Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.9

Question 1. In a \( \Delta \) ABC, if \( \frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)} \) prove that \( a^2, b^2, c^2 \) are in Arithmetic progression.
Answer:We are given the relation: \( \frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)} \)
Now, we cross-multiply:
\( \sin A \cdot \sin (B - C) = \sin C \cdot \sin (A - B) \)
We know that in a triangle, \( A+B+C = 180^\circ \), so \( A = 180^\circ - (B+C) \) and \( C = 180^\circ - (A+B) \).
Therefore, \( \sin A = \sin (180^\circ - (B+C)) = \sin (B+C) \) and \( \sin C = \sin (180^\circ - (A+B)) = \sin (A+B) \).
Substitute these into the equation:
\( \sin (B+C) \sin (B - C) = \sin (A+B) \sin (A - B) \)
We use the trigonometric identity: \( \sin(X+Y) \sin(X-Y) = \sin^2 X - \sin^2 Y \).
Applying this identity to both sides:
\( \sin^2 B - \sin^2 C = \sin^2 A - \sin^2 B \)
Now, rearrange the terms to gather \( \sin^2 B \) on one side:
\( \sin^2 B + \sin^2 B = \sin^2 A + \sin^2 C \)
This simplifies to:
\( 2 \sin^2 B = \sin^2 A + \sin^2 C \)
From the sine rule, we know that \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \), where R is the circumradius.
So, \( \sin A = \frac{a}{2R} \), \( \sin B = \frac{b}{2R} \), and \( \sin C = \frac{c}{2R} \).
Substitute these into the equation \( 2 \sin^2 B = \sin^2 A + \sin^2 C \):
\( 2 \left( \frac{b}{2R} \right)^2 = \left( \frac{a}{2R} \right)^2 + \left( \frac{c}{2R} \right)^2 \)
\( 2 \frac{b^2}{4R^2} = \frac{a^2}{4R^2} + \frac{c^2}{4R^2} \)
Multiply both sides by \( 4R^2 \) to clear the denominators:
\( 2b^2 = a^2 + c^2 \)
This equation shows that \( a^2, b^2, c^2 \) are in Arithmetic Progression (AP). In an AP, the middle term is the average of its neighbors. This relationship holds true for the squares of the sides.
In simple words: We used angle properties and a trigonometry rule to change the original equation. Then, using the sine rule, we replaced the sines of angles with the sides of the triangle. This led us to show that \( 2b^2 = a^2 + c^2 \), which is the definition of numbers being in an Arithmetic Progression.

🎯 Exam Tip: Remember key trigonometric identities like \( \sin(X+Y) \sin(X-Y) = \sin^2 X - \sin^2 Y \) and the sine rule \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \) as they are fundamental in solving triangle-related proofs.

Question 2. The angles of a triangle A B C, are in Arithmetic Progression and if \( b : c = \sqrt{3}:\sqrt{2} \) find \( \angle A \).
Answer:Given that the angles A, B, C are in Arithmetic Progression (AP).
For angles in AP, the middle angle (B) is the average of the other two, so \( 2B = A+C \).
We also know that the sum of angles in a triangle is \( A+B+C = 180^\circ \).
Substitute \( A+C = 2B \) into the sum equation:
\( B + (A+C) = 180^\circ \)
\( B + 2B = 180^\circ \)
\( 3B = 180^\circ \)
\( \implies B = \frac{180^\circ}{3} = 60^\circ \)
Now, we know \( A+C = 2B = 2 \times 60^\circ = 120^\circ \).
We are given the ratio of sides \( b:c = \sqrt{3}:\sqrt{2} \).
From the sine rule, \( \frac{b}{\sin B} = \frac{c}{\sin C} \).
So, \( \frac{b}{c} = \frac{\sin B}{\sin C} \).
Substitute the given ratio and the value of B:
\( \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin 60^\circ}{\sin C} \)
We know \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
\( \frac{\sqrt{3}}{\sqrt{2}} = \frac{\frac{\sqrt{3}}{2}}{\sin C} \)
Now, solve for \( \sin C \):
\( \sin C = \frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{2} \)
\( \sin C = \frac{\sqrt{2}}{2} \)
\( \sin C = \frac{1}{\sqrt{2}} \)
This means \( C = 45^\circ \) (since C is an angle in a triangle, it must be acute).
Finally, we find angle A using \( A+C = 120^\circ \):
\( A + 45^\circ = 120^\circ \)
\( A = 120^\circ - 45^\circ \)
\( A = 75^\circ \)
The unknown angle A is found to be 75 degrees. The property of angles in AP simplifies finding one angle, which then helps find others using the sine rule.
In simple words: First, we used the rule for angles in Arithmetic Progression to find angle B. Then, we used the sine rule along with the given side ratio to find angle C. Finally, we used the sum of angles in a triangle to calculate angle A.

🎯 Exam Tip: When angles are in Arithmetic Progression, remember the sum of angles \( A+B+C = 180^\circ \) immediately leads to \( B = 60^\circ \). This is a common shortcut to save time.

Question 3. In a \( \Delta \) ABC, if \( \cos C = \frac{\sin A}{2 \sin B} \) show that the triangle is isosceles.
Answer:Given the relation: \( \cos C = \frac{\sin A}{2 \sin B} \)
From the cosine rule, we know that \( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \).
From the sine rule, we know that \( \sin A = \frac{a}{2R} \) and \( \sin B = \frac{b}{2R} \), where R is the circumradius.
Substitute the sine rule expressions into the given relation:
\( \cos C = \frac{\frac{a}{2R}}{2 \cdot \frac{b}{2R}} \)
\( \cos C = \frac{\frac{a}{2R}}{\frac{2b}{2R}} \)
\( \cos C = \frac{a}{2R} \cdot \frac{2R}{2b} \)
\( \cos C = \frac{a}{2b} \)
Now, equate this with the cosine rule expression for \( \cos C \):
\( \frac{a^2 + b^2 - c^2}{2ab} = \frac{a}{2b} \)
We can cancel \( 2b \) from both sides (since \( 2b \neq 0 \)):
\( \frac{a^2 + b^2 - c^2}{a} = a \)
Multiply both sides by \( a \):
\( a^2 + b^2 - c^2 = a^2 \)
Subtract \( a^2 \) from both sides:
\( b^2 - c^2 = 0 \)
\( b^2 = c^2 \)
Since b and c represent lengths, they must be positive. Therefore, \( b = c \).
When two sides of a triangle are equal, the triangle is an isosceles triangle. This means triangle ABC is an isosceles triangle.
In simple words: We used both the cosine rule and the sine rule to rewrite the given equation. After simplifying, we found that side 'b' is equal to side 'c'. When a triangle has two equal sides, it is called an isosceles triangle.

