2$ where $a = (\cos \theta + \sin \theta)$ and $b = 1$ :

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

Answer

(ii) (C) $2$

**(iii) $(\sec A + \tan A) (1 - \sin A) =$

(A) $\sec A$ (B) $\sin A$ (C) $\csc A$ (D) $\cos A$ **

Step 1 · Convert to Sine and Cosine and Simplify

Express $\sec A$ and $\tan A$ in terms of $\sin A$ and $\cos A$ :

\begin{aligned} (\sec A + \tan A) (1 - \sin A) &= \left(\dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A}\right) (1 - \sin A) \\[0.6em] &= \left(\dfrac{1 + \sin A}{\cos A}\right) (1 - \sin A) \\[0.6em] &= \dfrac{1^2 - \sin^2 A}{\cos A} \\[0.6em] &= \dfrac{\cos^2 A}{\cos A} \\[0.6em] &= \cos A \end{aligned}

Answer

(iii) (D) $\cos A$

**(iv) $\dfrac{1 + \tan^2 A}{1 + \cot^2 A} =$

(A) $\sec^2 A$ (B) $-1$ (C) $\cot^2 A$ (D) $\tan^2 A$ **

Step 1 · Apply Pythagorean Identities and Simplify

Using $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \csc^2 A$ :

\begin{aligned} \dfrac{1 + \tan^2 A}{1 + \cot^2 A} &= \dfrac{\sec^2 A}{\csc^2 A} \\[0.6em] &= \dfrac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} \\[1.1em] &= \dfrac{1}{\cos^2 A} \times \dfrac{\sin^2 A}{1} \\[0.6em] &= \dfrac{\sin^2 A}{\cos^2 A} \\[0.6em] &= \tan^2 A \end{aligned}

Answer

(iv) (D) $\tan^2 A$

Common Mistakes

Sign Error in Identities: Misremembering $\tan^2 A - \sec^2 A$ as $1$ instead of $-1$ . Remember $\sec^2 A - \tan^2 A = 1$ .
Algebraic Expansion in Part (ii): Failing to group $(\cos \theta + \sin \theta)$ as a single term when applying the difference of squares identity $(a+b)(a-b) = a^2 - b^2$ .
Reciprocal Division in Part (iv): Incorrectly inverting fractions when simplifying $\dfrac{\sec^2 A}{\csc^2 A} = \dfrac{1/\cos^2 A}{1/\sin^2 A}$ .

Question 3

More questions in Exercise 8.3