Introduction to Trigonometry Ex 8.4 - NCERT Solutions & Concepts

Understanding Trigonometric Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angles involved. In CBSE Class 10, we deal with acute angles where 0° ≤ A ≤ 90°. The entire Exercise 8.4 revolves around transforming complex algebraic expressions into simpler forms. To do this, we rely heavily on three fundamental Pythagorean identities. These identities are derived directly using Pythagoras' Theorem on a right-angled triangle. By dividing the theorem's equation by different sides (hypotenuse, base, or perpendicular), we get the three core formulas. Memorizing these in their alternative algebraic structures (e.g., recognizing that 1 - sin²A is cos²A) is the secret to solving identities quickly.

Core Formulas & Identities

First Identity

sin²A + cos²A = 1. (Alternative forms: sin²A = 1 - cos²A; cos²A = 1 - sin²A)

Second Identity

1 + tan²A = sec²A. (Alternative forms: sec²A - tan²A = 1; tan²A = sec²A - 1)

Third Identity

1 + cot²A = cosec²A. (Alternative forms: cosec²A - cot²A = 1; cot²A = cosec²A - 1)

Reciprocal Relations

cosec A = 1/sin A; sec A = 1/cos A; tan A = sin A/cos A; cot A = cos A/sin A

Step-by-Step Strategy to Prove Identities

1. Analyze and Choose a Side — Start with the more complex side (usually LHS) because it is easier to simplify a complicated expression than to construct a complex one from a simple term.
2. Convert to Sine and Cosine — If you are stuck and do not see an immediate identity, convert all terms (tan, cot, sec, cosec) into their basic sin and cos equivalents.
3. Perform Algebraic Operations — Take common denominators (LCM), expand using algebraic formulas like (a+b)² or (a²-b²), and group terms to spot familiar identities.
4. Apply Pythagorean Substitutions — Look for terms like 1 - sin²A or sec²A - 1 and substitute them with their single-term equivalents to collapse the equation.

High-Yield Worked Examples

• Example 1: Express the ratio cos A in terms of sin A. Step 1: Use the identity sin²A + cos²A = 1. Step 2: Isolate cos²A: cos²A = 1 - sin²A. Step 3: Take the square root: cos A = √(1 - sin²A) (taking the positive root for acute angle A).
• Example 2: Prove that sec A(1 - sin A)(sec A + tan A) = 1. Step 1: Convert all terms to sin and cos: sec A = 1/cos A, tan A = sin A/cos A. Step 2: Substitute: (1/cos A) (1 - sin A) (1/cos A + sin A/cos A). Step 3: Combine fractions: [ (1 - sin A) / cos A ] * [ (1 + sin A) / cos A ]. Step 4: Multiply numerators and denominators: (1 - sin A)(1 + sin A) / cos²A = (1 - sin²A) / cos²A. Step 5: Apply identity 1 - sin²A = cos²A: cos²A / cos²A = 1. LHS = RHS.

Common Mistakes to Avoid in CBSE Exams

1. Incorrect Algebraic Squares: Never write (sin A + cos A)² as sin²A + cos²A. It must be expanded as sin²A + cos²A + 2 sin A cos A, which simplifies to 1 + 2 sin A cos A.
2. Ignoring the Domain Limits: Remember that tan 90° and sec 90° are not defined, while cot 0° and cosec 0° are undefined. Keep track of these constraints during numerical steps.
3. Failing to Show Steps: CBSE marking schemes reward every step. Clearly write down which identity you are using in brackets on the right side of your proof sheet.

Practice Questions with Solutions

• Q: Prove the identity: (cosec θ - cot θ)² = (1 - cos θ) / (1 + cos θ) A: Step 1: Start with the LHS: (cosec θ - cot θ)² Step 2: Convert to sine and cosine: (1/sin θ - cos θ/sin θ)² Step 3: Combine inside the bracket: [ (1 - cos θ) / sin θ ]² Step 4: Distribute the square: (1 - cos θ)² / sin²θ Step 5: Substitute sin²θ = 1 - cos²θ: (1 - cos θ)² / (1 - cos²θ) Step 6: Factorize the denominator using (a² - b²) = (a - b)(a + b): (1 - cos θ)² / [ (1 - cos θ)(1 + cos θ) ] Step 7: Cancel the common factor (1 - cos θ) from numerator and denominator: (1 - cos θ) / (1 + cos θ) Final answer: LHS = RHS. Proved.
• Q: Prove: cos A / (1 + sin A) + (1 + sin A) / cos A = 2 sec A A: Step 1: Write down LHS: cos A / (1 + sin A) + (1 + sin A) / cos A Step 2: Take the LCM of denominators: [ cos²A + (1 + sin A)² ] / [ cos A (1 + sin A) ] Step 3: Expand the numerator: [ cos²A + 1 + sin²A + 2 sin A ] / [ cos A (1 + sin A) ] Step 4: Group sin²A + cos²A to get 1: [ 1 + 1 + 2 sin A ] / [ cos A (1 + sin A) ] Step 5: Simplify: [ 2 + 2 sin A ] / [ cos A (1 + sin A) ] = 2(1 + sin A) / [ cos A (1 + sin A) ] Step 6: Cancel out (1 + sin A): 2 / cos A = 2 sec A Final answer: LHS = RHS. Proved.
• Q: Prove: √[ (1 + sin A) / (1 - sin A) ] = sec A + tan A A: Step 1: Start with LHS: √[ (1 + sin A) / (1 - sin A) ] Step 2: Rationalize the denominator inside the square root by multiplying numerator and denominator by (1 + sin A): √[ (1 + sin A)(1 + sin A) / (1 - sin A)(1 + sin A) ] Step 3: Simplify inside root: √[ (1 + sin A)² / (1 - sin²A) ] Step 4: Use identity 1 - sin²A = cos²A: √[ (1 + sin A)² / cos²A ] Step 5: Eliminate the square root: (1 + sin A) / cos A Step 6: Split the fraction: 1/cos A + sin A/cos A = sec A + tan A Final answer: LHS = RHS. Proved.
• Q: Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A. A: Step 1: To find sin A: We know cosec²A = 1 + cot²A. Therefore, cosec A = √(1 + cot²A). Since sin A = 1/cosec A, we get sin A = 1 / √(1 + cot²A). Step 2: To find tan A: Reciprocal relation gives tan A = 1/cot A. Step 3: To find sec A: Use identity sec²A = 1 + tan²A. Substitute tan A: sec²A = 1 + (1/cot A)² = 1 + 1/cot²A = (cot²A + 1)/cot²A. Taking square root: sec A = √(cot²A + 1) / cot A. Final answer: sin A = 1 / √(1 + cot²A), sec A = √(cot²A + 1) / cot A, tan A = 1/cot A.

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