🎯 Exam Tip: Always remember to link the final mathematical result (like \( b=c \)) back to the geometric property requested in the question (e.g., "triangle is isosceles"). This ensures you fully answer the prompt.

Question 4. In a \( \Delta \) ABC, prove that \( \frac{\sin B}{\sin C} = \frac{c - a \cos B}{b - a \cos C} \)
Answer:We need to prove that \( \frac{\sin B}{\sin C} = \frac{c - a \cos B}{b - a \cos C} \).
Let's start with the Right Hand Side (RHS).
From the cosine rule, we have:
\( \cos B = \frac{a^2 + c^2 - b^2}{2ac} \implies a \cos B = a \left( \frac{a^2 + c^2 - b^2}{2ac} \right) = \frac{a^2 + c^2 - b^2}{2c} \)
And \( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \implies a \cos C = a \left( \frac{a^2 + b^2 - c^2}{2ab} \right) = \frac{a^2 + b^2 - c^2}{2b} \)
Substitute these into the RHS expression:
\( RHS = \frac{c - \left( \frac{a^2 + c^2 - b^2}{2c} \right)}{b - \left( \frac{a^2 + b^2 - c^2}{2b} \right)} \)
Find a common denominator for the numerator and the denominator parts:
\( RHS = \frac{\frac{2c^2 - (a^2 + c^2 - b^2)}{2c}}{\frac{2b^2 - (a^2 + b^2 - c^2)}{2b}} \)
\( RHS = \frac{\frac{2c^2 - a^2 - c^2 + b^2}{2c}}{\frac{2b^2 - a^2 - b^2 + c^2}{2b}} \)
\( RHS = \frac{\frac{c^2 - a^2 + b^2}{2c}}{\frac{b^2 - a^2 + c^2}{2b}} \)
Notice that the numerator of the top fraction is \( b^2 + c^2 - a^2 \), and the numerator of the bottom fraction is also \( b^2 + c^2 - a^2 \).
So, \( RHS = \frac{b^2 + c^2 - a^2}{2c} \times \frac{2b}{b^2 + c^2 - a^2} \)
Since \( b^2 + c^2 - a^2 \) is a common term (and not zero for a non-degenerate triangle), we can cancel it:
\( RHS = \frac{2b}{2c} \)
\( RHS = \frac{b}{c} \)
Now, from the sine rule, \( \frac{b}{\sin B} = \frac{c}{\sin C} \)
This implies \( \frac{b}{c} = \frac{\sin B}{\sin C} \).
So, \( RHS = \frac{\sin B}{\sin C} \).
Thus, \( \frac{\sin B}{\sin C} = \frac{c - a \cos B}{b - a \cos C} \) is proven. This identity demonstrates the relationship between the sines of angles and a combination of sides and cosines in a triangle.
In simple words: We started with the right side of the equation and used the cosine rule to replace \( a \cos B \) and \( a \cos C \) with expressions involving sides a, b, and c. After simplifying the complex fractions, we ended up with \( \frac{b}{c} \). We then used the sine rule to show that \( \frac{b}{c} \) is equal to \( \frac{\sin B}{\sin C} \), which matches the left side of the equation.

🎯 Exam Tip: When proving identities involving sides and angles of a triangle, it's often effective to start from the more complex side (usually the RHS) and simplify it using the sine and cosine rules until it matches the simpler side (LHS).

Question 5. In an \( \Delta \)ABC, prove that \( a \cos A + b \cos B + c \cos C = 2a \sin B \sin C \).
Answer:Let's start with the Left Hand Side (LHS): \( a \cos A + b \cos B + c \cos C \).
Using the sine rule, \( a = 2R \sin A \), \( b = 2R \sin B \), \( c = 2R \sin C \).
Substitute these into the LHS:
\( LHS = (2R \sin A) \cos A + (2R \sin B) \cos B + (2R \sin C) \cos C \)
Factor out \( R \):
\( LHS = R (2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C) \)
We know the double angle identity: \( 2 \sin x \cos x = \sin 2x \).
\( LHS = R (\sin 2A + \sin 2B + \sin 2C) \)
Now, use the sum-to-product formula for \( \sin 2A + \sin 2B \):
\( \sin 2A + \sin 2B = 2 \sin (A+B) \cos (A-B) \)
Since \( A+B+C = 180^\circ \), \( A+B = 180^\circ - C \).
So, \( \sin (A+B) = \sin (180^\circ - C) = \sin C \).
\( LHS = R (2 \sin C \cos (A-B) + \sin 2C) \)
We also know \( \sin 2C = 2 \sin C \cos C \).
\( LHS = R (2 \sin C \cos (A-B) + 2 \sin C \cos C) \)
Factor out \( 2R \sin C \):
\( LHS = 2R \sin C (\cos (A-B) + \cos C) \)
Again, using \( C = 180^\circ - (A+B) \), so \( \cos C = \cos (180^\circ - (A+B)) = -\cos (A+B) \).
\( LHS = 2R \sin C (\cos (A-B) - \cos (A+B)) \)
Use the identity \( \cos(X-Y) - \cos(X+Y) = 2 \sin X \sin Y \).
Here, \( X=A \) and \( Y=B \).
\( LHS = 2R \sin C (2 \sin A \sin B) \)
\( LHS = 4R \sin A \sin B \sin C \)
We want to show this is equal to \( 2a \sin B \sin C \).
From the sine rule, \( \sin A = \frac{a}{2R} \). Substitute this back in:
\( LHS = 4R \left( \frac{a}{2R} \right) \sin B \sin C \)
\( LHS = 2a \sin B \sin C \)
This is equal to the RHS. The proof shows a fundamental identity for triangles, linking sums of cosine terms with products of sine terms.
In simple words: We started with the left side of the equation and replaced the sides 'a', 'b', 'c' using the sine rule. Then, we used double angle formulas and sum-to-product identities for sines and cosines. After several steps of simplification, we substituted back the sine rule expression for \( \sin A \) to match the right side of the equation.

🎯 Exam Tip: Problems asking to prove relations involving sums of \( \cos A \) or \( \sin A \) terms often require converting sides to sines using the sine rule, then applying sum-to-product and double angle formulas. Remember to substitute back to the required form at the end.

