Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.7

Question 1. If A + B + C = 180° prove that
(i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
Answer:
We start by considering the left-hand side of the equation:
\( \sin 2A + \sin 2B + \sin 2C \)
Apply the sum-to-product formula for \( \sin 2A + \sin 2B \):
\( = 2 \sin \left(\frac{2A+2B}{2}\right) \cos \left(\frac{2A-2B}{2}\right) + \sin 2C \)
\( = 2 \sin (A+B) \cos (A-B) + \sin 2C \)
Since \( A+B+C = 180° \), we know \( A+B = 180° - C \). Substitute this into the expression:
\( = 2 \sin (180° - C) \cos (A-B) + \sin 2C \)
Because \( \sin (180° - C) = \sin C \), and using \( \sin 2C = 2 \sin C \cos C \):
\( = 2 \sin C \cos (A-B) + 2 \sin C \cos C \)
Now, factor out \( 2 \sin C \):
\( = 2 \sin C [\cos (A-B) + \cos C] \)
We again use \( C = 180° - (A+B) \), so \( \cos C = \cos (180° - (A+B)) = -\cos (A+B) \):
\( = 2 \sin C [\cos (A-B) - \cos (A+B)] \)
Apply the identity \( \cos X - \cos Y = 2 \sin \left(\frac{X+Y}{2}\right) \sin \left(\frac{Y-X}{2}\right) \), with \( X = A-B \) and \( Y = A+B \):
\( = 2 \sin C \left[2 \sin \left(\frac{(A-B)+(A+B)}{2}\right) \sin \left(\frac{(A+B)-(A-B)}{2}\right)\right] \)
\( = 2 \sin C [2 \sin A \sin B] \)
\( = 4 \sin A \sin B \sin C \)
Understanding sum-to-product and product-to-sum identities is key to simplifying these trigonometric expressions.
In simple words: We start by changing sums of sines into products. Because \( A \), \( B \), \( C \) add up to \( 180° \), we can replace \( A+B \) with \( 180° - C \). We then use a special rule for \( \cos(X) - \cos(Y) \) to simplify the expression further, leading us to the final answer.

🎯 Exam Tip: Remember the key trigonometric identities, especially sum-to-product formulas, and how to use the \( A+B+C = 180° \) condition effectively.

Question 1. (ii) cos A + cos B - cos C = -1 + 4 cos \( \frac{A}{2} \) cos \( \frac{B}{2} \) sin \( \frac{C}{2} \)
Answer:
We begin with the left-hand side of the equation:
\( \cos A + \cos B - \cos C \)
Apply the sum-to-product formula for \( \cos A + \cos B \) and the double angle formula for \( \cos C = 1 - 2 \sin^2 \frac{C}{2} \):
\( = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) - \left(1 - 2 \sin^2 \frac{C}{2}\right) \)
Since \( A+B+C = 180° \), we know \( \frac{A+B}{2} = 90° - \frac{C}{2} \).
So, \( \cos \left(\frac{A+B}{2}\right) = \cos \left(90° - \frac{C}{2}\right) = \sin \frac{C}{2} \).
Substitute this into the expression:
\( = 2 \sin \frac{C}{2} \cos \left(\frac{A-B}{2}\right) - 1 + 2 \sin^2 \frac{C}{2} \)
Rearrange the terms and factor out \( 2 \sin \frac{C}{2} \):
\( = -1 + 2 \sin \frac{C}{2} \left[\cos \left(\frac{A-B}{2}\right) + \sin \frac{C}{2}\right] \)
Again, use the relationship \( \sin \frac{C}{2} = \sin \left(90° - \left(\frac{A+B}{2}\right)\right) = \cos \left(\frac{A+B}{2}\right) \):
\( = -1 + 2 \sin \frac{C}{2} \left[\cos \left(\frac{A-B}{2}\right) + \cos \left(\frac{A+B}{2}\right)\right] \)
Apply the sum-to-product formula for \( \cos X + \cos Y \), with \( X = \frac{A-B}{2} \) and \( Y = \frac{A+B}{2} \):
\( = -1 + 2 \sin \frac{C}{2} \left[2 \cos \left(\frac{\frac{A-B}{2} + \frac{A+B}{2}}{2}\right) \cos \left(\frac{\frac{A+B}{2} - \frac{A-B}{2}}{2}\right)\right] \)
\( = -1 + 2 \sin \frac{C}{2} \left[2 \cos \frac{A}{2} \cos \frac{B}{2}\right] \)
\( = -1 + 4 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} \)
Trigonometric identities like sum-to-product and double angle formulas are essential tools for simplifying complex expressions and proving relationships.
In simple words: We start by changing \( \cos A + \cos B \) into a product and \( \cos C \) into a form with \( \sin^2(C/2) \). Then, using the fact that \( A \), \( B \), \( C \) add up to \( 180° \), we substitute and simplify terms. This involves changing angles and using another product formula to get the final answer.

