Samacheer Kalvi Class 11 Business Maths Solutions Chapter 4 Trigonometry Exercise 4.3

Question 1. Express each of the following as the sum or difference of sine or cosine:
(i) \( \sin \frac{A}{8} \sin \frac{3A}{8} \)
(ii) \( \cos(60^\circ + A) \sin(120^\circ + A) \)
(iii) \( \cos \frac{7 A}{3} \sin \frac{5 A}{3} \)
(iv) \( \cos 7\theta \sin 3\theta \)
Answer:
(i) We need to express \( \sin \frac{A}{8} \sin \frac{3A}{8} \) as a sum or difference. We use the identity \( 2 \sin A \sin B = \cos(A - B) - \cos(A + B) \).
First, multiply and divide the expression by 2:
\( \sin \frac{A}{8} \sin \frac{3A}{8} = \frac{1}{2} \left[ 2 \sin \frac{A}{8} \sin \frac{3A}{8} \right] \)
Now, apply the identity:
\( = \frac{1}{2} \left[ \cos \left(\frac{A}{8} - \frac{3A}{8}\right) - \cos \left(\frac{A}{8} + \frac{3A}{8}\right) \right] \)
\( = \frac{1}{2} \left[ \cos \left(-\frac{2A}{8}\right) - \cos \left(\frac{4A}{8}\right) \right] \)
\( = \frac{1}{2} \left[ \cos \left(-\frac{A}{4}\right) - \cos \left(\frac{A}{2}\right) \right] \)
Since \( \cos(-\theta) = \cos \theta \), we get:
\( = \frac{1}{2} \left[ \cos \frac{A}{4} - \cos \frac{A}{2} \right] \)

(ii) We need to express \( \cos(60^\circ + A) \sin(120^\circ + A) \) as a sum or difference. We use the identity \( 2 \cos A \sin B = \sin(A + B) - \sin(A - B) \).
First, multiply and divide the expression by 2:
\( \cos(60^\circ + A) \sin(120^\circ + A) = \frac{1}{2} \left[ 2 \cos(60^\circ + A) \sin(120^\circ + A) \right] \)
Now, apply the identity:
\( = \frac{1}{2} \left[ \sin((60^\circ + A) + (120^\circ + A)) - \sin((60^\circ + A) - (120^\circ + A)) \right] \)
\( = \frac{1}{2} \left[ \sin(180^\circ + 2A) - \sin(60^\circ + A - 120^\circ - A) \right] \)
\( = \frac{1}{2} \left[ \sin(180^\circ + 2A) - \sin(-60^\circ) \right] \)
Using identities \( \sin(180^\circ + \theta) = -\sin \theta \) and \( \sin(-\theta) = -\sin \theta \):
\( = \frac{1}{2} \left[ -\sin 2A - (-\sin 60^\circ) \right] \)
\( = \frac{1}{2} \left[ -\sin 2A + \sin 60^\circ \right] \)
Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\( = \frac{1}{2} \left[ -\sin 2A + \frac{\sqrt{3}}{2} \right] \)

(iii) We need to express \( \cos \frac{7 A}{3} \sin \frac{5 A}{3} \) as a sum or difference. We use the identity \( 2 \cos A \sin B = \sin(A + B) - \sin(A - B) \).
First, multiply and divide the expression by 2:
\( \cos \frac{7 A}{3} \sin \frac{5 A}{3} = \frac{1}{2} \left[ 2 \cos \frac{7 A}{3} \sin \frac{5 A}{3} \right] \)
Now, apply the identity:
\( = \frac{1}{2} \left[ \sin \left(\frac{7 A}{3} + \frac{5 A}{3}\right) - \sin \left(\frac{7 A}{3} - \frac{5 A}{3}\right) \right] \)
\( = \frac{1}{2} \left[ \sin \left(\frac{12 A}{3}\right) - \sin \left(\frac{2 A}{3}\right) \right] \)
\( = \frac{1}{2} \left[ \sin 4A - \sin \frac{2 A}{3} \right] \)

(iv) We need to express \( \cos 7\theta \sin 3\theta \) as a sum or difference. We use the identity \( 2 \cos A \sin B = \sin(A + B) - \sin(A - B) \).
First, multiply and divide the expression by 2:
\( \cos 7\theta \sin 3\theta = \frac{1}{2} \left[ 2 \cos 7\theta \sin 3\theta \right] \)
Now, apply the identity:
\( = \frac{1}{2} \left[ \sin(7\theta + 3\theta) - \sin(7\theta - 3\theta) \right] \)
\( = \frac{1}{2} (sin 10\theta - sin 4\theta) \)
In simple words: To change a product of sine and cosine into a sum or difference, we use special formulas. We often need to multiply and divide by 2 first to match the formula's structure. These formulas help simplify complex trigonometric expressions.