Question 6. In a \( \Delta \) ABC, \( \angle A = 60^\circ \). Prove that \( b + c = 2a \cos \left(\frac{B-C}{2}\right) \)
Answer:Given that \( \angle A = 60^\circ \).
In a triangle, the sum of angles is \( A+B+C = 180^\circ \).
Substitute \( A = 60^\circ \):
\( 60^\circ + B + C = 180^\circ \)
\( B+C = 180^\circ - 60^\circ \)
\( B+C = 120^\circ \)
We need to prove \( b + c = 2a \cos \left(\frac{B-C}{2}\right) \).
Let's start with the Left Hand Side (LHS): \( b+c \).
From the sine rule, \( b = 2R \sin B \) and \( c = 2R \sin C \).
\( LHS = 2R \sin B + 2R \sin C \)
Factor out \( 2R \):
\( LHS = 2R (\sin B + \sin C) \)
Use the sum-to-product formula: \( \sin X + \sin Y = 2 \sin \left(\frac{X+Y}{2}\right) \cos \left(\frac{X-Y}{2}\right) \).
\( LHS = 2R \left( 2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \right) \)
\( LHS = 4R \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \)
We know \( B+C = 120^\circ \), so \( \frac{B+C}{2} = \frac{120^\circ}{2} = 60^\circ \).
\( LHS = 4R \sin (60^\circ) \cos \left(\frac{B-C}{2}\right) \)
We know \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
\( LHS = 4R \left( \frac{\sqrt{3}}{2} \right) \cos \left(\frac{B-C}{2}\right) \)
\( LHS = 2R\sqrt{3} \cos \left(\frac{B-C}{2}\right) \)
This doesn't seem to directly match the RHS. Let's try expressing \( 2a \) in terms of R and A.
From the sine rule, \( a = 2R \sin A \).
Since \( A = 60^\circ \), \( a = 2R \sin 60^\circ = 2R \frac{\sqrt{3}}{2} = R\sqrt{3} \).
So, \( R\sqrt{3} = a \).
Substitute \( a \) back into our LHS expression:
\( LHS = (R\sqrt{3}) \cdot 2 \cos \left(\frac{B-C}{2}\right) \)
\( LHS = a \cdot 2 \cos \left(\frac{B-C}{2}\right) \)
\( LHS = 2a \cos \left(\frac{B-C}{2}\right) \)
This is equal to the RHS. The proof connects the sum of two sides to the third side and the cosine of half the difference of the other two angles, specifically when one angle is 60 degrees.
In simple words: We used the given angle A to find the sum of angles B and C. Then, we started with the left side of the equation, used the sine rule to replace 'b' and 'c', and applied a sum-to-product formula. Finally, we used the sine rule again for 'a' to show that the expression simplifies to the right side of the equation.

🎯 Exam Tip: When a specific angle (like \( A = 60^\circ \)) is given, use it immediately to find relationships between other angles (e.g., \( B+C = 120^\circ \)) and also to simplify terms involving the sine or cosine of that angle (e.g., \( a = 2R \sin A \)).

Question 7. In an \( \Delta \) ABC, prove the following,
(i) \( a \sin \left(\frac{A}{2}+B\right) = (b + c) \cdot \sin \frac{A}{2} \)
Answer:
(i) We want to prove \( a \sin \left(\frac{A}{2}+B\right) = (b + c) \sin \frac{A}{2} \).
Let's start with the Right Hand Side (RHS): \( (b+c) \sin \frac{A}{2} \).
From the sine rule, \( b = 2R \sin B \) and \( c = 2R \sin C \).
\( RHS = (2R \sin B + 2R \sin C) \sin \frac{A}{2} \)
\( RHS = 2R (\sin B + \sin C) \sin \frac{A}{2} \)
Use the sum-to-product formula \( \sin B + \sin C = 2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \).
\( RHS = 2R \left( 2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \right) \sin \frac{A}{2} \)
\( RHS = 4R \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \sin \frac{A}{2} \)
In a triangle, \( A+B+C = 180^\circ \), so \( \frac{B+C}{2} = \frac{180^\circ - A}{2} = 90^\circ - \frac{A}{2} \).
Therefore, \( \sin \left(\frac{B+C}{2}\right) = \sin \left(90^\circ - \frac{A}{2}\right) = \cos \frac{A}{2} \).
Substitute this into the RHS:
\( RHS = 4R \cos \frac{A}{2} \cos \left(\frac{B-C}{2}\right) \sin \frac{A}{2} \)
Rearrange the terms:
\( RHS = 2R (2 \sin \frac{A}{2} \cos \frac{A}{2}) \cos \left(\frac{B-C}{2}\right) \)
Use the double angle identity \( 2 \sin X \cos X = \sin 2X \). Here \( X = \frac{A}{2} \), so \( 2 \sin \frac{A}{2} \cos \frac{A}{2} = \sin A \).
\( RHS = 2R \sin A \cos \left(\frac{B-C}{2}\right) \)
From the sine rule, \( a = 2R \sin A \).
\( RHS = a \cos \left(\frac{B-C}{2}\right) \)
This result doesn't directly match the LHS \( a \sin \left(\frac{A}{2}+B\right) \). Let's re-examine the LHS.
Consider the LHS: \( a \sin \left(\frac{A}{2}+B\right) \).
We know \( A+B+C = 180^\circ \). So, \( B = 180^\circ - (A+C) \).
\( \frac{A}{2}+B = \frac{A}{2} + 180^\circ - (A+C) = 180^\circ - \left( \frac{A}{2}+C \right) \)
This substitution does not simplify easily.
Let's try to express \( \frac{A}{2}+B \) in terms of \( \frac{C-B}{2} \) or similar. We know \( \frac{B+C}{2} = 90^\circ - \frac{A}{2} \).
So, \( \frac{A}{2} = 90^\circ - \frac{B+C}{2} \).
\( \frac{A}{2} + B = (90^\circ - \frac{B+C}{2}) + B = 90^\circ + B - \frac{B}{2} - \frac{C}{2} = 90^\circ + \frac{B}{2} - \frac{C}{2} = 90^\circ + \left(\frac{B-C}{2}\right) \)
Now, substitute this into the LHS:
\( LHS = a \sin \left(90^\circ + \left(\frac{B-C}{2}\right)\right) \)
Using the identity \( \sin(90^\circ + X) = \cos X \):
\( LHS = a \cos \left(\frac{B-C}{2}\right) \)
This matches the simplified RHS. Thus, the identity is proven. This shows a useful relationship between one side, an angle, and the cosine of half the difference of the other two angles.
In simple words: For the first part, we started by simplifying the right side of the equation using the sine rule and sum-to-product formulas. We used the fact that angles in a triangle add up to 180 degrees. Then, we simplified the left side of the equation using the same angle sum property. Both sides simplified to the same expression, proving the identity.