🎯 Exam Tip: Be careful with signs when applying \( \cos C = 1 - 2 \sin^2(C/2) \) and \( \cos(90° - X) = \sin X \).

Question 1. (iii) sin 2A + sin 2B + sin 2C = 2 + 2 cos A cos B cos C
Answer:
We are asked to prove the given identity. Based on common trigonometric identities for a triangle where \( A+B+C = 180° \), this specific equality as stated is not a standard identity for \( \sin 2A + \sin 2B + \sin 2C \). However, the detailed solution provided in the source PDF works through the identity for \( \sin^2 A + \sin^2 B + \sin^2 C \). We will present the solution for \( \sin^2 A + \sin^2 B + \sin^2 C \) which proves to be equal to \( 2 + 2 \cos A \cos B \cos C \) under the condition \( A+B+C = 180° \).
Let's start with the expression for \( \sin^2 A + \sin^2 B + \sin^2 C \):
\( \sin^2 A + \sin^2 B + \sin^2 C \)
First, use the identity \( \sin^2 X = \frac{1 - \cos 2X}{2} \) for each term:
\( = \frac{1 - \cos 2A}{2} + \frac{1 - \cos 2B}{2} + \frac{1 - \cos 2C}{2} \)
\( = \frac{1}{2} [1 - \cos 2A + 1 - \cos 2B + 1 - \cos 2C] \)
\( = \frac{1}{2} [3 - (\cos 2A + \cos 2B + \cos 2C)] \)
Now, apply the sum-to-product formula for \( \cos 2A + \cos 2B \):
\( = \frac{1}{2} \left[3 - \left(2 \cos \left(\frac{2A+2B}{2}\right) \cos \left(\frac{2A-2B}{2}\right) + \cos 2C\right)\right] \)
\( = \frac{3}{2} - \cos (A+B) \cos (A-B) - \frac{1}{2} \cos 2C \)
Substitute \( \cos 2C = 2 \cos^2 C - 1 \):
\( = \frac{3}{2} - \cos (A+B) \cos (A-B) - \frac{1}{2} (2 \cos^2 C - 1) \)
\( = \frac{3}{2} - \cos (A+B) \cos (A-B) - \cos^2 C + \frac{1}{2} \)
Combine constant terms and substitute \( A+B = 180° - C \) so \( \cos (A+B) = \cos (180° - C) = -\cos C \):
\( = 2 - (-\cos C) \cos (A-B) - \cos^2 C \)
\( = 2 + \cos C \cos (A-B) - \cos^2 C \)
Factor out \( \cos C \):
\( = 2 + \cos C [\cos (A-B) - \cos C] \)
Replace \( \cos C \) with \( \cos (180° - (A+B)) = -\cos (A+B) \):
\( = 2 + \cos C [\cos (A-B) + \cos (A+B)] \)
Apply the sum-to-product formula for \( \cos X + \cos Y \), with \( X = A-B \) and \( Y = A+B \):
\( = 2 + \cos C \left[2 \cos \left(\frac{(A-B)+(A+B)}{2}\right) \cos \left(\frac{(A+B)-(A-B)}{2}\right)\right] \)
\( = 2 + \cos C [2 \cos A \cos B] \)
\( = 2 + 2 \cos A \cos B \cos C \)
Identities involving \( \sin^2 \) terms often simplify using \( \cos 2X \) formulas, making \( A+B+C = 180° \) substitutions crucial.
In simple words: We start by changing all \( \sin^2 \) terms using a double angle identity. Then, we group terms and apply sum-to-product formulas. We repeatedly use the fact that \( A \), \( B \), \( C \) add up to \( 180° \) to change angles and simplify the expression until we reach the right side.

🎯 Exam Tip: Carefully track the minus signs when expanding \( \cos 2X \) and using \( \cos(180°-X) \) identities.