🎯 Exam Tip: Remember the product-to-sum identities: \( 2 \sin A \sin B \), \( 2 \cos A \cos B \), \( 2 \sin A \cos B \), and \( 2 \cos A \sin B \). Practice applying them to ensure you choose the correct one for each product.

Question 2. Express each of the following as the product of sine and cosine:
(i) \( \sin A + \sin 2A \)
(ii) \( \cos 2A + \cos 4A \)
(iii) \( \sin 6\theta - \sin 2\theta \)
(iv) \( \cos 2\theta - \cos \theta \)
Answer:
(i) We need to express \( \sin A + \sin 2A \) as a product. We use the identity \( \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \).
Here, \( C=A \) and \( D=2A \).
\( \sin A + \sin 2A = 2 \sin\left(\frac{A+2A}{2}\right) \cos\left(\frac{A-2A}{2}\right) \)
\( = 2 \sin\left(\frac{3A}{2}\right) \cos\left(\frac{-A}{2}\right) \)
Since \( \cos(-\theta) = \cos \theta \), we get:
\( = 2 \sin \frac{3A}{2} \cos \frac{A}{2} \)

(ii) We need to express \( \cos 2A + \cos 4A \) as a product. We use the identity \( \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \).
Here, \( C=2A \) and \( D=4A \).
\( \cos 2A + \cos 4A = 2 \cos\left(\frac{2A+4A}{2}\right) \cos\left(\frac{2A-4A}{2}\right) \)
\( = 2 \cos\left(\frac{6A}{2}\right) \cos\left(\frac{-2A}{2}\right) \)
\( = 2 \cos 3A \cos(-A) \)
Since \( \cos(-\theta) = \cos \theta \), we get:
\( = 2 \cos 3A \cos A \)

(iii) We need to express \( \sin 6\theta - \sin 2\theta \) as a product. We use the identity \( \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \).
Here, \( C=6\theta \) and \( D=2\theta \).
\( \sin 6\theta - \sin 2\theta = 2 \cos\left(\frac{6\theta+2\theta}{2}\right) \sin\left(\frac{6\theta-2\theta}{2}\right) \)
\( = 2 \cos\left(\frac{8\theta}{2}\right) \sin\left(\frac{4\theta}{2}\right) \)
\( = 2 \cos 4\theta \sin 2\theta \)

(iv) We need to express \( \cos 2\theta - \cos \theta \) as a product. We use the identity \( \cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \).
Here, \( C=2\theta \) and \( D=\theta \).
\( \cos 2\theta - \cos \theta = -2 \sin\left(\frac{2\theta+\theta}{2}\right) \sin\left(\frac{2\theta-\theta}{2}\right) \)
\( = -2 \sin\left(\frac{3\theta}{2}\right) \sin\left(\frac{\theta}{2}\right) \)
In simple words: To turn sums or differences of sine and cosine functions into products, we use different set of formulas. These are called sum-to-product identities and help in simplifying expressions or solving equations.

🎯 Exam Tip: Pay close attention to the signs in the sum-to-product formulas, especially for \( \cos C - \cos D \), which has a negative sign in front.