🎯 Exam Tip: When dealing with expressions like \( \frac{A}{2}+B \), try to use the angle sum property (\( A+B+C = 180^\circ \)) to rewrite them in terms of \( 90^\circ \pm \) combinations, as this often leads to a simpler trigonometric function (e.g., \( \sin(90^\circ + X) = \cos X \)).

Question 7. In an \( \Delta \) ABC, prove the following,
(ii) \( a (\cos B + \cos C) = 2(b + c) \sin^2 \frac{A}{2} \)
Answer:
(ii) We need to prove \( a (\cos B + \cos C) = 2(b + c) \sin^2 \frac{A}{2} \).
Let's start with the Left Hand Side (LHS): \( a (\cos B + \cos C) \).
From the cosine rule, \( \cos B = \frac{a^2 + c^2 - b^2}{2ac} \) and \( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \).
Substitute these into the LHS:
\( LHS = a \left( \frac{a^2 + c^2 - b^2}{2ac} + \frac{a^2 + b^2 - c^2}{2ab} \right) \)
Find a common denominator \( 2abc \):
\( LHS = a \left( \frac{b(a^2 + c^2 - b^2) + c(a^2 + b^2 - c^2)}{2abc} \right) \)
Cancel 'a' from numerator and denominator:
\( LHS = \frac{b(a^2 + c^2 - b^2) + c(a^2 + b^2 - c^2)}{2bc} \)
Expand the terms:
\( LHS = \frac{a^2b + bc^2 - b^3 + a^2c + b^2c - c^3}{2bc} \)
Now, let's work on the Right Hand Side (RHS): \( 2(b+c) \sin^2 \frac{A}{2} \).
From the sine rule, \( b = 2R \sin B \) and \( c = 2R \sin C \). Also \( a = 2R \sin A \).
RHS \( = 2(2R \sin B + 2R \sin C) \sin^2 \frac{A}{2} \)
RHS \( = 2 \cdot 2R (\sin B + \sin C) \sin^2 \frac{A}{2} \)
RHS \( = 4R (\sin B + \sin C) \sin^2 \frac{A}{2} \)
Use sum-to-product formula for \( \sin B + \sin C = 2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \).
RHS \( = 4R \left( 2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \right) \sin^2 \frac{A}{2} \)
RHS \( = 8R \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \sin^2 \frac{A}{2} \)
In a triangle, \( A+B+C = 180^\circ \), so \( \frac{B+C}{2} = 90^\circ - \frac{A}{2} \).
Thus, \( \sin \left(\frac{B+C}{2}\right) = \sin \left(90^\circ - \frac{A}{2}\right) = \cos \frac{A}{2} \).
RHS \( = 8R \cos \frac{A}{2} \cos \left(\frac{B-C}{2}\right) \sin^2 \frac{A}{2} \)
Rearrange:
RHS \( = 2R \sin A \cdot 2 \sin \frac{A}{2} \cos \left(\frac{B-C}{2}\right) \)
This is not simplifying as expected. Let's try starting with the projection formula for 'a':
We know the projection formula: \( a = b \cos C + c \cos B \).
We need to prove \( a(\cos B + \cos C) = 2(b+c) \sin^2 \frac{A}{2} \).
Let's consider \( b+c \). From the sine rule, \( b=2R \sin B \) and \( c=2R \sin C \).
\( b+c = 2R (\sin B + \sin C) = 2R \cdot 2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \)
Since \( \frac{B+C}{2} = 90^\circ - \frac{A}{2} \), we have \( \sin \left(\frac{B+C}{2}\right) = \cos \frac{A}{2} \).
So, \( b+c = 4R \cos \frac{A}{2} \cos \left(\frac{B-C}{2}\right) \).
Now consider the RHS of the identity we need to prove: \( 2(b+c) \sin^2 \frac{A}{2} \).
\( RHS = 2 \left( 4R \cos \frac{A}{2} \cos \left(\frac{B-C}{2}\right) \right) \sin^2 \frac{A}{2} \)
\( RHS = 8R \cos \frac{A}{2} \cos \left(\frac{B-C}{2}\right) \sin^2 \frac{A}{2} \)
From the LHS of the identity to prove: \( a (\cos B + \cos C) \).
Use sum-to-product for \( \cos B + \cos C \): \( 2 \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \).
\( \cos B + \cos C = 2 \cos \left(90^\circ - \frac{A}{2}\right) \cos \left(\frac{B-C}{2}\right) \)
\( \cos B + \cos C = 2 \sin \frac{A}{2} \cos \left(\frac{B-C}{2}\right) \).
Now, substitute this into the LHS of the identity we need to prove:
\( LHS = a \left( 2 \sin \frac{A}{2} \cos \left(\frac{B-C}{2}\right) \right) \)
From the sine rule, \( a = 2R \sin A \). Also, \( \sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2} \).
So, \( a = 2R (2 \sin \frac{A}{2} \cos \frac{A}{2}) = 4R \sin \frac{A}{2} \cos \frac{A}{2} \).
Substitute this 'a' into the LHS expression:
\( LHS = \left( 4R \sin \frac{A}{2} \cos \frac{A}{2} \right) \left( 2 \sin \frac{A}{2} \cos \left(\frac{B-C}{2}\right) \right) \)
\( LHS = 8R \sin^2 \frac{A}{2} \cos \frac{A}{2} \cos \left(\frac{B-C}{2}\right) \)
This matches the RHS that we derived earlier.
Therefore, \( a (\cos B + \cos C) = 2(b + c) \sin^2 \frac{A}{2} \) is proven. This identity connects a side and the sum of two cosines to the sum of the other two sides and the square of the sine of half the first angle.
In simple words: For the second part, we used the sum-to-product formula for \( \cos B + \cos C \) and for \( b+c \) using sine rule. We also used the angle sum property of a triangle. By substituting these simplified forms and the sine rule for side 'a', we showed that both sides of the equation become identical, thus proving the identity.

🎯 Exam Tip: When dealing with sum of cosines, remember the sum-to-product formula \( \cos X + \cos Y = 2 \cos \left(\frac{X+Y}{2}\right) \cos \left(\frac{X-Y}{2}\right) \). Also, don't forget the half-angle formulas like \( \sin^2 \frac{A}{2} = \frac{1 - \cos A}{2} \) if direct substitution doesn't work out.

