Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 21 Trigonometric Identities

Exercise 21.1

Answer 5.
Answer:
(i) \( (\sec\theta - \tan\theta)^2 = \frac{1 - \sin\theta}{1 + \sin\theta} \)

\( (\sec\theta - \tan\theta)^2 \)
\[ = \left(\frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta}\right)^2 \]
\[ = \left(\frac{1 - \sin\theta}{\cos\theta}\right)^2 = \frac{(1 - \sin\theta)^2}{\cos^2\theta} \]
\[ = \frac{(1 - \sin\theta)^2}{(1 - \sin\theta)(1 + \sin\theta)} \] (∵ \( 1 - \sin^2\theta = \cos^2\theta \))
\[ = \frac{1 - \sin\theta}{1 + \sin\theta} \]

(ii) \( \frac{1}{\sin A - \cos A} - \frac{1}{\sin A + \cos A} = \frac{2\sin A}{1 - 2\cos^2 A} \)

\( \frac{1}{\sin A - \cos A} - \frac{1}{\sin A + \cos A} \)
\[ = \frac{\sin A + \cos A - \sin A + \cos A}{\sin^2 A - \cos^2 A} \]
\[ = \frac{2\sin A}{1 - \cos^2 A - \cos^2 A} = \frac{2\sin A}{1 - 2\cos^2 A} \]

(iii) \( \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2\sin^2 A - 1} \)

\( \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} \)
\[ = \frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{(\sin A + \cos A)(\sin A - \cos A)} \]
\[ = \frac{\sin^2 A + \cos^2 A + 2\sin A \cos A + \sin^2 A + \cos^2 A - 2\sin \cos A}{\sin^2 A - \cos^2 A} \]
\[ = \frac{2(\sin^2 A + \cos^2 A)}{\sin^2 A - \cos^2 A} \]
\[ = \frac{2}{\sin^2 A - \cos^2 A} \] (∵ \( \sin^2A + \cos^2A = 1 \))
\[ = \frac{2}{\sin^2 A - \cos^2 A} = \frac{2}{\sin^2 A - (1 - \sin^2 A)} \]
\[ = \frac{2}{2\sin^2 A - 1} \]

(iv) \( \tan^2 A - \tan^2 B = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} \)

L.H.S = \( \tan^2 A - \tan^2 B \)
\[ = \frac{\sin^2 A}{\cos^2 A} - \frac{\sin^2 B}{\cos^2 B} \]
\[ = \frac{\sin^2 A \cos^2 B - \cos^2 A \sin^2 B}{\cos^2 A \cos^2 B} \]
\[ = \frac{(1-\cos^2 A)\cos^2 B - \cos^2 A (1-\cos^2 B)}{\cos^2 A \cos^2 B} \]
\[ = \frac{\cos^2 B - \cos^2 A \cos^2 B - \cos^2 A + \cos^2 A \cos^2 B}{\cos^2 A \cos^2 B} \]
\[ = \frac{\cos^2 B - \cos^2 A}{\cos^2 A \cos^2 B} \]
\[ = \frac{(1-\sin^2 B) - (1-\sin^2 A)}{\cos^2 A \cos^2 B} \]
\[ = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} \]
Hence \( \tan^2 A - \tan^2 B = \frac{\cos^2 B - \cos^2 A}{\cos^2 A \cos^2 B} = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} \)

In simple words: These are identities that show different ways to write the same trigonometric expression. We use basic identities like \( \sin^2\theta + \cos^2\theta = 1 \) to prove them.

📝 Teacher's Note: Start with basic identities. Show students how to convert sec and tan to sin and cos first. Practice with simple examples before complex ones.

🎯 Exam Tip: Always write "L.H.S" and "R.H.S" clearly. Show each step. Use the identity \( \sin^2\theta + \cos^2\theta = 1 \) in most proofs.

(v) \( \frac{\cos A}{1 - \tan A} + \frac{\sin^2 A}{\sin A - \cos A} = \cos A + \sin A \)

Answer:
\( \frac{\cos A}{1 - \tan A} + \frac{\sin^2 A}{\sin A - \cos A} \)
\[ = \frac{\cos A}{\frac{\cos A - \sin A}{\cos A}} + \frac{\sin^2 A}{\sin A - \cos A} \]
\[ = \frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{\sin A - \cos A} \]
\[ = \frac{\cos^2 A}{\cos A - \sin A} - \frac{\sin^2 A}{\cos A - \sin A} \]
\[ = \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} = \frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos A - \sin A} \]
\[ = \cos A + \sin A \]