Question 3. Prove that
(i) \( \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8} \)
(ii) \( \tan 20^\circ \tan 40^\circ \tan 80^\circ = \sqrt{3} \)
Answer:
(i) We need to prove that \( \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8} \).
Let's take the Left Hand Side (LHS):
\( LHS = \cos 20^\circ \cos 40^\circ \cos 80^\circ \)
We multiply and divide by \( 2 \sin 20^\circ \) to use the identity \( 2 \sin A \cos A = \sin 2A \):
\( LHS = \frac{1}{2 \sin 20^\circ} (2 \sin 20^\circ \cos 20^\circ) \cos 40^\circ \cos 80^\circ \)
\( = \frac{1}{2 \sin 20^\circ} (\sin (2 \times 20^\circ)) \cos 40^\circ \cos 80^\circ \)
\( = \frac{1}{2 \sin 20^\circ} \sin 40^\circ \cos 40^\circ \cos 80^\circ \)
Again, multiply and divide by 2 for \( \sin 40^\circ \cos 40^\circ \):
\( = \frac{1}{2 \sin 20^\circ} \frac{1}{2} (2 \sin 40^\circ \cos 40^\circ) \cos 80^\circ \)
\( = \frac{1}{4 \sin 20^\circ} (\sin (2 \times 40^\circ)) \cos 80^\circ \)
\( = \frac{1}{4 \sin 20^\circ} \sin 80^\circ \cos 80^\circ \)
Once more, multiply and divide by 2 for \( \sin 80^\circ \cos 80^\circ \):
\( = \frac{1}{4 \sin 20^\circ} \frac{1}{2} (2 \sin 80^\circ \cos 80^\circ) \)
\( = \frac{1}{8 \sin 20^\circ} (\sin (2 \times 80^\circ)) \)
\( = \frac{1}{8 \sin 20^\circ} \sin 160^\circ \)
Using the identity \( \sin(180^\circ - \theta) = \sin \theta \):
\( = \frac{1}{8 \sin 20^\circ} \sin(180^\circ - 20^\circ) \)
\( = \frac{1}{8 \sin 20^\circ} \sin 20^\circ \)
\( = \frac{1}{8} \)
This equals the Right Hand Side (RHS). Hence proved.

(ii) We need to prove that \( \tan 20^\circ \tan 40^\circ \tan 80^\circ = \sqrt{3} \).
Let's take the Left Hand Side (LHS):
\( LHS = \tan 20^\circ \tan 40^\circ \tan 80^\circ \)
We can rewrite tangent in terms of sine and cosine:
\( = \frac{\sin 20^\circ}{\cos 20^\circ} \times \frac{\sin 40^\circ}{\cos 40^\circ} \times \frac{\sin 80^\circ}{\cos 80^\circ} \)
\( = \frac{\sin 20^\circ \sin 40^\circ \sin 80^\circ}{\cos 20^\circ \cos 40^\circ \cos 80^\circ} \)
From part (i), we know the denominator: \( \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8} \).
Now let's find the value of the numerator: \( \sin 20^\circ \sin 40^\circ \sin 80^\circ \).
We can use the identity \( \sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta \).
Here, \( \theta = 20^\circ \). So, \( \sin 20^\circ \sin(60^\circ - 20^\circ) \sin(60^\circ + 20^\circ) = \sin 20^\circ \sin 40^\circ \sin 80^\circ \).
This expression equals \( \frac{1}{4} \sin(3 \times 20^\circ) = \frac{1}{4} \sin 60^\circ \).
Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\( \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8} \).
Now substitute the values of the numerator and denominator back into the LHS expression:
\( LHS = \frac{\frac{\sqrt{3}}{8}}{\frac{1}{8}} \)
\( = \sqrt{3} \)
This equals the Right Hand Side (RHS). Hence proved.
In simple words: For proving these identities, we start from one side (usually LHS) and use known trigonometric formulas to transform it step-by-step until it matches the other side (RHS). Breaking down complex terms helps simplify the proof.

🎯 Exam Tip: When proving products of sines or cosines, remember to multiply and divide by terms strategically to create expressions that fit the \( \sin 2A \) or \( \sin 3A \) formulas. This is a common technique.