Question 7. In an \( \Delta \) ABC, prove the following,
(iii) \( \frac{a^2 - c^2}{b^2} = \frac{\sin (A-C)}{\sin (A + C)} \)
Answer:
(iii) We need to prove \( \frac{a^2 - c^2}{b^2} = \frac{\sin (A-C)}{\sin (A + C)} \).
Let's start with the Left Hand Side (LHS): \( \frac{a^2 - c^2}{b^2} \).
From the sine rule, \( a = 2R \sin A \), \( b = 2R \sin B \), \( c = 2R \sin C \).
Substitute these into the LHS:
\( LHS = \frac{(2R \sin A)^2 - (2R \sin C)^2}{(2R \sin B)^2} \)
\( LHS = \frac{4R^2 \sin^2 A - 4R^2 \sin^2 C}{4R^2 \sin^2 B} \)
Factor out \( 4R^2 \) from the numerator:
\( LHS = \frac{4R^2 (\sin^2 A - \sin^2 C)}{4R^2 \sin^2 B} \)
Cancel \( 4R^2 \):
\( LHS = \frac{\sin^2 A - \sin^2 C}{\sin^2 B} \)
Use the trigonometric identity \( \sin^2 X - \sin^2 Y = \sin(X+Y) \sin(X-Y) \).
\( LHS = \frac{\sin(A+C) \sin(A-C)}{\sin^2 B} \)
In a triangle, \( A+B+C = 180^\circ \), so \( A+C = 180^\circ - B \).
Therefore, \( \sin(A+C) = \sin(180^\circ - B) = \sin B \).
Substitute \( \sin(A+C) = \sin B \) into the LHS:
\( LHS = \frac{\sin B \sin(A-C)}{\sin^2 B} \)
Cancel one \( \sin B \) from the numerator and denominator:
\( LHS = \frac{\sin(A-C)}{\sin B} \)
Again, use \( \sin B = \sin(180^\circ - (A+C)) = \sin(A+C) \).
So, \( LHS = \frac{\sin(A-C)}{\sin(A+C)} \).
This is equal to the Right Hand Side (RHS). The identity is proven. This shows how the difference of squares of sides relates to the sines of the sum and difference of the corresponding angles.
In simple words: For the third part, we started with the left side of the equation and replaced the sides 'a', 'b', 'c' using the sine rule. We then used a trigonometry identity for the difference of sines squared. Finally, we used the angle sum property of a triangle to simplify the sines of sums of angles, leading to the right side of the equation.

🎯 Exam Tip: The identity \( \sin^2 X - \sin^2 Y = \sin(X+Y) \sin(X-Y) \) is very useful for problems involving differences of squares of sines. Also, remember to constantly use the \( A+B+C = 180^\circ \) relation to simplify angle sums.

Question 7. In an \( \Delta \) ABC, prove the following,
(iv) \( \frac{a \sin (B - C)}{b^2 - c^2} = \frac{b \sin (C-A)}{c^2 - a^2} = \frac{c \sin (A-B)}{a^2 - b^2} \)
Answer:
(iv) We need to prove \( \frac{a \sin (B - C)}{b^2 - c^2} = \frac{b \sin (C-A)}{c^2 - a^2} = \frac{c \sin (A-B)}{a^2 - b^2} \).
Let's prove the first part: \( \frac{a \sin (B - C)}{b^2 - c^2} \).
From the sine rule, \( a = 2R \sin A \), \( b = 2R \sin B \), \( c = 2R \sin C \).
Substitute these into the expression:
\( \frac{2R \sin A \sin (B - C)}{(2R \sin B)^2 - (2R \sin C)^2} \)
\( = \frac{2R \sin A \sin (B - C)}{4R^2 \sin^2 B - 4R^2 \sin^2 C} \)
\( = \frac{2R \sin A \sin (B - C)}{4R^2 (\sin^2 B - \sin^2 C)} \)
\( = \frac{\sin A \sin (B - C)}{2R (\sin^2 B - \sin^2 C)} \)
Use the identity \( \sin^2 X - \sin^2 Y = \sin(X+Y) \sin(X-Y) \).
\( = \frac{\sin A \sin (B - C)}{2R \sin(B+C) \sin(B-C)} \)
Since \( \sin(B-C) \) is common (assuming \( B \neq C \)), we can cancel it:
\( = \frac{\sin A}{2R \sin(B+C)} \)
In a triangle, \( A+B+C = 180^\circ \), so \( B+C = 180^\circ - A \).
Therefore, \( \sin(B+C) = \sin(180^\circ - A) = \sin A \).
Substitute this into the expression:
\( = \frac{\sin A}{2R \sin A} \)
Cancel \( \sin A \) (assuming \( A \neq 0, 180^\circ \)):
\( = \frac{1}{2R} \)
Now, let's prove the second part: \( \frac{b \sin (C-A)}{c^2 - a^2} \).
Substitute sine rule expressions:
\( \frac{2R \sin B \sin (C-A)}{(2R \sin C)^2 - (2R \sin A)^2} \)
\( = \frac{2R \sin B \sin (C-A)}{4R^2 (\sin^2 C - \sin^2 A)} \)
\( = \frac{\sin B \sin (C-A)}{2R \sin(C+A) \sin(C-A)} \)
Cancel \( \sin(C-A) \):
\( = \frac{\sin B}{2R \sin(C+A)} \)
Since \( C+A = 180^\circ - B \), \( \sin(C+A) = \sin B \).
\( = \frac{\sin B}{2R \sin B} \)
\( = \frac{1}{2R} \)
Similarly, for the third part: \( \frac{c \sin (A-B)}{a^2 - b^2} \).
\( = \frac{2R \sin C \sin (A-B)}{(2R \sin A)^2 - (2R \sin B)^2} \)
\( = \frac{2R \sin C \sin (A-B)}{4R^2 (\sin^2 A - \sin^2 B)} \)
\( = \frac{\sin C \sin (A-B)}{2R \sin(A+B) \sin(A-B)} \)
\( = \frac{\sin C}{2R \sin(A+B)} \)
Since \( A+B = 180^\circ - C \), \( \sin(A+B) = \sin C \).
\( = \frac{\sin C}{2R \sin C} \)
\( = \frac{1}{2R} \)
Since all three expressions simplify to \( \frac{1}{2R} \), they are all equal. This set of identities demonstrates a cyclic relationship between the sides, sines of angle differences, and the circumradius.
In simple words: For the fourth part, we took each of the three fractions separately. For each one, we replaced the sides with their sine rule equivalents. Then, we used the identity for the difference of sines squared and the triangle's angle sum property. Each fraction simplified to \( \frac{1}{2R} \), which means all three are equal to each other.

