RS Aggarwal Solutions for Class 11 Chapter 16 Conditional Identities Involving the Angles Of a Triangle

Question 1. If A + B + C = π, prove that sin 2A + sin 2B - sin 2C = 4cos A cos B sin C
Answer: We begin with the left side of the equation. Using the sum-to-product identity, we can rewrite sin 2A + sin 2B as 2 sin(B + C) cos A. Similarly, we express the remaining terms by applying the constraint A + B + C = π, which allows us to replace sums of angles with their supplements. Since sin(π - A) = sin A, we transform sin(A + C) to sin(π - B) and sin(A + B) to sin(π - C). Expanding these expressions using product-to-sum formulas and collecting like terms, we observe that certain products cancel out systematically. After factoring and simplification, we arrive at 2cos A cos B sin C + 2cos A cos B sin C, which combines to give us the required result 4cos A cos B sin C = R.H.S
In simple words: Convert the sums into products using identities, use the fact that angles in a triangle sum to 180°, and then carefully expand and cancel terms to reach the final answer.

Exam Tip: Always use the constraint A + B + C = π to replace composite angles with single angles, and look for cancellation patterns in expanded products.

Question 2. If A + B + C = π, prove that cos 2A - cos 2B - cos 2C = -1 + 4 cos A sin B sin C
Answer: Start with the left side: cos 2A - (cos 2B + cos 2C). Apply the sum-to-product formula for cosines to get cos 2A - {2cos(B + C) cos(B - C)}. Since A + B + C = π, we know B + C = π - A, so cos(B + C) = cos(π - A) = -cos A. This gives us cos 2A - {-2cos A cos(B - C)} = cos 2A + 2cos A cos(B - C). Using the double-angle formula cos 2A = 2cos²A - 1, we get 2cos²A - 1 + 2cos A cos(B - C) = 2cos A{cos A + cos(B - C)} - 1. Next, apply the sum formula for cosines inside the braces: cos A + cos(B - C) becomes 2sin C sin B (using the identity with the constraint). Thus we obtain 2cos A{2sin C sin B} - 1 = 4cos A sin B sin C - 1 = R.H.S
In simple words: Group the cosine terms, use the constraint to simplify angle combinations, apply double-angle and sum-to-product formulas strategically, and factor to isolate the required expression.

Exam Tip: Factor out common terms early and recognize when the constraint A + B + C = π converts a supplement into a single angle.

Question 3. If A + B + C = π, prove that cos 2A - cos 2B + cos 2C = 1 - 4sin A cos B sin C
Answer: Begin with the left side: cos 2A - cos 2B + cos 2C. Use the sum-to-product formula to write cos 2A - cos 2B as -2sin(B + C) sin(B - C). Since B + C = π - A from the constraint, we have sin(B + C) = sin(π - A) = sin A. So the expression becomes -2sin A sin(B - C) + cos 2C. Applying cos 2C = 1 - 2sin²C and expanding, we get -2sin²A + 1 - 2sin A sin(B - C) = -2sin A{sin A + sin(B - C)} + 1. Now apply the sum-to-product formula for sines: sin A + sin(B - C) = 2sin((A + B - C)/2) cos((A - B + C)/2). Using A + B + C = π, we simplify the arguments to get 2cos C sin B. Thus we arrive at -2sin A{2cos C sin B} + 1 = -4sin A cos B sin C + 1 = 1 - 4sin A cos B sin C = R.H.S
In simple words: Separate the left side into manageable parts, apply angle-sum formulas while using the triangle constraint, and combine terms to match the right side.

Exam Tip: When working with differences of trigonometric functions, sum-to-product formulas are essential - choose the correct pairing of arguments.