Question 4. Prove that
(i) \( (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2 = 4 \sin^2 \left(\frac{\alpha-\beta}{2}\right) \)
(ii) \( \sin A \sin(60^\circ + A) \sin(60^\circ - A) = \frac{1}{4} \sin 3A \)
Answer:
(i) We need to prove that \( (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2 = 4 \sin^2 \left(\frac{\alpha-\beta}{2}\right) \).
Let's take the Left Hand Side (LHS):
\( LHS = (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2 \)
We use the sum-to-product formulas:
\( \cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \)
\( \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \)
Substitute these into the LHS:
\( LHS = \left(-2 \sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right)\right)^2 + \left(2 \cos\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right)\right)^2 \)
\( = 4 \sin^2\left(\frac{\alpha+\beta}{2}\right) \sin^2\left(\frac{\alpha-\beta}{2}\right) + 4 \cos^2\left(\frac{\alpha+\beta}{2}\right) \sin^2\left(\frac{\alpha-\beta}{2}\right) \)
Factor out the common term \( 4 \sin^2\left(\frac{\alpha-\beta}{2}\right) \):
\( = 4 \sin^2\left(\frac{\alpha-\beta}{2}\right) \left( \sin^2\left(\frac{\alpha+\beta}{2}\right) + \cos^2\left(\frac{\alpha+\beta}{2}\right) \right) \)
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\( = 4 \sin^2\left(\frac{\alpha-\beta}{2}\right) (1) \)
\( = 4 \sin^2\left(\frac{\alpha-\beta}{2}\right) \)
This equals the Right Hand Side (RHS). Hence proved.

(ii) We need to prove that \( \sin A \sin(60^\circ + A) \sin(60^\circ - A) = \frac{1}{4} \sin 3A \).
Let's take the Left Hand Side (LHS):
\( LHS = \sin A \sin(60^\circ + A) \sin(60^\circ - A) \)
We use the identity \( \sin(X+Y) \sin(X-Y) = \sin^2 X - \sin^2 Y \). Here, \( X=60^\circ \) and \( Y=A \):
\( LHS = \sin A \left( \sin^2 60^\circ - \sin^2 A \right) \)
Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), then \( \sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \):
\( = \sin A \left( \frac{3}{4} - \sin^2 A \right) \)
\( = \frac{1}{4} (3 \sin A - 4 \sin^3 A) \)
Using the triple angle identity \( \sin 3A = 3 \sin A - 4 \sin^3 A \):
\( = \frac{1}{4} \sin 3A \)
This equals the Right Hand Side (RHS). Hence proved.
In simple words: Trigonometric identities help us prove complex relationships. For the first part, we use sum-to-product formulas and the Pythagorean identity. For the second part, we use a product-to-difference formula and the triple angle identity for sine.

🎯 Exam Tip: Recognizing identities like \( \sin(X+Y) \sin(X-Y) \) is key to simplifying expressions efficiently. Always keep a list of common trigonometric formulas handy for proofs.

Question 5. Prove that
(i) \( \sin (A - B) \sin C + \sin (B - C) \sin A + \sin(C - A) \sin B = 0 \)
(ii) \( 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0 \)
Answer:
(i) We need to prove that \( \sin (A - B) \sin C + \sin (B - C) \sin A + \sin(C - A) \sin B = 0 \).
Let's expand each term using the identity \( \sin(X-Y) = \sin X \cos Y - \cos X \sin Y \):
Term 1: \( \sin (A - B) \sin C = (\sin A \cos B - \cos A \sin B) \sin C \)
\( = \sin A \cos B \sin C - \cos A \sin B \sin C \) (1)
Term 2: \( \sin (B - C) \sin A = (\sin B \cos C - \cos B \sin C) \sin A \)
\( = \sin B \cos C \sin A - \cos B \sin C \sin A \) (2)
Term 3: \( \sin (C - A) \sin B = (\sin C \cos A - \cos C \sin A) \sin B \)
\( = \sin C \cos A \sin B - \cos C \sin A \sin B \) (3)
Now, add equations (1), (2), and (3):
\( (\sin A \cos B \sin C - \cos A \sin B \sin C) \)
\( + (\sin B \cos C \sin A - \cos B \sin C \sin A) \)
\( + (\sin C \cos A \sin B - \cos C \sin A \sin B) \)
Let's look for terms that cancel each other out:
\( \sin A \cos B \sin C \) and \( - \cos B \sin C \sin A \) cancel.
\( - \cos A \sin B \sin C \) and \( \sin C \cos A \sin B \) cancel.
\( \sin B \cos C \sin A \) and \( - \cos C \sin A \sin B \) cancel.
Thus, the sum is \( 0 \).
This equals the Right Hand Side (RHS). Hence proved.