Question 1. (iv) sin 2A + sin 2B - sin 2C = 2 sin A sin B cos C
Answer:
We need to simplify the expression \( \sin 2A + \sin 2B - \sin 2C \) given that \( A+B+C = 180° \).
Start with the left-hand side (LHS):
\( \sin 2A + \sin 2B - \sin 2C \)
Apply the sum-to-product formula for \( \sin 2A + \sin 2B \):
\( = 2 \sin \left(\frac{2A+2B}{2}\right) \cos \left(\frac{2A-2B}{2}\right) - \sin 2C \)
\( = 2 \sin (A+B) \cos (A-B) - \sin 2C \)
Since \( A+B+C = 180° \), we have \( A+B = 180° - C \). Substitute this:
\( = 2 \sin (180° - C) \cos (A-B) - \sin 2C \)
Because \( \sin (180° - C) = \sin C \), and using the double angle formula \( \sin 2C = 2 \sin C \cos C \):
\( = 2 \sin C \cos (A-B) - 2 \sin C \cos C \)
Factor out \( 2 \sin C \):
\( = 2 \sin C [\cos (A-B) - \cos C] \)
Replace \( \cos C \) using \( C = 180° - (A+B) \), so \( \cos C = \cos (180° - (A+B)) = -\cos (A+B) \):
\( = 2 \sin C [\cos (A-B) + \cos (A+B)] \)
Apply the sum-to-product formula for \( \cos X + \cos Y \), with \( X = A-B \) and \( Y = A+B \):
\( = 2 \sin C \left[2 \cos \left(\frac{(A-B)+(A+B)}{2}\right) \cos \left(\frac{(A+B)-(A-B)}{2}\right)\right] \)
\( = 2 \sin C [2 \cos A \cos B] \)
\( = 4 \sin C \cos A \cos B \)
This identity showcases how the sum of angles in a triangle helps simplify trigonometric expressions.
In simple words: We start by adding \( \sin 2A \) and \( \sin 2B \) using a special formula. Then, we use the fact that \( A + B + C \) equals \( 180° \) to change angles like \( A+B \) to \( 180°-C \). We also change \( \sin 2C \) using a double angle formula. We then factor out common terms and apply another cosine sum formula to get the final simplified expression for the left side.

🎯 Exam Tip: When simplifying sums or differences of sines, always look for opportunities to use sum-to-product identities and substitute \( A+B = 180°-C \) where possible.

Question 1. (v) tan \( \frac{A}{2} \) tan \( \frac{B}{2} \) + tan \( \frac{B}{2} \) tan \( \frac{C}{2} \) + tan \( \frac{C}{2} \) tan \( \frac{A}{2} \) = 1
Answer:
Given that \( A+B+C = 180° \).
Divide the entire equation by 2:
\( \frac{A+B+C}{2} = 90° \)
\( \implies \frac{A+B}{2} = 90° - \frac{C}{2} \)
Now, take the tangent of both sides:
\( \tan \left(\frac{A+B}{2}\right) = \tan \left(90° - \frac{C}{2}\right) \)
Using the identity \( \tan (90° - X) = \cot X \):
\( \tan \left(\frac{A+B}{2}\right) = \cot \frac{C}{2} \)
Expand \( \tan \left(\frac{A+B}{2}\right) \) using the sum formula for tangent, and replace \( \cot \frac{C}{2} \) with \( \frac{1}{\tan \frac{C}{2}} \):
\( \frac{\tan \frac{A}{2} + \tan \frac{B}{2}}{1 - \tan \frac{A}{2} \tan \frac{B}{2}} = \frac{1}{\tan \frac{C}{2}} \)
Cross-multiply to rearrange the terms:
\( \left(\tan \frac{A}{2} + \tan \frac{B}{2}\right) \tan \frac{C}{2} = 1 - \tan \frac{A}{2} \tan \frac{B}{2} \)
Distribute \( \tan \frac{C}{2} \) on the left side:
\( \tan \frac{A}{2} \tan \frac{C}{2} + \tan \frac{B}{2} \tan \frac{C}{2} = 1 - \tan \frac{A}{2} \tan \frac{B}{2} \)
Move the term \( \tan \frac{A}{2} \tan \frac{B}{2} \) to the left side to complete the proof:
\( \tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1 \)
This identity is a classic result for triangles, often used in geometric proofs involving angles.
In simple words: We start by dividing \( A + B + C = 180° \) by two, giving us a useful relationship between half angles. We then take the tangent of both sides and use the formula for \( \tan(90° - X) \). Expanding the \( \tan((A+B)/2) \) term and cross-multiplying helps us rearrange the equation to prove the identity.