🎯 Exam Tip: This type of cyclic identity often requires proving one part and then stating "Similarly, the other expressions can be shown to be equal to \( \frac{1}{2R} \)" to save time, unless explicitly asked to prove each one in detail. The sine rule and \( \sin^2 X - \sin^2 Y \) identity are key.

Question 7. In an \( \Delta \) ABC, prove the following,
(iv) \( \frac{\text{a sin (B - C)}}{\text{b}^2 - \text{c}^2} = \frac{\text{b sin (C-A)}}{\text{c}^2 - \text{a}^2} = \frac{\text{c sin (A-B)}}{\text{a}^2 - \text{b}^2} \)
Answer: We need to show that each expression is equal to a constant value, \( \frac{1}{2R} \).
Using the Sine Rule, we know that \( a = 2R \sin A \), \( b = 2R \sin B \), and \( c = 2R \sin C \), where \( R \) is the circumradius of the triangle.
We also use the identity: \( \sin^2 X - \sin^2 Y = \sin(X+Y) \sin(X-Y) \).
Also, in any triangle \( A+B+C = 180^\circ \). This means \( \sin A = \sin(180^\circ - (B+C)) = \sin(B+C) \), and similarly for other angles.
Let's prove the first part:
\( \frac{\text{a sin (B - C)}}{\text{b}^2 - \text{c}^2} \)
Substitute \( a, b, c \) using the Sine Rule:
\( = \frac{2R \sin A \sin (B - C)}{(2R \sin B)^2 - (2R \sin C)^2} \)
\( = \frac{2R \sin A \sin (B - C)}{4R^2 (\sin^2 B - \sin^2 C)} \)
Now, apply the identity \( \sin^2 B - \sin^2 C = \sin(B+C) \sin(B-C) \):
\( = \frac{2R \sin A \sin (B - C)}{4R^2 \sin(B+C) \sin(B-C)} \)
\( = \frac{\sin A}{2R \sin(B+C)} \)
Since \( \sin A = \sin(B+C) \), we can cancel these terms:
\( = \frac{1}{2R} \)
Next, prove the second part:
\( \frac{\text{b sin (C-A)}}{\text{c}^2 - \text{a}^2} \)
Substitute \( b, c, a \) using the Sine Rule:
\( = \frac{2R \sin B \sin (C - A)}{(2R \sin C)^2 - (2R \sin A)^2} \)
\( = \frac{2R \sin B \sin (C - A)}{4R^2 (\sin^2 C - \sin^2 A)} \)
Apply the identity \( \sin^2 C - \sin^2 A = \sin(C+A) \sin(C-A) \):
\( = \frac{2R \sin B \sin (C - A)}{4R^2 \sin(C+A) \sin(C-A)} \)
\( = \frac{\sin B}{2R \sin(C+A)} \)
Since \( \sin B = \sin(C+A) \), we can cancel these terms:
\( = \frac{1}{2R} \)
Finally, prove the third part:
\( \frac{\text{c sin (A-B)}}{\text{a}^2 - \text{b}^2} \)
Substitute \( c, a, b \) using the Sine Rule:
\( = \frac{2R \sin C \sin (A - B)}{(2R \sin A)^2 - (2R \sin B)^2} \)
\( = \frac{2R \sin C \sin (A - B)}{4R^2 (\sin^2 A - \sin^2 B)} \)
Apply the identity \( \sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B) \):
\( = \frac{2R \sin C \sin (A - B)}{4R^2 \sin(A+B) \sin(A-B)} \)
\( = \frac{\sin C}{2R \sin(A+B)} \)
Since \( \sin C = \sin(A+B) \), we can cancel these terms:
\( = \frac{1}{2R} \)
Since all three expressions simplify to \( \frac{1}{2R} \), they are equal to each other. This shows a beautiful relationship between sides and angles in a triangle.
In simple words: We used special rules for triangles, like the Sine Rule and a sine identity, to show that all three complex fractions simplify to the same simple value, \( \frac{1}{2R} \). This proves they are all equal to each other.

🎯 Exam Tip: Remember the Sine Rule (\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \)) and the identity \( \sin^2 X - \sin^2 Y = \sin(X+Y)\sin(X-Y) \) as these are key for many triangle proofs.

Question 7. In an \( \Delta \) ABC, prove the following,
(v) \( \frac{\text{a + b}}{\text{a - b}} = \tan\left(\frac{\text{A + B}}{2}\right) \cot\left(\frac{\text{A - B}}{2}\right) \)
Answer: We need to prove the given identity relating the sides and angles of a triangle.
Let's start with the left-hand side (LHS) of the equation:
\( \text{LHS} = \frac{\text{a + b}}{\text{a - b}} \)
Using the Sine Rule, we know that \( a = 2R \sin A \) and \( b = 2R \sin B \). Substitute these into the LHS:
\( = \frac{2R \sin A + 2R \sin B}{2R \sin A - 2R \sin B} \)
Factor out \( 2R \) from both the numerator and the denominator:
\( = \frac{2R (\sin A + \sin B)}{2R (\sin A - \sin B)} \)
\( = \frac{\sin A + \sin B}{\sin A - \sin B} \)
Now, we use the sum-to-product trigonometric identities:
\( \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \)
\( \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \)
Substitute these identities into the expression:
\( = \frac{2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)}{2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)} \)
Cancel out the \( 2 \)s and rearrange the terms:
\( = \left(\frac{\sin\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A+B}{2}\right)}\right) \left(\frac{\cos\left(\frac{A-B}{2}\right)}{\sin\left(\frac{A-B}{2}\right)}\right) \)
Recognize that \( \frac{\sin X}{\cos X} = \tan X \) and \( \frac{\cos X}{\sin X} = \cot X \):
\( = \tan\left(\frac{A+B}{2}\right) \cot\left(\frac{A-B}{2}\right) \)
This matches the right-hand side (RHS) of the equation.
\( = \text{RHS} \)
Hence proved. This identity is a form of Napier's Analogy, useful in solving oblique triangles.
In simple words: We used the Sine Rule to change the sides 'a' and 'b' into sine terms. Then, we used special formulas that change sums and differences of sines into products. After simplifying, we found that the expression becomes exactly what was on the other side of the equation, proving it true.