(ii) We need to prove that \( 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0 \).
Let's take the Left Hand Side (LHS):
\( LHS = 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \)
First, apply the product-to-sum formula \( 2 \cos X \cos Y = \cos(X+Y) + \cos(X-Y) \) to the first term:
\( 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} = \cos\left(\frac{\pi}{13} + \frac{9 \pi}{13}\right) + \cos\left(\frac{\pi}{13} - \frac{9 \pi}{13}\right) \)
\( = \cos\left(\frac{10 \pi}{13}\right) + \cos\left(-\frac{8 \pi}{13}\right) \)
Since \( \cos(-\theta) = \cos \theta \):
\( = \cos\left(\frac{10 \pi}{13}\right) + \cos\left(\frac{8 \pi}{13}\right) \)
So, the LHS becomes:
\( LHS = \cos\left(\frac{10 \pi}{13}\right) + \cos\left(\frac{8 \pi}{13}\right) + \cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \)
Now, we know that \( \cos(\pi - \theta) = -\cos \theta \).
So, \( \cos\left(\frac{10 \pi}{13}\right) = \cos\left(\pi - \frac{3 \pi}{13}\right) = -\cos\left(\frac{3 \pi}{13}\right) \)
And \( \cos\left(\frac{8 \pi}{13}\right) = \cos\left(\pi - \frac{5 \pi}{13}\right) = -\cos\left(\frac{5 \pi}{13}\right) \)
Substitute these values back into the LHS:
\( LHS = -\cos\left(\frac{3 \pi}{13}\right) - \cos\left(\frac{5 \pi}{13}\right) + \cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \)
\( = 0 \)
This equals the Right Hand Side (RHS). Hence proved.
In simple words: For the first proof, expanding each term and carefully checking for cancellations leads to the result. For the second, we transform the product into a sum and then use properties of cosine to show that terms cancel out.

🎯 Exam Tip: When terms involve \( \pi \) and a denominator, look for ways to rewrite them using \( \pi - \theta \) or \( \pi + \theta \) to find complementary or supplementary angles that might cancel out.

Question 6. Prove that
(i) \( \frac{\cos 2 A-\cos 3 A}{\sin 2 A+\sin 3 A}=\tan \frac{A}{2} \)
(ii) \( \frac{\cos 7 A+\cos 5 A}{\sin 7 A-\sin 5 A}=\cot A \)
Answer:
(i) We need to prove that \( \frac{\cos 2 A-\cos 3 A}{\sin 2 A+\sin 3 A}=\tan \frac{A}{2} \).
Let's take the Left Hand Side (LHS):
\( LHS = \frac{\cos 2 A-\cos 3 A}{\sin 2 A+\sin 3 A} \)
Use sum-to-product formulas for the numerator and denominator:
Numerator: \( \cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \)
Here \( C=2A, D=3A \). So, \( \cos 2A - \cos 3A = -2 \sin\left(\frac{2A+3A}{2}\right) \sin\left(\frac{2A-3A}{2}\right) = -2 \sin\left(\frac{5A}{2}\right) \sin\left(-\frac{A}{2}\right) \).
Denominator: \( \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \)
Here \( C=2A, D=3A \). So, \( \sin 2A + \sin 3A = 2 \sin\left(\frac{2A+3A}{2}\right) \cos\left(\frac{2A-3A}{2}\right) = 2 \sin\left(\frac{5A}{2}\right) \cos\left(-\frac{A}{2}\right) \).
Now substitute these back into the LHS:
\( LHS = \frac{-2 \sin\left(\frac{5A}{2}\right) \sin\left(-\frac{A}{2}\right)}{2 \sin\left(\frac{5A}{2}\right) \cos\left(-\frac{A}{2}\right)} \)
Cancel out \( 2 \sin\left(\frac{5A}{2}\right) \):
\( = \frac{-\sin\left(-\frac{A}{2}\right)}{\cos\left(-\frac{A}{2}\right)} \)
Using identities \( \sin(-\theta) = -\sin \theta \) and \( \cos(-\theta) = \cos \theta \):
\( = \frac{-(-\sin\left(\frac{A}{2}\right))}{\cos\left(\frac{A}{2}\right)} \)
\( = \frac{\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right)} \)
\( = \tan \frac{A}{2} \)
This equals the Right Hand Side (RHS). Hence proved.