🎯 Exam Tip: Remember the \( \tan(X+Y) \) formula and \( \tan(90°-X) = \cot X = 1/\tan X \) are crucial for this type of problem.

Question 1. (vi) sin A + sin B + sin C = 4 cos \( \frac{A}{2} \) cos \( \frac{B}{2} \) cos \( \frac{C}{2} \)
Answer:
Given that \( A+B+C = 180° \).
This implies \( \frac{A+B}{2} = 90° - \frac{C}{2} \).
Start with the left-hand side:
\( \sin A + \sin B + \sin C \)
Apply the sum-to-product formula for \( \sin A + \sin B \) and the double angle formula for \( \sin C = 2 \sin \frac{C}{2} \cos \frac{C}{2} \):
\( = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) + 2 \sin \frac{C}{2} \cos \frac{C}{2} \)
Substitute \( \frac{A+B}{2} = 90° - \frac{C}{2} \):
\( = 2 \sin \left(90° - \frac{C}{2}\right) \cos \left(\frac{A-B}{2}\right) + 2 \sin \frac{C}{2} \cos \frac{C}{2} \)
Since \( \sin (90° - X) = \cos X \):
\( = 2 \cos \frac{C}{2} \cos \left(\frac{A-B}{2}\right) + 2 \sin \frac{C}{2} \cos \frac{C}{2} \)
Factor out \( 2 \cos \frac{C}{2} \):
\( = 2 \cos \frac{C}{2} \left[\cos \left(\frac{A-B}{2}\right) + \sin \frac{C}{2}\right] \)
Again, use \( \sin \frac{C}{2} = \sin \left(90° - \left(\frac{A+B}{2}\right)\right) = \cos \left(\frac{A+B}{2}\right) \):
\( = 2 \cos \frac{C}{2} \left[\cos \left(\frac{A-B}{2}\right) + \cos \left(\frac{A+B}{2}\right)\right] \)
Apply the sum-to-product formula for \( \cos X + \cos Y \), with \( X = \frac{A-B}{2} \) and \( Y = \frac{A+B}{2} \):
\( = 2 \cos \frac{C}{2} \left[2 \cos \left(\frac{\frac{A-B}{2} + \frac{A+B}{2}}{2}\right) \cos \left(\frac{\frac{A+B}{2} - \frac{A-B}{2}}{2}\right)\right] \)
\( = 2 \cos \frac{C}{2} \left[2 \cos \frac{A}{2} \cos \frac{B}{2}\right] \)
\( = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \)
This identity, known as a sum-to-product identity for triangles, is widely used in competitive mathematics.
In simple words: We start by combining \( \sin A + \sin B \) using a sum formula. We also change \( \sin C \) using its double angle identity. Then, using the fact that \( A \), \( B \), \( C \) add up to \( 180° \), we replace \( (A+B)/2 \) with \( 90° - C/2 \). We continue to simplify by factoring and applying another sum formula for cosines until we reach the desired right-hand side.

🎯 Exam Tip: Be mindful of using the correct half-angle identities and how \( \sin(90°-X) = \cos X \) helps simplify expressions.

Question 1. (vii) sin(B + C – A) + sin(C + A – B) + sin(A + B – C) = 4 sin A sin B sin C
Answer:
Given that \( A+B+C = 180° \).
We can simplify each term in the left-hand side:
\( B+C-A = (A+B+C) - 2A = 180° - 2A \)
So, \( \sin(B+C-A) = \sin(180° - 2A) = \sin 2A \)

\( C+A-B = (A+B+C) - 2B = 180° - 2B \)
So, \( \sin(C+A-B) = \sin(180° - 2B) = \sin 2B \)

\( A+B-C = (A+B+C) - 2C = 180° - 2C \)
So, \( \sin(A+B-C) = \sin(180° - 2C) = \sin 2C \)
Now, substitute these simplified forms back into the left-hand side:
\( \text{LHS} = \sin 2A + \sin 2B + \sin 2C \)
From part (i), we know that if \( A+B+C = 180° \), then \( \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \).
Therefore,
\( \text{LHS} = 4 \sin A \sin B \sin C \)
\( \text{LHS} = \text{RHS} \)
Rewriting angles in terms of \( 180° \) simplifies complex trigonometric expressions into more manageable forms.
In simple words: First, we change each angle in the sine terms, like \( B + C - A \), by using the fact that \( A + B + C = 180° \). This transforms \( B + C - A \) into \( 180° - 2A \), and \( \sin(180° - 2A) \) becomes \( \sin 2A \). After simplifying all three terms, the problem turns into proving \( \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \), which is a known identity for triangles.