(vi) \( (1 - \tan^2 A) = \left(1 - \frac{1}{\tan^2 A}\right) = \frac{1}{\sin^2 A - \sin^2 A} \)

\( (1 - \tan^2 A) - \left(1 - \frac{1}{\tan^2 A}\right) \)
\[ = \left(1 - \frac{\sin^2 A}{\cos^2 A}\right) - \left(1 - \frac{1}{\sin^2 A}\right) \frac{\cos^2 A}{\sin^2 A} \]
\[ = \left(\frac{\cos^2 A - \sin^2 A}{\cos^2 A}\right) - \left(\frac{\cos^2 A - \sin^2 A}{\sin^2 A}\right) \]
\[ = \frac{1}{1 - \sin^2 A} - \frac{1}{\sin^2 A} \] (∵ \( \cos^2 A - \sin^2 A = 1 \))
\[ = \frac{\sin^2 A - 1 + \sin^2 A}{\sin^2 A (1 - \sin^2 A)} = \frac{1}{\sin^2 A - \sin^2 A} \]

(vii) \( \frac{\cos^2 A - \sin^2 A}{\cos A + \sin A} + \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} = 2 \)

\( \frac{\cos^2 A - \sin^2 A}{\cos A + \sin A} + \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} \)
\[ = \frac{(\cos^2 A - \sin^2 A)(\cos A - \sin A) + (\cos^2 A - \sin^2 A)(\cos A + \sin A)}{(\cos A + \sin A)(\cos A - \sin A)} \]
\[ = \frac{\cos^4 A - \cos^2 A \sin A - \sin^2 A \cos A + \sin^4 A + \cos^4 A + \cos^2 A \sin A - \sin^2 A \cos A - \sin^4 A}{\cos^2 A - \sin^2 A} \]
\[ = \frac{2(\cos^4 A - \sin^4 A)}{(\cos^2 A - \sin^2 A)} = \frac{2(\cos^2 A - \sin^2 A)(\cos^2 A + \sin^2 A)}{(\cos^2 A - \sin^2 A)} = 2(\cos^2 A + \sin^2 A) \]
\[ = 2 \] (∵ \( \cos^2 A + \sin^2 A = 1 \))
OR
\( \frac{\cos^2 A - \sin^2 A}{\cos A + \sin A} + \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} \)
\[ = \frac{(\cos A - \sin A)(\cos^2 A + \sin^2 A + \cos A \sin A)}{(\cos A + \sin A)} + \frac{(\cos A + \sin A)(\cos^2 A + \sin^2 A - \cos A \sin A)}{(\cos A - \sin A)} \]
\[ = (\cos A - \sin A) + (\cos A + \sin A) = 2\cos A \]
\[ = 2 \]

In simple words: We use factoring and the basic identity to simplify these expressions. The key is to look for common factors.

📝 Teacher's Note: Teach students to look for difference of squares patterns. Show how \( a^2 - b^2 = (a+b)(a-b) \) helps in factoring.

🎯 Exam Tip: When you see \( \cos^2A - \sin^2A \), try to factor it. Always use the identity \( \cos^2A + \sin^2A = 1 \) to simplify final answers.

(viii) \( \tan\theta - \frac{1}{\cos\theta} = \left(\tan\theta - \frac{1}{\cos\theta}\right)^2 = 2\frac{1 - \sin^2\theta}{1 + \sin^2\theta} \)

Answer:
\( \left(\tan\theta - \frac{1}{\cos\theta}\right) - \left(\tan\theta - \frac{1}{\cos\theta}\right)^2 \)
\[ = \left(\frac{\sin\theta}{\cos\theta} - \frac{1}{\cos\theta}\right) - \left(\frac{\sin\theta}{\cos\theta} - \frac{1}{\cos\theta}\right)^2 \]
\[ = \left(\frac{\sin\theta - 1}{\cos\theta}\right) - \left(\frac{\sin\theta - 1}{\cos\theta}\right)^2 \]
\[ = \frac{(\sin\theta - 1)^2 - (\sin\theta - 1)^2}{\cos^2\theta} \]
\[ = \frac{(\sin\theta - 1)^2 - (\sin\theta - 1)^2}{\cos^2\theta} \]
\[ = \frac{\sin^2\theta - 1 - 2\sin\theta + \sin^2\theta - 1 + 2\sin\theta}{1 - \sin^2\theta} \]
\[ = \frac{2(1 - \sin^2\theta)}{1 - \sin^2\theta} \]

(ix) \( \frac{\sin A - \sin B}{\cos A + \cos B} + \frac{\cos A - \cos B}{\sin A + \sin B} = 0 \)