(ii) We need to prove that \( \frac{\cos 7 A+\cos 5 A}{\sin 7 A-\sin 5 A}=\cot A \).
Let's take the Left Hand Side (LHS):
\( LHS = \frac{\cos 7 A+\cos 5 A}{\sin 7 A-\sin 5 A} \)
Use sum-to-product formulas for the numerator and denominator:
Numerator: \( \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \)
Here \( C=7A, D=5A \). So, \( \cos 7A + \cos 5A = 2 \cos\left(\frac{7A+5A}{2}\right) \cos\left(\frac{7A-5A}{2}\right) = 2 \cos\left(\frac{12A}{2}\right) \cos\left(\frac{2A}{2}\right) = 2 \cos 6A \cos A \).
Denominator: \( \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \)
Here \( C=7A, D=5A \). So, \( \sin 7A - \sin 5A = 2 \cos\left(\frac{7A+5A}{2}\right) \sin\left(\frac{7A-5A}{2}\right) = 2 \cos\left(\frac{12A}{2}\right) \sin\left(\frac{2A}{2}\right) = 2 \cos 6A \sin A \).
Now substitute these back into the LHS:
\( LHS = \frac{2 \cos 6A \cos A}{2 \cos 6A \sin A} \)
Cancel out \( 2 \cos 6A \):
\( = \frac{\cos A}{\sin A} \)
\( = \cot A \)
This equals the Right Hand Side (RHS). Hence proved.
In simple words: These proofs involve using specific formulas to change sums and differences into products. Simplifying the expressions by canceling common terms often leads to the final identity.

🎯 Exam Tip: Carefully identify the correct sum-to-product formula for each part of the fraction. Make sure to simplify trigonometric terms like \( \sin(-\theta) \) and \( \cos(-\theta) \) correctly.

Question 7. Prove that \( \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16} \).
Answer:
We need to prove that \( \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16} \).
Let's take the Left Hand Side (LHS):
\( LHS = \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ \)
We know that \( \cos 60^\circ = \frac{1}{2} \). Substitute this value:
\( LHS = \cos 20^\circ \cos 40^\circ \left(\frac{1}{2}\right) \cos 80^\circ \)
\( = \frac{1}{2} (\cos 20^\circ \cos 40^\circ \cos 80^\circ) \)
Now, let's focus on the term \( \cos 20^\circ \cos 40^\circ \cos 80^\circ \).
Multiply and divide by \( 2 \sin 20^\circ \) to use the identity \( 2 \sin A \cos A = \sin 2A \):
\( \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{2 \sin 20^\circ} (2 \sin 20^\circ \cos 20^\circ) \cos 40^\circ \cos 80^\circ \)
\( = \frac{1}{2 \sin 20^\circ} (\sin 40^\circ) \cos 40^\circ \cos 80^\circ \)
Again, multiply and divide by 2 for \( \sin 40^\circ \cos 40^\circ \):
\( = \frac{1}{2 \sin 20^\circ} \frac{1}{2} (2 \sin 40^\circ \cos 40^\circ) \cos 80^\circ \)
\( = \frac{1}{4 \sin 20^\circ} (\sin 80^\circ) \cos 80^\circ \)
Once more, multiply and divide by 2 for \( \sin 80^\circ \cos 80^\circ \):
\( = \frac{1}{4 \sin 20^\circ} \frac{1}{2} (2 \sin 80^\circ \cos 80^\circ) \)
\( = \frac{1}{8 \sin 20^\circ} (\sin 160^\circ) \)
Using the identity \( \sin(180^\circ - \theta) = \sin \theta \):
\( = \frac{1}{8 \sin 20^\circ} \sin(180^\circ - 20^\circ) \)
\( = \frac{1}{8 \sin 20^\circ} \sin 20^\circ \)
\( = \frac{1}{8} \)
Now, substitute this value back into the original LHS expression:
\( LHS = \frac{1}{2} \times \frac{1}{8} \)
\( = \frac{1}{16} \)
This equals the Right Hand Side (RHS). Hence proved.
In simple words: We used a known value for \( \cos 60^\circ \) and then repeatedly applied the \( 2 \sin A \cos A = \sin 2A \) identity to simplify the product of cosines until we reached the final value.