🎯 Exam Tip: The trick here is recognizing how to simplify \( B+C-A \) and similar expressions using the \( A+B+C=180° \) property.

Question 2. If A + B + C = 2s, then prove that sin (s – A) sin (s- B ) + sin s . sin(s – C) = sin A sin B
Answer:
We are given that \( A+B+C = 2s \).
Start with the left-hand side (LHS):
\( \sin (s – A) \sin (s – B) + \sin s . \sin(s – C) \)
Apply the product-to-sum identity \( \sin X \sin Y = \frac{1}{2}[\cos(X-Y) - \cos(X+Y)] \) to both pairs of sine terms:
\( = \frac{1}{2}[\cos((s-A)-(s-B)) - \cos((s-A)+(s-B))] + \frac{1}{2}[\cos(s-(s-C)) - \cos(s+s-C)] \)
\( = \frac{1}{2}[\cos(B-A) - \cos(2s-A-B)] + \frac{1}{2}[\cos C - \cos(2s-C)] \)
Since \( 2s = A+B+C \), we can make the following substitutions:
\( 2s-A-B = C \)
\( 2s-C = A+B \)
Substitute these into the expression:
\( = \frac{1}{2}[\cos(B-A) - \cos C] + \frac{1}{2}[\cos C - \cos(A+B)] \)
Combine the terms:
\( = \frac{1}{2}[\cos(B-A) - \cos C + \cos C - \cos(A+B)] \)
The \( \cos C \) terms cancel out:
\( = \frac{1}{2}[\cos(B-A) - \cos(A+B)] \)
Since \( \cos(B-A) = \cos(A-B) \), we have:
\( = \frac{1}{2}[\cos(A-B) - \cos(A+B)] \)
Apply the identity \( \cos X - \cos Y = 2 \sin \left(\frac{X+Y}{2}\right) \sin \left(\frac{Y-X}{2}\right) \), with \( X=A-B \) and \( Y=A+B \):
\( = \frac{1}{2}[2 \sin A \sin B] \)
\( = \sin A \sin B \)
\( \text{LHS} = \text{RHS} \)
This identity is a form of Cagnoli's formula relating sides and angles of a triangle.
In simple words: We start by using a product-to-sum rule for \( \sin(X)\sin(Y) \) on both parts. Then, we use the fact that \( A + B + C = 2s \) to replace \( 2s - A - B \) with \( C \) and \( 2s - C \) with \( A + B \). After simplifying, we use another cosine identity to turn the expression into \( \sin A \sin B \).

🎯 Exam Tip: Remember the product-to-sum identity for \( \sin X \sin Y \) and how to effectively use the given \( A+B+C = 2s \) condition.

Question 3. If x + y + z = xyz prove that \( \frac{2x}{1-x^2} + \frac{2y}{1-y^2} + \frac{2z}{1-z^2} = \frac{2x}{1-x^2} \frac{2y}{1-y^2} \frac{2z}{1-z^2} \)
Answer:
Given the condition \( x+y+z = xyz \).
Let us substitute \( x = \tan A \), \( y = \tan B \), and \( z = \tan C \).
The given condition then becomes:
\( \tan A + \tan B + \tan C = \tan A \tan B \tan C \)
This is a standard trigonometric identity that holds true if and only if \( A+B+C = n\pi \) for some integer \( n \) (e.g., \( 180° \) or \( \pi \) radians).
If \( A+B+C = n\pi \), then also \( 2A+2B+2C = 2n\pi \).
For angles \( 2A, 2B, 2C \), the same tangent identity applies:
\( \tan 2A + \tan 2B + \tan 2C = \tan 2A \tan 2B \tan 2C \)
Now, we use the double angle formula for tangent: \( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \).
Substitute back for each angle:
\( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} = \frac{2x}{1-x^2} \)
\( \tan 2B = \frac{2 \tan B}{1 - \tan^2 B} = \frac{2y}{1-y^2} \)
\( \tan 2C = \frac{2 \tan C}{1 - \tan^2 C} = \frac{2z}{1-z^2} \)
Substitute these expressions into the identity \( \tan 2A + \tan 2B + \tan 2C = \tan 2A \tan 2B \tan 2C \):
\( \frac{2x}{1-x^2} + \frac{2y}{1-y^2} + \frac{2z}{1-z^2} = \frac{2x}{1-x^2} \frac{2y}{1-y^2} \frac{2z}{1-z^2} \)
This proves the required identity.
The identity \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \) holds true when the sum of the angles \( A+B+C \) is an integer multiple of \( \pi \), a fundamental property in trigonometry.
In simple words: First, we replace \( x \), \( y \), \( z \) with \( \tan A \), \( \tan B \), \( \tan C \). The given condition then means \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \). This is a special rule that tells us \( A + B + C \) must be \( 180° \) (or a multiple of \( 180° \)). If this is true, then \( 2A + 2B + 2C \) must be \( 360° \) (or a multiple of \( 360° \)). This leads to the same tangent identity for \( 2A \), \( 2B \), \( 2C \). Finally, we replace \( \tan 2A \) with its formula involving \( \tan A \), and similarly for \( B \) and \( C \), to get the required proof in terms of \( x \), \( y \), \( z \).