L.H.S = \( \frac{\sin A + \sin B}{\cos A + \cos B} + \frac{\cos A - \cos B}{\sin A - \sin B} \)
\[ = \frac{(\sin A + \sin B)(\sin A - \sin B) + (\cos A + \cos B)(\cos A - \cos B)}{(\cos A + \cos B)(\sin A - \sin B)} \]
\[ = \frac{\sin^2 A - \sin^2 B + \cos^2 A - \cos^2 B}{(\cos A + \cos B)(\sin A - \sin B)} \]
\[ = \frac{(\sin^2 A + \cos^2 A) - (\sin^2 B + \cos^2 B)}{(\cos A + \cos B)(\sin A - \sin B)} \]
\[ = \frac{1 - 1}{(\cos A + \cos B)(\sin A - \sin B)} \]
\[ = \frac{0}{(\cos A + \cos B)(\sin A - \sin B)} \]

\( \frac{\sin A + \sin B}{\cos A + \cos B} + \frac{\cos A - \cos B}{\sin A - \sin B} = 0 \)

Hence proved.

In simple words: When we expand and simplify using the basic identity, everything cancels out to give zero.

📝 Teacher's Note: Show students how \( \sin^2A + \cos^2A = 1 \) makes the numerator become zero. This is a common pattern in trigonometry proofs.

🎯 Exam Tip: Look for opportunities to use \( \sin^2A + \cos^2A = 1 \). When proving something equals zero, try to get 1-1 in the numerator.

Answer 6.
Answer:
(i) \( (1 - \cot A)^2 - (1 + \cot A)^2 = 2\cos\text{ec}^2A \)

\( (1 - \cot A)^2 - (1 + \cot A)^2 \)
\[ = 1 - \cot^2 A - 2\cot A - 1 + \cot^2 A + 2\cot A \]
\[ = 2 - 2\cot^2 A = 2(1 - \cot^2 A) \]
\[ = 2\cos\text{ec}^2A \]

(ii) \( \frac{\cos\text{ec}\theta}{\tan\theta - \cot\theta} = \cos\theta \)

\( \frac{\cos\text{ec}\theta}{\tan\theta - \cot\theta} \)
\[ = \frac{\frac{1}{\sin\theta}}{\frac{\sin\theta}{\cos\theta} - \frac{\cos\theta}{\sin\theta}} \]
\[ = \frac{\frac{1}{\sin\theta}}{\frac{\sin^2\theta - \cos^2\theta}{\sin\theta \cos\theta}} = \frac{\frac{1}{\sin\theta}}{\frac{1}{\cos\theta \sin\theta}} \]
\[ = \frac{1}{\sin\theta} \times \frac{\cos\theta \sin\theta}{1} = \cos\theta \]

(iii) \( (1 - \tan^2\theta)\sin\theta \cos\theta = \tan\theta \)

\( (1 - \tan^2\theta)\sin\theta \cos\theta \)
\[ = \left(1 - \frac{\sin^2\theta}{\cos^2\theta}\right)\sin\theta \cos\theta \]
\[ = \left(\frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta}\right)\sin\theta \cos\theta \]
\[ = \frac{1}{\cos\theta} \times \sin\theta \cos\theta \] (∵ \( \cos^2\theta - \sin^2\theta = 1 \))
\[ = \frac{\sin\theta}{\cos\theta} = \tan\theta \]

In simple words: We substitute the definitions of trigonometric functions and use basic identities to simplify.

📝 Teacher's Note: Practice converting all functions to sin and cos first. Students find this easier than working with multiple function types.

🎯 Exam Tip: Always convert cot, tan, sec, cosec to sin and cos when solving. Write each substitution step clearly for full marks.

Question xi. \( \frac{\cos^4 \theta - \sin^4 \theta}{\cos \theta - \sin \theta} = \frac{\cos^4 \theta - \sin^4 \theta}{\cos \theta - \sin \theta} = 2 \)
Answer:
\( \frac{\cos^4 \theta - \sin^4 \theta}{\cos \theta - \sin \theta} = \frac{\cos^4 \theta - \sin^4 \theta}{\cos \theta - \sin \theta} \)

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

\( = \frac{\cos^2 \theta - \cos^2 \theta \sin^2 \theta - \sin^2 \theta \cos^2 \theta + \cos^2 \theta \cos^2 \theta \sin^2 \theta - \sin^2 \theta \cos^2 \theta - \sin^4 \theta}{\cos \theta - \sin \theta} \)

\( = \frac{2 \cos^2 \theta - 2 \sin^2 \theta}{2(\cos^2 \theta - \sin^2 \theta)} \cdot \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta - \sin^2 \theta} \)