🎯 Exam Tip: When evaluating trigonometric products with angles like \( \theta, 60^\circ - \theta, 60^\circ + \theta \), remember the general formula \( \cos \theta \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \frac{1}{4} \cos 3\theta \). This can greatly shorten your steps.

Question 8. Evaluate:
(i) \( \cos 20^\circ + \cos 100^\circ + \cos 140^\circ \)
(ii) \( \sin 50^\circ - \sin 70^\circ + \sin 10^\circ \)
Answer:
(i) We need to evaluate \( \cos 20^\circ + \cos 100^\circ + \cos 140^\circ \).
Let's group the first two terms and use the sum-to-product formula \( \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \):
\( LHS = (\cos 20^\circ + \cos 100^\circ) + \cos 140^\circ \)
\( = 2 \cos\left(\frac{20^\circ+100^\circ}{2}\right) \cos\left(\frac{20^\circ-100^\circ}{2}\right) + \cos 140^\circ \)
\( = 2 \cos\left(\frac{120^\circ}{2}\right) \cos\left(\frac{-80^\circ}{2}\right) + \cos 140^\circ \)
\( = 2 \cos 60^\circ \cos(-40^\circ) + \cos 140^\circ \)
Since \( \cos 60^\circ = \frac{1}{2} \) and \( \cos(-\theta) = \cos \theta \):
\( = 2 \left(\frac{1}{2}\right) \cos 40^\circ + \cos 140^\circ \)
\( = \cos 40^\circ + \cos 140^\circ \)
Now, we know that \( \cos(180^\circ - \theta) = -\cos \theta \). So, \( \cos 140^\circ = \cos(180^\circ - 40^\circ) = -\cos 40^\circ \).
\( = \cos 40^\circ + (-\cos 40^\circ) \)
\( = 0 \)

(ii) We need to evaluate \( \sin 50^\circ - \sin 70^\circ + \sin 10^\circ \).
Let's group the first two terms and use the sum-to-product formula \( \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \):
\( LHS = (\sin 50^\circ - \sin 70^\circ) + \sin 10^\circ \)
\( = 2 \cos\left(\frac{50^\circ+70^\circ}{2}\right) \sin\left(\frac{50^\circ-70^\circ}{2}\right) + \sin 10^\circ \)
\( = 2 \cos\left(\frac{120^\circ}{2}\right) \sin\left(\frac{-20^\circ}{2}\right) + \sin 10^\circ \)
\( = 2 \cos 60^\circ \sin(-10^\circ) + \sin 10^\circ \)
Since \( \cos 60^\circ = \frac{1}{2} \) and \( \sin(-\theta) = -\sin \theta \):
\( = 2 \left(\frac{1}{2}\right) (-\sin 10^\circ) + \sin 10^\circ \)
\( = -\sin 10^\circ + \sin 10^\circ \)
\( = 0 \)
In simple words: To evaluate these sums, we group terms and apply sum-to-product formulas. Using trigonometric identities like \( \cos(-\theta) = \cos \theta \) or \( \sin(-\theta) = -\sin \theta \) and angle relations like \( \cos(180^\circ - \theta) = -\cos \theta \) helps simplify the expression until it becomes zero.

🎯 Exam Tip: When evaluating expressions with three terms, try to combine two terms using a sum-to-product formula. Look for angles that might simplify to common values like \( 60^\circ \) or that are supplementary or complementary.