🎯 Exam Tip: Recognizing the condition \( x+y+z=xyz \) as \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \) (implying \( A+B+C = n\pi \)) is the key to solving this problem quickly.

Question 4. (i) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
Answer:
Given that \( A+B+C = \frac{\pi}{2} \) (or \( 90° \)).
This implies \( A+B = \frac{\pi}{2} - C \).
Start with the left-hand side (LHS):
\( \sin 2A + \sin 2B + \sin 2C \)
Apply the sum-to-product formula for \( \sin 2A + \sin 2B \) and the double angle formula for \( \sin 2C = 2 \sin C \cos C \):
\( = 2 \sin \left(\frac{2A+2B}{2}\right) \cos \left(\frac{2A-2B}{2}\right) + 2 \sin C \cos C \)
\( = 2 \sin (A+B) \cos (A-B) + 2 \sin C \cos C \)
Substitute \( A+B = \frac{\pi}{2} - C \):
\( = 2 \sin \left(\frac{\pi}{2} - C\right) \cos (A-B) + 2 \sin C \cos C \)
Since \( \sin \left(\frac{\pi}{2} - C\right) = \cos C \):
\( = 2 \cos C \cos (A-B) + 2 \sin C \cos C \)
Factor out \( 2 \cos C \):
\( = 2 \cos C [\cos (A-B) + \sin C] \)
Replace \( \sin C \) using \( C = \frac{\pi}{2} - (A+B) \), so \( \sin C = \sin \left(\frac{\pi}{2} - (A+B)\right) = \cos (A+B) \):
\( = 2 \cos C [\cos (A-B) + \cos (A+B)] \)
Apply the sum-to-product formula for \( \cos X + \cos Y \), with \( X = A-B \) and \( Y = A+B \):
\( = 2 \cos C \left[2 \cos \left(\frac{(A-B)+(A+B)}{2}\right) \cos \left(\frac{(A+B)-(A-B)}{2}\right)\right] \)
\( = 2 \cos C [2 \cos A \cos B] \)
\( = 4 \cos A \cos B \cos C \)
\( \text{LHS} = \text{RHS} \)
When angles sum to \( 90° \), \( \sin \) and \( \cos \) terms swap, which is a key tool in these types of proofs.
In simple words: We start by combining \( \sin 2A \) and \( \sin 2B \) using a sum formula, and \( \sin 2C \) using a double angle formula. Then, because \( A + B + C = 90° \), we replace \( A+B \) with \( 90° - C \). This helps us change \( \sin(A+B) \) to \( \cos C \). We factor out \( 2 \cos C \) and use the angle relationship again to simplify the remaining \( \sin C \) term. Finally, another sum-to-product rule for cosines gives us the desired \( 4 \cos A \cos B \cos C \).

🎯 Exam Tip: For \( A+B+C = 90° \) conditions, remember that \( A+B = 90°-C \), which leads to \( \sin(A+B) = \cos C \) and \( \cos(A+B) = \sin C \).