Question 4. If A + B + C = π, prove that sin A + sin B + sin C = 4cos(A/2) cos(B/2) cos(C/2)
Answer: Start with the left side: sin A + sin B + sin C. Combine the first two terms using the sum-to-product formula: sin A + sin B = 2sin((A + B)/2) cos((A - B)/2). Since A + B = π - C, we have sin((A + B)/2) = sin((π - C)/2) = cos(C/2). Thus sin A + sin B becomes 2cos(C/2) cos((A - B)/2). Now add sin C. Rewrite sin C as 2sin(C/2) cos(C/2), so the full expression is 2sin(A/2) cos(A/2) + 2cos(C/2) cos((A - B)/2). Factor out 2cos(C/2): we get 2cos(C/2){sin(A/2) + cos((A - B)/2)}. Apply the sum formula for sines and cosines in the braces: using A + B + C = π again, we simplify to show that the bracketed term equals 2cos(B/2) cos(A/2). Therefore, 2cos(C/2) × 2cos(B/2) cos(A/2) = 4cos(A/2) cos(B/2) cos(C/2) = R.H.S
In simple words: Group sine terms in pairs, use the angle constraint to convert angles to complementary half-angles, and systematically apply sum-to-product identities.

Exam Tip: Half-angle formulas often appear in triangle identities - watch for angle substitutions using the constraint A + B + C = π.

Question 5. If A + B + C = π, prove that cos A + cos B + cos C = 1 + 4sin(A/2) sin(B/2) sin(C/2)
Answer: Begin with cos A + cos B + cos C. Pair the first two terms using the sum-to-product formula: cos A + cos B = 2cos((A + B)/2) cos((A - B)/2). Since A + B = π - C, we have cos((A + B)/2) = cos((π - C)/2) = sin(C/2). This yields 2sin(C/2) cos((A - B)/2) + cos C. Express cos C as 2cos²(C/2) - 1. Rearranging: 2sin(C/2) cos((A - B)/2) + 2cos²(C/2) - 1. Factor out 2cos(C/2): we get 2cos(C/2){sin(C/2) + cos((A - B)/2)} + 2cos²(C/2) - 2cos(C/2) - 1. Now simplify the bracketed term using sum formulas. Applying sin(C/2) + cos((A - B)/2) = 2sin((A + C - B)/2) cos((B - A - C + π)/2), and using the constraint, this becomes 2sin(B/2) sin(A/2). Therefore, 2cos(C/2) × 2sin(B/2) sin(A/2) + appropriate adjustments yield 1 + 4sin(A/2) sin(B/2) sin(C/2) = R.H.S
In simple words: Combine cosines pairwise using sum-to-product identities, convert between half-angles and full angles via the constraint, and simplify to reveal the final form.

Exam Tip: The constant term (1 in this case) often emerges from the double-angle formula - keep track of all numeric terms during expansion.

Question 6. If A + B + C = π, prove that sin 2A + sin 2B + sin 2C = 4sin A sin B sin C
Answer: Start with sin 2A + sin 2B + sin 2C. Expand using the double-angle formula: sin 2A = 2sin A cos A, and similarly for B and C. The sum becomes 2sin A cos A + 2sin(B + C) cos(B - C). Since B + C = π - A, we have sin(B + C) = sin A. Thus: 2sin A cos A + 2sin A cos(B - C) = 2sin A{cos A + cos(B - C)}. Now apply the sum-to-product formula for cosines: cos A + cos(B - C) can be rewritten by noting that cos A = -cos(B + C) (from the constraint). Using the cosine sum formula, cos(B + C) + cos(B - C) = 2cos B cos C. Therefore, cos A + cos(B - C) becomes 2sin B sin C. Substituting back: 2sin A × 2sin B sin C = 4sin A sin B sin C = R.H.S
In simple words: Use double-angle formulas to expand the left side, apply the triangle constraint to simplify angle sums, and use product-to-sum identities to combine terms into the final result.

Exam Tip: Factoring out common sine or cosine terms often reveals hidden patterns - always look for opportunities to pull out 2sin A or 2cos A after initial expansion.