\( = \frac{2(\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta - \sin^2 \theta)}{(\cos^2 \theta - \sin^2 \theta)} = 2(\cos^2 \theta - \sin^2 \theta) \)

\( = 2 \cdot \{ \because (\cos^2 \theta - \sin^2 \theta) = 1\} \)

OR

\( \frac{\cos^4 \theta - \sin^4 \theta}{\cos \theta - \sin \theta} = \frac{\cos^4 \theta - \sin^4 \theta}{\cos \theta - \sin \theta} \)

\( = \frac{(\cos \theta - \sin \theta)(\cos^3 \theta + \sin^2 \theta + \cos \theta \sin \theta)}{(\cos \theta - \sin \theta)} = \frac{(\cos \theta - \sin \theta)(\cos^2 \theta + \sin^2 \theta + \cos \theta \sin \theta)}{(\cos \theta - \sin \theta)} \)

\( \{ \because a^4 - b^4 = (a - b)(a^3 + b^2 + ab)\} \)

\( = (\cos^2 \theta + \sin^2 \theta + \cos \theta \sin \theta) - (\cos^2 \theta - \sin^2 \theta - \cos \theta \sin \theta) \)

\( = 1 - \cos \theta \sin \theta - 1 - \cos \theta \sin \theta \{ \because \cos^2 \theta + \sin^2 \theta = 1\} \)

\( = 2 \)
In simple words: We used the formula \( a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) \) to break down the numerator. Then we simplified using \( \cos^2 \theta + \sin^2 \theta = 1 \) to get 2.

📝 Teacher's Note: Teach students the difference of squares formula first. Then show how \( a^4 - b^4 \) can be factored step by step. Practice with simple numbers before using trigonometric functions.

🎯 Exam Tip: Always write the factorization step clearly. Use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \) at the right place. Show all steps to get full marks.

Question xii. \( \frac{\tan \theta - \sin \theta}{\tan \theta + \sin \theta} = \frac{\sec \theta - 1}{\sec \theta + 1} \)
Answer:
\( \frac{\tan \theta - \sin \theta}{\tan \theta + \sin \theta} \)

\( = \frac{\sin \theta - \sin \theta}{\cos \theta} \cdot \frac{\sin \theta + \sin \theta \cos \theta}{\sin \theta - \sin \theta \cos \theta} \)

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

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

\( = \frac{1}{\sec \theta} - \frac{\sec \theta}{\sec \theta - 1} = \frac{\sec \theta - 1}{\sec \theta + 1} \)

\( = \frac{1}{\sec \theta} - \frac{\sec \theta}{\sec \theta - 1} = \frac{\sec \theta - 1}{\sec \theta + 1} \)
In simple words: We wrote tan and sin in terms of sin and cos. Then we factored and simplified using the identity \( \sec \theta = \frac{1}{\cos \theta} \). The left side equals the right side.

📝 Teacher's Note: Start by writing \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Then factor out \( \sin \theta \) from numerator and denominator. Students often forget this step.

🎯 Exam Tip: Write \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) first. Factor carefully and use \( \sec \theta = \frac{1}{\cos \theta} \) at the end. Show the final form clearly.

Question xiii. \( \frac{1}{\sec^2 \theta + \cos^2 \theta} \cdot \frac{1}{\cos \sec \theta - \sin^2 \theta} \cdot (\sin^4 \cos^2 \theta) = \frac{1 - \sin^4 \theta \cos^2 \theta}{2 - \sin^4 \theta \cos^2 \theta} \)
Answer:
\( \frac{1}{\sec^2 \theta + \cos^2 \theta} \cdot \frac{1}{\cos \sec \theta - \sin^2 \theta} \)

\( = \frac{1}{\cos^2 \theta} - \frac{1}{\cos^2 \theta} \cdot (\sin^4 \cos^2 \theta) \)

\( = \frac{1}{\cos^2 \theta} - \frac{1 - \sin^4 \theta}{\sin^2 \theta} \cdot (\sin^4 \cos^2 \theta) \)

\( = \frac{1}{1 - \cos^2 \theta} \cdot \frac{1}{1 - \sin^4 \theta} \cdot (\sin^4 \cos^2 \theta) \)

\( = \frac{\cos^2 \theta}{\cos^2 \theta} \cdot \frac{\sin^2 \theta}{1 - \sin^4 \theta} \cdot (\sin^4 \cos^2 \theta) \)

\( = \frac{\cos^2 \theta - \cos^2 \theta \sin^4 \theta + \sin^2 \theta - \sin^2 \theta \cos^2 \theta}{(1 - \cos^2 \theta)(1 - \sin^4 \theta)} \cdot (\sin^4 \cos^2 \theta) \)