Question 4. (ii) cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C
Answer:
Given that \( A+B+C = \frac{\pi}{2} \) (or \( 90° \)).
This implies \( A+B = \frac{\pi}{2} - C \).
Start with the left-hand side (LHS):
\( \cos 2A + \cos 2B + \cos 2C \)
Apply the sum-to-product formula for \( \cos 2A + \cos 2B \):
\( = 2 \cos \left(\frac{2A+2B}{2}\right) \cos \left(\frac{2A-2B}{2}\right) + \cos 2C \)
\( = 2 \cos (A+B) \cos (A-B) + \cos 2C \)
Substitute \( A+B = \frac{\pi}{2} - C \) and use the double angle formula \( \cos 2C = 1 - 2 \sin^2 C \):
\( = 2 \cos \left(\frac{\pi}{2} - C\right) \cos (A-B) + (1 - 2 \sin^2 C) \)
Since \( \cos \left(\frac{\pi}{2} - C\right) = \sin C \):
\( = 2 \sin C \cos (A-B) + 1 - 2 \sin^2 C \)
Rearrange terms and factor out \( 2 \sin C \):
\( = 1 + 2 \sin C \cos (A-B) - 2 \sin^2 C \)
\( = 1 + 2 \sin C [\cos (A-B) - \sin C] \)
Replace \( \sin C \) using \( C = \frac{\pi}{2} - (A+B) \), so \( \sin C = \sin \left(\frac{\pi}{2} - (A+B)\right) = \cos (A+B) \):
\( = 1 + 2 \sin C [\cos (A-B) - \cos (A+B)] \)
Apply the identity \( \cos X - \cos Y = 2 \sin \left(\frac{X+Y}{2}\right) \sin \left(\frac{Y-X}{2}\right) \), with \( X=A-B \) and \( Y=A+B \):
\( = 1 + 2 \sin C [2 \sin A \sin B] \)
\( = 1 + 4 \sin A \sin B \sin C \)
\( \text{LHS} = \text{RHS} \)
This identity, specific to triangles where angles sum to \( 90° \), elegantly connects sums of double cosines to products of sines.
In simple words: We begin by using a sum formula for \( \cos 2A + \cos 2B \) and a double angle formula for \( \cos 2C \). Because \( A + B + C \) adds up to \( 90° \), we can replace \( \cos(A+B) \) with \( \sin C \). We then factor out common terms and apply another cosine difference formula. This simplifies the expression to \( 1 + 4 \sin A \sin B \sin C \).

🎯 Exam Tip: Be mindful of which double angle formula for cosine ( \( 1-2\sin^2 X \) or \( 2\cos^2 X - 1 \) ) to use for the most direct path to the solution.

Question 5. If ∆ABC is a right triangle and if ∠A = \( \frac{\pi}{2} \) then prove that
(i) cos² B + cos² C = 1
Answer:
Given that \( \Delta ABC \) is a right triangle and \( \angle A = 90° \).
We know that the sum of angles in any triangle is \( 180° \).
So, \( A+B+C = 180° \).
Since \( A = 90° \), we have \( 90° + B + C = 180° \).
This implies \( B+C = 180° - 90° = 90° \).
From this, we can write \( C = 90° - B \).
Now, let's consider the left-hand side (LHS):
\( \cos^2 B + \cos^2 C \)
Substitute \( C = 90° - B \) into the expression:
\( = \cos^2 B + \cos^2 (90° - B) \)
Using the trigonometric identity \( \cos(90° - X) = \sin X \):
\( = \cos^2 B + \sin^2 B \)
Finally, apply the fundamental identity \( \cos^2 X + \sin^2 X = 1 \):
\( = 1 \)
\( \text{LHS} = \text{RHS} \)

The relationship \( B+C = 90° \) is fundamental in right-angled triangles, enabling easy conversion between \( \sin \) and \( \cos \) of angles \( B \) and \( C \).
In simple words: In a right triangle, one angle is \( 90° \). Since all angles add up to \( 180° \), the other two angles (\( B \) and \( C \)) must add up to \( 90° \). This means \( C \) is \( 90° \) minus \( B \). We replace \( C \) in \( \cos^2 C \) with \( 90° - B \), which makes it \( \sin^2 B \). Then, using the basic rule \( \cos^2 B + \sin^2 B = 1 \), we prove the identity.

🎯 Exam Tip: For right-angled triangle problems, always use the \( B+C = 90° \) (or \( A+C = 90° \) etc.) relationship to simplify expressions.