🎯 Exam Tip: When dealing with identities involving ratios of sides and sums/differences of angles, the Sine Rule and sum-to-product/product-to-sum identities are often very useful.

Question 8. In a triangle \( \Delta \) ABC, prove that \( (a^2 - b^2 + c^2) \tan B = (a^2 + b^2 - c^2) \tan C \)
Answer: We need to prove the given identity in triangle \( \Delta \) ABC.
Let's use the Law of Cosines and the relationship \( \tan X = \frac{\sin X}{\cos X} \).
From the Law of Cosines, we know:
\( \cos B = \frac{a^2 + c^2 - b^2}{2ac} \)
\( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \)
From these, we can write:
\( a^2 + c^2 - b^2 = 2ac \cos B \)
\( a^2 + b^2 - c^2 = 2ab \cos C \)
Now, let's substitute these expressions into the given equation:
\( (2ac \cos B) \tan B = (2ab \cos C) \tan C \)
Replace \( \tan B \) with \( \frac{\sin B}{\cos B} \) and \( \tan C \) with \( \frac{\sin C}{\cos C} \):
\( (2ac \cos B) \left(\frac{\sin B}{\cos B}\right) = (2ab \cos C) \left(\frac{\sin C}{\cos C}\right) \)
Simplify by cancelling \( \cos B \) and \( \cos C \):
\( 2ac \sin B = 2ab \sin C \)
Now, divide both sides by \( 2a \) (since \( a \neq 0 \)):
\( c \sin B = b \sin C \)
This is a well-known result from the Sine Rule. The Sine Rule states that \( \frac{b}{\sin B} = \frac{c}{\sin C} \), which means \( c \sin B = b \sin C \). This step confirms the equality, proving the original identity. The identity shows how sides and angles are balanced in a triangle.
In simple words: We used the cosine rule to rewrite parts of the equation involving squares of sides and then replaced 'tan' with 'sin' over 'cos'. After simplifying, we got a known relationship from the Sine Rule, which means the original statement is true.

🎯 Exam Tip: For proofs involving both sides and angles (especially powers of sides like \( a^2 \)), the Law of Cosines is often very useful. Combine it with the definition of tangent or the Law of Sines for a complete solution.

Question 9. An Engineer has to develop a triangular shaped park with a perimeter 120m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park.
Answer: To achieve the maximum area for a triangle with a fixed perimeter, the triangle must be an equilateral triangle. This means all its sides must be equal.
Given perimeter of the park \( = 120 \) m.
Let the lengths of the sides of the triangle be \( a, b, c \).
So, \( a + b + c = 120 \).
For maximum area, we set \( a = b = c \).
So, \( 3a = 120 \)
\( \implies \) \( a = \frac{120}{3} \)
\( \implies \) \( a = 40 \) m.
Therefore, the dimensions of the park are 40 m, 40 m, and 40 m.
Now, let's find the maximum area of this equilateral triangle.
The semi-perimeter, \( s = \frac{\text{Perimeter}}{2} = \frac{120}{2} = 60 \) m.
The area of an equilateral triangle in terms of its semi-perimeter \( s \) is given by the formula \( \text{Area} = \frac{s^2}{3\sqrt{3}} \).
\( \text{Area} = \frac{(60)^2}{3\sqrt{3}} \)
\( = \frac{3600}{3\sqrt{3}} \)
\( = \frac{1200}{\sqrt{3}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \):
\( = \frac{1200\sqrt{3}}{3} \)
\( = 400\sqrt{3} \) sq.m.
This shows how the shape of a given perimeter affects its enclosed area.
In simple words: To get the biggest possible area for a triangle with a set perimeter, all its sides must be equal. Since the total length of the sides is 120 meters, each side will be 40 meters. Using a formula for the area of such a triangle, the largest area comes out to be \( 400\sqrt{3} \) square meters.

🎯 Exam Tip: Always remember that for a fixed perimeter, an equilateral triangle encloses the maximum area. This is a common property applied in optimization problems.

Question 10. A rope of length 42m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed.
Answer: The length of the rope represents the perimeter of the triangle. To form a triangle with the largest possible area using a fixed length of rope (fixed perimeter), the triangle must be an equilateral triangle.
Given perimeter of the triangle \( = 42 \) m.
Let the lengths of the sides of the triangle be \( a, b, c \).
So, \( a + b + c = 42 \).
For the largest area, the triangle must be equilateral, meaning \( a = b = c \).
\( 3a = 42 \)
\( \implies \) \( a = \frac{42}{3} \)
\( \implies \) \( a = 14 \) m.
So, the dimensions of the triangle are 14 m, 14 m, and 14 m.
Now, let's calculate the largest area.
The semi-perimeter, \( s = \frac{\text{Perimeter}}{2} = \frac{42}{2} = 21 \) m.
Using the formula for the area of an equilateral triangle in terms of its semi-perimeter \( s \), which is \( \text{Area} = \frac{s^2}{3\sqrt{3}} \):
\( \text{Area} = \frac{(21)^2}{3\sqrt{3}} \)
\( = \frac{441}{3\sqrt{3}} \)
\( = \frac{147}{\sqrt{3}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \):
\( = \frac{147\sqrt{3}}{3} \)
\( = 49\sqrt{3} \) sq.m.
This demonstrates the geometric principle that equilateral shapes maximize area for a given perimeter.
In simple words: The rope's length is the triangle's edge total. To make the biggest triangle area with this rope, all three sides must be the same length. So, each side will be 14 meters. The largest area you can get is \( 49\sqrt{3} \) square meters.

🎯 Exam Tip: Recognize that "largest area" with a "given perimeter" always points to an equilateral triangle. Using this property simplifies the problem significantly.

Question 11. Derive Projection formula
(i) Law of sines,
(ii) Law of cosines.
Answer: We will derive the projection formulas for a triangle using both the Law of Sines and the Law of Cosines. The projection formulas describe a relationship between the sides and cosines of angles in a triangle.

(i) Derivation using the Law of Sines:
The Law of Sines states that \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \), where \( R \) is the circumradius. From this, we get:
\( a = 2R \sin A \)
\( b = 2R \sin B \)
\( c = 2R \sin C \)
Also, for any triangle, the sum of angles is \( A+B+C = 180^\circ \). This means \( A = 180^\circ - (B+C) \), \( B = 180^\circ - (A+C) \), and \( C = 180^\circ - (A+B) \).

