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📖 Trigonometry📚 Class 10🎯 Hard🏷️ CBSE Class 10🏷️ ICSE Class 10

Q: Prove: (cosecA – sinA)(secA – cosA) = 1/(tanA + cotA)

✅ Step-by-Step Solution

1LHS = (cosecA – sinA)(secA – cosA)

2cosecA = 1/sinA, secA = 1/cosA → (1/sinA – sinA)(1/cosA – cosA)

3Combine: ((1 – sin²A)/sinA)((1 – cos²A)/cosA)

4Identity: 1 – sin²A = cos²A, 1 – cos²A = sin²A

5LHS = (cos²A/sinA)(sin²A/cosA) = sinA·cosA

6RHS = 1/(tanA + cotA) = 1/(sinA/cosA + cosA/sinA)

7Combine: 1/((sin²A + cos²A)/(sinA·cosA))

8Identity: sin²A + cos²A = 1 → RHS = 1/(1/(sinA·cosA)) = sinA·cosA

9LHS = RHS = sinA·cosA ✅ Hence proved

✅ Answer: Hence proved

🎥 Video Solution

Trigonometry Class 10 🔥 Prove (cosecA – sinA)(secA – cosA) = 1/(tanA + cotA) | ExamClever

#trigonometry#class-10-maths#cbse#trigonometric-identities

📝 Key Formulas Used

Trigonometric Identities Used:

• sin²A + cos²A = 1 (Pythagorean identity)

• cosecA = 1/sinA (Reciprocal)

• secA = 1/cosA (Reciprocal)

• tanA = sinA/cosA (Ratio)

• cotA = cosA/sinA (Ratio)

⚠️ Common Mistakes Students Make

Wrong sign in LHS expansion — 1/sinA – sinA = (1 – sin²A)/sinA, not (1 – sinA)/sinA
Forgetting sin²A + cos²A = 1 — needed twice (once in LHS, once in RHS)
Doing LHS and RHS simultaneously — simplify each side separately and show they both equal sinA·cosA
Skipping the rationalisation step on RHS — 1/(sinA/cosA + cosA/sinA) needs a common denominator first

💡 Alternate Method

Start from RHS: 1/(tanA + cotA) = 1/(sinA/cosA + cosA/sinA) = sinA·cosA. Then show LHS also simplifies to sinA·cosA independently. Both meet at the same value.

Verification: A = 30°: LHS = sin30·cos30 = (1/2)(√3/2) = √3/4. RHS = 1/(1/√3 + √3) = 1/(4/√3) = √3/4 ✅

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Teacher's Note

This solution has been carefully prepared, but we recommend showing it to your Maths teacher once — marking schemes can vary slightly between boards (CBSE, ICSE, State Boards). Your teacher knows exactly what your examiner expects.

📝 Practice Questions (Try Yourself)

Q1. Prove: (1 + cot²A) / (1 + tan²A) = cot²A

Show Answer

LHS = cosec²A/sec²A = (1/sin²A)/(1/cos²A) = cos²A/sin²A = cot²A

Q2. Prove: (sinA – cosecA)² + (cosA – secA)² = tan²A + cot²A – 1

Show Answer

Expand: sin²A – 2 + cosec²A + cos²A – 2 + sec²A = 1 – 4 + (1+tan²A) + (1+cot²A) = tan²A + cot²A – 1

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