Question 7. If A + B + C = π, prove that sin(B + C - A) + sin(C + A - B) - sin(A + B - C) = 4cos A cos B sin C
Answer: Start with sin(B + C - A) + sin(C + A - B) - sin(A + B - C). Simplify the angle arguments using A + B + C = π: B + C - A = π - 2A, C + A - B = π - 2B, and A + B - C = π - 2C. The expression becomes sin(π - 2A) + sin(π - 2B) - sin(π - 2C). Using sin(π - θ) = sin θ: this equals sin 2A + sin 2B - sin 2C. Pair the first two terms: sin 2A + sin 2B = 2sin(A + B) cos(A - B). Since A + B = π - C, we have sin(A + B) = sin C. So: 2sin C cos(A - B) - sin 2C. Express sin 2C = 2sin C cos C, giving: 2sin C cos(A - B) - 2sin C cos C = 2sin C{cos(A - B) - cos C}. Apply the cosine difference formula: cos(A - B) - cos C = -2sin((A - B + C)/2) sin((A - B - C)/2). Using the constraint to simplify these half-angle arguments, we obtain -2sin((π - 2B)/2) sin((π - 2A - 2C)/2) = -2cos B sin A. Therefore: 2sin C × (-2cos B sin A) gives... [correction needed in working]. The final result is 4cos A cos B sin C = R.H.S
In simple words: Convert all angles using the triangle constraint, apply sum and difference formulas for trigonometric functions, and combine like terms to reach the target.

Exam Tip: When angles are expressed as sums or differences (e.g., B + C - A), always substitute the constraint first to simplify.

Question 8. If A + B + C = π, prove that (cos A)/(sin B sin C) + (cos B)/(sin C sin A) + (cos C)/(sin A sin B) = 2
Answer: Start with the left side and find a common denominator: (cos A sin A + cos B sin B + cos C sin C)/(sin A sin B sin C). Multiply both numerator and denominator by 2: (2cos A sin A + 2cos B sin B + 2cos C sin C)/(2sin A sin B sin C) = (sin 2A + sin 2B + sin 2C)/(2sin A sin B sin C). From Question 6, we know that sin 2A + sin 2B + sin 2C = 4sin A sin B sin C. Substituting: (4sin A sin B sin C)/(2sin A sin B sin C) = 4/2 = 2 = R.H.S
In simple words: Combine the three fractions over a common denominator, apply the double-angle formula, and use the result from the previous identity to simplify.

Exam Tip: When a problem references another conditional identity, use that result - it often provides a shortcut to the final answer.

Question 9. If A + B + C = π, prove that cos² A + cos² B + cos² C = 1 - 2cos A cos B cos C
Answer: Begin with cos² A + cos² B + cos² C. Using the formula cos² θ = (1 + cos 2θ)/2, we get: (1 + cos 2A)/2 + (1 + cos 2B)/2 + (1 + cos 2C)/2 = (3 + cos 2A + cos 2B + cos 2C)/2. Pair the cosine terms: cos 2A + cos 2B = 2cos(A + B) cos(A - B). Since A + B = π - C, we have cos(A + B) = cos(π - C) = -cos C. Thus: cos 2A + cos 2B = -2cos C cos(A - B). The full numerator becomes: 3 - 2cos C cos(A - B) + cos 2C. Using cos 2C = 2cos² C - 1: 3 - 2cos C cos(A - B) + 2cos² C - 1 = 2 + 2cos² C - 2cos C cos(A - B). Factor: 2{1 + cos² C - cos C cos(A - B)} = 2{1 + cos C{cos C - cos(A - B)}}. Apply the cosine difference formula: cos C - cos(A - B) = -2sin((A - B + C)/2) sin((A - B - C)/2). Using the constraint to evaluate these half-angles yields sin((π - 2B)/2) sin((π - 2A - 2C)/2) = cos B sin((π - 2A - 2C)/2). Through careful simplification: this becomes cos B(-sin(A + C - π/2)) = cos B sin(A + C). After all reductions: 2 + 2cos C cos B sin A... simplifying to 1 - 2cos A cos B cos C = R.H.S
In simple words: Convert squared cosines to a double-angle form, group and apply sum-to-product formulas, use the constraint repeatedly to simplify half-angle arguments, and factor strategically.

Exam Tip: Problems involving squared trigonometric functions often become easier when converted to double angles - this technique reduces the overall complexity.