(a) Prove \( a = b \cos C + c \cos B \)
Start with the Right-Hand Side (RHS):
\( \text{RHS} = b \cos C + c \cos B \)
Substitute \( b = 2R \sin B \) and \( c = 2R \sin C \):
\( = 2R \sin B \cos C + 2R \sin C \cos B \)
Factor out \( 2R \):
\( = 2R (\sin B \cos C + \cos B \sin C) \)
Using the trigonometric identity for \( \sin(X+Y) = \sin X \cos Y + \cos X \sin Y \):
\( = 2R \sin(B+C) \)
Since \( B+C = 180^\circ - A \), then \( \sin(B+C) = \sin(180^\circ - A) = \sin A \):
\( = 2R \sin A \)
From the Law of Sines, we know \( a = 2R \sin A \):
\( = a \)
\( = \text{LHS} \). Thus, the first projection formula is proved.

(b) Prove \( b = c \cos A + a \cos C \)
Start with the Right-Hand Side (RHS):
\( \text{RHS} = c \cos A + a \cos C \)
Substitute \( c = 2R \sin C \) and \( a = 2R \sin A \):
\( = 2R \sin C \cos A + 2R \sin A \cos C \)
Factor out \( 2R \):
\( = 2R (\sin C \cos A + \cos C \sin A) \)
Using the trigonometric identity for \( \sin(X+Y) \):
\( = 2R \sin(C+A) \)
Since \( C+A = 180^\circ - B \), then \( \sin(C+A) = \sin(180^\circ - B) = \sin B \):
\( = 2R \sin B \)
From the Law of Sines, we know \( b = 2R \sin B \):
\( = b \)
\( = \text{LHS} \). Thus, the second projection formula is proved.

(c) Prove \( c = a \cos B + b \cos A \)
Start with the Right-Hand Side (RHS):
\( \text{RHS} = a \cos B + b \cos A \)
Substitute \( a = 2R \sin A \) and \( b = 2R \sin B \):
\( = 2R \sin A \cos B + 2R \sin B \cos A \)
Factor out \( 2R \):
\( = 2R (\sin A \cos B + \cos A \sin B) \)
Using the trigonometric identity for \( \sin(X+Y) \):
\( = 2R \sin(A+B) \)
Since \( A+B = 180^\circ - C \), then \( \sin(A+B) = \sin(180^\circ - C) = \sin C \):
\( = 2R \sin C \)
From the Law of Sines, we know \( c = 2R \sin C \):
\( = c \)
\( = \text{LHS} \). Thus, the third projection formula is proved.

(ii) Derivation using the Law of Cosines:
The Law of Cosines states the following relationships between sides and angles:
\( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \)
\( \cos B = \frac{a^2 + c^2 - b^2}{2ac} \)
\( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \)
From these, we can express the terms in the projection formulas:
\( b^2 + c^2 - a^2 = 2bc \cos A \)
\( a^2 + c^2 - b^2 = 2ac \cos B \)
\( a^2 + b^2 - c^2 = 2ab \cos C \)

(a) Prove \( a = b \cos C + c \cos B \)
Start with the Right-Hand Side (RHS):
\( \text{RHS} = b \cos C + c \cos B \)
Substitute the Law of Cosines expressions for \( \cos C \) and \( \cos B \):
\( = b \left(\frac{a^2 + b^2 - c^2}{2ab}\right) + c \left(\frac{a^2 + c^2 - b^2}{2ac}\right) \)
Cancel out common terms (b in the first term, c in the second term):
\( = \frac{a^2 + b^2 - c^2}{2a} + \frac{a^2 + c^2 - b^2}{2a} \)
Combine the fractions since they have a common denominator:
\( = \frac{(a^2 + b^2 - c^2) + (a^2 + c^2 - b^2)}{2a} \)
Simplify the numerator by cancelling opposite terms ( \( b^2 \) and \( -b^2 \), \( -c^2 \) and \( c^2 \)):
\( = \frac{a^2 + a^2}{2a} \)
\( = \frac{2a^2}{2a} \)
\( = a \)
\( = \text{LHS} \). Thus, the first projection formula is proved.

(b) Prove \( b = c \cos A + a \cos C \)
Start with the Right-Hand Side (RHS):
\( \text{RHS} = c \cos A + a \cos C \)
Substitute the Law of Cosines expressions for \( \cos A \) and \( \cos C \):
\( = c \left(\frac{b^2 + c^2 - a^2}{2bc}\right) + a \left(\frac{a^2 + b^2 - c^2}{2ab}\right) \)
Cancel out common terms (c in the first term, a in the second term):
\( = \frac{b^2 + c^2 - a^2}{2b} + \frac{a^2 + b^2 - c^2}{2b} \)
Combine the fractions:
\( = \frac{(b^2 + c^2 - a^2) + (a^2 + b^2 - c^2)}{2b} \)
Simplify the numerator:
\( = \frac{b^2 + b^2}{2b} \)
\( = \frac{2b^2}{2b} \)
\( = b \)
\( = \text{LHS} \). Thus, the second projection formula is proved.

(c) Prove \( c = a \cos B + b \cos A \)
Start with the Right-Hand Side (RHS):
\( \text{RHS} = a \cos B + b \cos A \)
Substitute the Law of Cosines expressions for \( \cos B \) and \( \cos A \):
\( = a \left(\frac{a^2 + c^2 - b^2}{2ac}\right) + b \left(\frac{b^2 + c^2 - a^2}{2bc}\right) \)
Cancel out common terms (a in the first term, b in the second term):
\( = \frac{a^2 + c^2 - b^2}{2c} + \frac{b^2 + c^2 - a^2}{2c} \)
Combine the fractions:
\( = \frac{(a^2 + c^2 - b^2) + (b^2 + c^2 - a^2)}{2c} \)
Simplify the numerator:
\( = \frac{c^2 + c^2}{2c} \)
\( = \frac{2c^2}{2c} \)
\( = c \)
\( = \text{LHS} \). Thus, the third projection formula is proved.
These derivations show that the projection formulas are fundamental relationships in trigonometry, consistently derived from the basic laws of sines and cosines.
In simple words: We proved the projection formulas in two ways. First, using the Sine Rule, we turned sides into sines of angles and used angle sum identities. Second, using the Cosine Rule, we replaced cosine terms with side lengths and simplified them. Both methods correctly showed the formulas.

🎯 Exam Tip: When proving projection formulas, be clear whether you are using the Law of Sines or the Law of Cosines. Both methods are valid, but ensure all intermediate trigonometric identities and algebraic steps are correct and clearly shown.