Question 5. (ii) sin² B + sin² C = 1
Answer:
Given that \( \Delta ABC \) is a right triangle and \( \angle A = 90° \).
From part (i), we know that in a right triangle with \( \angle A = 90° \), the sum of the other two angles is \( B+C = 90° \).
This implies that \( C = 90° - B \).
Now, consider the left-hand side (LHS):
\( \sin^2 B + \sin^2 C \)
Substitute \( C = 90° - B \) into the expression:
\( = \sin^2 B + \sin^2 (90° - B) \)
Using the trigonometric identity \( \sin(90° - X) = \cos X \):
\( = \sin^2 B + \cos^2 B \)
Finally, apply the fundamental identity \( \sin^2 X + \cos^2 X = 1 \):
\( = 1 \)
\( \text{LHS} = \text{RHS} \)
This identity, similar to the previous one, highlights the complementary nature of angles in a right triangle.
In simple words: Knowing that \( B \) and \( C \) add up to \( 90° \) in this right triangle, we can write \( C \) as \( 90° - B \). When we replace \( C \) in \( \sin^2 C \) with this, it becomes \( \cos^2 B \). Then, using the basic rule \( \sin^2 B + \cos^2 B = 1 \), the identity is proved.

🎯 Exam Tip: Similar to \( \cos^2 B + \cos^2 C = 1 \), this identity \( \sin^2 B + \sin^2 C = 1 \) also holds true for the acute angles of a right triangle.

Question 5. (iii) cos B - cos C = -1 + 2√2 cos \( \frac{B}{2} \) . sin \( \frac{C}{2} \)
Answer:
Given that \( \Delta ABC \) is a right triangle and \( \angle A = 90° \).
From previous parts, we know that \( B+C = 90° \).

Let's simplify the Left-Hand Side (LHS):
\( \text{LHS} = \cos B - \cos C \)
Since \( B+C = 90° \), we have \( B = 90° - C \).
\( \text{LHS} = \cos (90° - C) - \cos C \)
Using the identity \( \cos(90° - X) = \sin X \):
\( \text{LHS} = \sin C - \cos C \)

Now, let's simplify the Right-Hand Side (RHS):
\( \text{RHS} = -1 + 2\sqrt{2} \cos \frac{B}{2} \sin \frac{C}{2} \)
Since \( B+C = 90° \), dividing by 2 gives \( \frac{B+C}{2} = 45° \).
This implies \( \frac{B}{2} = 45° - \frac{C}{2} \).
Substitute this into the RHS expression:
\( \text{RHS} = -1 + 2\sqrt{2} \cos \left(45° - \frac{C}{2}\right) \sin \frac{C}{2} \)
Apply the angle difference formula for cosine: \( \cos(45° - X) = \cos 45° \cos X + \sin 45° \sin X = \frac{1}{\sqrt{2}} (\cos X + \sin X) \).
So, \( \cos \left(45° - \frac{C}{2}\right) = \frac{1}{\sqrt{2}} \left(\cos \frac{C}{2} + \sin \frac{C}{2}\right) \)
Substitute this back into the RHS:
\( \text{RHS} = -1 + 2\sqrt{2} \left[\frac{1}{\sqrt{2}} \left(\cos \frac{C}{2} + \sin \frac{C}{2}\right)\right] \sin \frac{C}{2} \)
\( = -1 + 2 \left(\cos \frac{C}{2} + \sin \frac{C}{2}\right) \sin \frac{C}{2} \)
\( = -1 + 2 \sin \frac{C}{2} \cos \frac{C}{2} + 2 \sin^2 \frac{C}{2} \)
Using the double angle identities \( 2 \sin \frac{C}{2} \cos \frac{C}{2} = \sin C \) and \( 2 \sin^2 \frac{C}{2} = 1 - \cos C \):
\( = -1 + \sin C + (1 - \cos C) \)
\( = \sin C - \cos C \)
Since \( \text{LHS} = \sin C - \cos C \) and \( \text{RHS} = \sin C - \cos C \), therefore \( \text{LHS} = \text{RHS} \).
Identities can often be proven by simplifying both sides independently until they meet a common form.
In simple words: First, we simplify the left side \( \cos B - \cos C \). Because \( B + C = 90° \), \( \cos B \) is the same as \( \sin C \), so the left side becomes \( \sin C - \cos C \). Next, we simplify the right side. We use the fact that \( B/2 \) can be written as \( 45° - C/2 \) and expand the cosine term. After careful multiplication and using other half-angle formulas, the right side also simplifies to \( \sin C - \cos C \). Since both sides are equal, the proof is complete.

🎯 Exam Tip: When a proof seems complex, try simplifying both the left-hand side and the right-hand side separately to a common, simpler expression.