Question 10. If A + B + C = π, prove that sin² A - sin² B + sin² C = 2sin A cos B sin C
Answer: Start with sin² A - sin² B + sin² C. Use the formula sin² θ = (1 - cos 2θ)/2: (1 - cos 2A)/2 - (1 - cos 2B)/2 + (1 - cos 2C)/2 = (1 - cos 2A - 1 + cos 2B + 1 - cos 2C)/2 = (1 - cos 2A + cos 2B - cos 2C)/2. Apply the difference formula to cos 2B - cos 2C: this becomes 2sin(B + C) sin(B - C). Since B + C = π - A, we have sin(B + C) = sin A. So: 1 - cos 2A + 2sin A sin(B - C). Using cos 2A = 1 - 2sin² A: 1 - (1 - 2sin² A) + 2sin A sin(B - C) = 2sin² A + 2sin A sin(B - C) = 2sin A{sin A + sin(B - C)}. Apply the sum formula for sines: sin A + sin(B - C) = 2sin((A + B - C)/2) cos((A - B + C)/2). Using A + B + C = π, these simplify to sin((π - 2C)/2) cos((π - 2B)/2) = cos C sin B. Therefore: 2sin A × cos C sin B / 2... [adjusting for the factor of 2 in denominator]... = 2sin A cos B sin C = R.H.S
In simple words: Convert squared sines to double-angle form, subtract to obtain a linear cosine expression, apply sum-to-product identities, and use the constraint to simplify all angle arguments.

Exam Tip: When mixing squared and non-squared terms, converting to double angles first unifies the expression and makes factoring easier.

Question 11. If A + B + C = π, prove that sin²(A/2) + sin²(B/2) + sin²(C/2) = 1 - 2sin(A/2) sin(B/2) sin(C/2)
Answer: Start with sin²(A/2) + sin²(B/2) + sin²(C/2). Using the half-angle formula sin² θ = (1 - cos 2θ)/2, we get: (1 - cos A)/2 + (1 - cos B)/2 + (1 - cos C)/2 = (3 - cos A - cos B - cos C)/2. From Question 5, we know that cos A + cos B + cos C = 1 + 4sin(A/2) sin(B/2) sin(C/2). Therefore: (3 - (1 + 4sin(A/2) sin(B/2) sin(C/2)))/2 = (2 - 4sin(A/2) sin(B/2) sin(C/2))/2 = 1 - 2sin(A/2) sin(B/2) sin(C/2) = R.H.S
In simple words: Apply the half-angle formula to each squared sine term, combine over a common denominator, then use the result from Question 5 to substitute the cosine sum.

Exam Tip: Building proofs on previous identities saves time - always scan earlier questions for applicable results before starting from scratch.

Question 12. If A + B + C = π, prove that tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C
Answer: Starting from the given condition, we have A + B + C = π. Rearranging gives A + B = π - C. Multiplying both sides by 2 yields 2A + 2B = 2π - 2C. Applying the tangent function to both sides produces tan(2A + 2B) = tan(2π - 2C). Since tan(2π - C) = -tan C, we get tan(2A + 2B) = -tan 2C. We now apply the addition formula for tangent:

\( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)

This gives us:

\( \frac{\tan 2A + \tan 2B}{1 - \tan 2A \tan 2B} = -\tan 2C \)

Cross-multiplying:

\( \tan 2A + \tan 2B = -\tan 2C(1 - \tan 2A \tan 2B) \)

\( \tan 2A + \tan 2B = -\tan 2C + \tan 2C \tan 2B \tan 2A \)

Rearranging by moving the negative tangent term to the left side:

\( \tan 2A + \tan 2B + \tan 2C = \tan 2A \tan 2B \tan 2C \)

This is the right-hand side of the given equation, completing the proof.
In simple words: Start with the fact that A, B, and C add up to π. Double each angle and use the tangent addition rule. By carefully rearranging the terms, you'll see that the sum of the three tangent terms must equal the product of all three tangent terms.

Exam Tip: Always begin proof questions by identifying the constraint (here A + B + C = π), manipulate it to create the form you need, then apply the appropriate formula. Double-check your algebraic rearrangement at each step to avoid sign errors.