\( = \frac{\cos^2 \theta + \sin^2 \theta - \cos^2 \theta \sin^4 \theta - \sin^2 \theta \cos^2 \theta}{(1 - \cos^2 \theta)(1 - \sin^4 \theta)} \cdot (\sin^4 \cos^2 \theta) \)

\( = \frac{1 - \cos^2 \theta \sin^4 \theta}{(1 - \cos^2 \theta)(1 - \sin^4 \theta)} \cdot (\sin^4 \cos^2 \theta) \)

\( \{ \because \cos^2 \theta + \sin^2 \theta = 1, (1 - \cos^2 \theta) = \sin^2 \theta, (1 - \sin^4 \theta) = 1, \cos^2 \theta = \cos^2 \theta \} \)

\( = \frac{1 - \cos^2 \theta \sin^4 \theta}{(1 - \cos^2 \theta)(1 - \sin^4 \theta)} \)

\( = \frac{1 - \cos^2 \theta \sin^4 \theta}{1 - \cos^2 \theta - \cos^2 \theta + \sin^4 \theta \cos^2 \theta} \)

\( = \frac{1 - \sin^4 \theta \cos^2 \theta}{1 - \sin^4 \theta \cos^2 \theta} \)

\( = \frac{1 - \sin^4 \theta \cos^2 \theta}{2 - \sin^4 \theta \cos^2 \theta} \)
In simple words: This is a complex trigonometric identity proof. We simplified step by step using basic identities like \( \cos^2 \theta + \sin^2 \theta = 1 \) and \( \sec \theta = \frac{1}{\cos \theta} \).

📝 Teacher's Note: This problem has many steps. Break it down into smaller parts. Make sure students know \( \sec \theta = \frac{1}{\cos \theta} \) and \( \cos^2 \theta + \sin^2 \theta = 1 \) very well before attempting.

🎯 Exam Tip: Write each step clearly. Don't skip any algebraic manipulation. Use the fundamental identities at each step. Check your work by substituting simple values like \( \theta = 45° \).

Question xiv. \( \frac{\cot^2 \theta(\sec \theta - 1)}{(1 - \sin \theta)} = \sec^2 \theta \left(\frac{1 - \sin \theta}{1 + \sec \theta}\right) \)
Answer:
\( \frac{\cot^2 \theta(\sec \theta - 1)}{(1 - \sin \theta)} \)

\( = \frac{\cot^2 \theta(\sec \theta - 1)(1 - \sin \theta)(\sec \theta - 1)}{(1 - \sin \theta)(1 - \sin \theta)(\sec \theta - 1)} \)

\( = \frac{\cot^2 \theta(\sec \theta - 1)(\sec \theta - 1)(1 - \sin \theta)}{(1 - \sin \theta)(1 - \sin \theta)(\sec \theta - 1)} \)

\( = \frac{\cot^2 \theta(\sec^2 \theta - 1)(1 - \sin \theta)}{(1 - \sin^2 \theta)(1 - \sec \theta)} \)

\( = \frac{\cot^2 \theta(\tan^2 \theta)(1 - \sin \theta)}{(\cos^2 \theta)(1 - \sec \theta)} \)

\( \{ \because \tan^2 \theta = \sec^2 \theta - 1, 1 - \sin^2 \theta = \cos^2 \theta \} \)

\( = \frac{(\cot \theta \tan \theta)^2(1 - \sin \theta)}{(\cos^2 \theta)(1 - \sec \theta)} \)

\( = \frac{1(1 - \sin \theta)}{(\cos^2 \theta)(1 - \sec \theta)} \)

\( \{ \because \cot \theta \tan \theta = 1 \} \)

\( = \frac{1}{(\cos^2 \theta)(1 - \sec \theta)} \left(\frac{1 - \sin \theta}{1 - \sec \theta}\right) \)

\( = \sec^2 \theta \left(\frac{1 - \sin \theta}{1 - \sec \theta}\right) \)
In simple words: We used the identities \( \sec^2 \theta - 1 = \tan^2 \theta \) and \( \cot \theta \cdot \tan \theta = 1 \). Step by step, we simplified both sides to show they are equal.

📝 Teacher's Note: Remind students that \( \cot \theta \times \tan \theta = 1 \) always. Also teach them \( \sec^2 \theta - 1 = \tan^2 \theta \). These are key identities for solving such problems.

🎯 Exam Tip: Write the main identities you will use at the top. Use \( \sec^2 \theta - 1 = \tan^2 \theta \) and \( 1 - \sin^2 \theta = \cos^2 \theta \). Show every algebraic step clearly.