Question 9. If \( \cos A + \cos B = \frac{1}{2} \) and \( \sin A + \sin B = \frac{1}{4} \), prove that \( \tan \left(\frac{A+B}{2}\right)=\frac{1}{2} \).
Answer:
We are given two equations:
1. \( \cos A + \cos B = \frac{1}{2} \)
2. \( \sin A + \sin B = \frac{1}{4} \)
We need to prove that \( \tan \left(\frac{A+B}{2}\right)=\frac{1}{2} \).
Let's apply the sum-to-product formulas to the given equations:
For equation (1), using \( \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \):
\( 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) = \frac{1}{2} \) (Equation 3)
For equation (2), using \( \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \):
\( 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) = \frac{1}{4} \) (Equation 4)
Now, divide Equation 4 by Equation 3:
\( \frac{2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)}{2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)} = \frac{\frac{1}{4}}{\frac{1}{2}} \)
The term \( 2 \cos\left(\frac{A-B}{2}\right) \) cancels from the numerator and denominator:
\( \frac{\sin\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A+B}{2}\right)} = \frac{1}{4} \times \frac{2}{1} \)
\( \tan\left(\frac{A+B}{2}\right) = \frac{2}{4} \)
\( \tan\left(\frac{A+B}{2}\right) = \frac{1}{2} \)
Hence proved. This shows the desired relationship between the given sums and the tangent of the half-angle.
In simple words: We used sum-to-product formulas for sine and cosine to change the given additions into multiplications. Then, by dividing the two new equations, we could easily find the value of the tangent of the half-angle.

🎯 Exam Tip: When you have sums of sines and cosines and need to find a tangent of a half-angle, converting to product forms and then dividing the equations is a standard and effective strategy.

Question 10. If \( \sin(y + z - x) \), \( \sin(z + x - y) \), \( \sin(x + y - z) \) are in A.P, then prove that \( \tan x, \tan y \text{ and } \tan z \) are in A.P.
Answer:
If three terms \( a, b, c \) are in Arithmetic Progression (A.P.), then the middle term is the average of the other two, i.e., \( 2b = a + c \).
Given that \( \sin(y + z - x) \), \( \sin(z + x - y) \), \( \sin(x + y - z) \) are in A.P.
So, \( 2 \sin(z + x - y) = \sin(y + z - x) + \sin(x + y - z) \)
Let's apply the sum-to-product formula \( \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \) to the RHS.
Let \( C = y + z - x \) and \( D = x + y - z \).
\( C+D = (y+z-x) + (x+y-z) = 2y \)
\( C-D = (y+z-x) - (x+y-z) = y+z-x-x-y+z = 2z - 2x = 2(z-x) \)
So, \( \sin(y + z - x) + \sin(x + y - z) = 2 \sin\left(\frac{2y}{2}\right) \cos\left(\frac{2(z-x)}{2}\right) = 2 \sin y \cos(z-x) \).
Therefore, the equation becomes:
\( 2 \sin(z + x - y) = 2 \sin y \cos(z-x) \)
\( \sin(z + x - y) = \sin y \cos(z-x) \)
Now, expand both sides using \( \sin(A-B) = \sin A \cos B - \cos A \sin B \) and \( \cos(A-B) = \cos A \cos B + \sin A \sin B \):
\( \sin((z+x) - y) = \sin y (\cos z \cos x + \sin z \sin x) \)
\( \sin(z+x)\cos y - \cos(z+x)\sin y = \sin y \cos z \cos x + \sin y \sin z \sin x \)
Expand \( \sin(z+x) \) and \( \cos(z+x) \):
\( (\sin z \cos x + \cos z \sin x)\cos y - (\cos z \cos x - \sin z \sin x)\sin y \)
\( = \sin y \cos z \cos x + \sin y \sin z \sin x \)
\( \sin z \cos x \cos y + \cos z \sin x \cos y - \cos z \cos x \sin y + \sin z \sin x \sin y \)
\( = \sin y \cos z \cos x + \sin y \sin z \sin x \)
Now, divide the entire equation by \( \cos x \cos y \cos z \) (assuming none of these are zero):
\( \frac{\sin z \cos x \cos y}{\cos x \cos y \cos z} + \frac{\cos z \sin x \cos y}{\cos x \cos y \cos z} - \frac{\cos z \cos x \sin y}{\cos x \cos y \cos z} + \frac{\sin z \sin x \sin y}{\cos x \cos y \cos z} \)
\( = \frac{\sin y \cos z \cos x}{\cos x \cos y \cos z} + \frac{\sin y \sin z \sin x}{\cos x \cos y \cos z} \)
Simplify each term:
\( \tan z + \tan x - \tan y + \frac{\sin z \sin x \sin y}{\cos x \cos y \cos z} = \tan y + \frac{\sin y \sin z \sin x}{\cos x \cos y \cos z} \)
Notice the last term on both sides is the same, so they cancel out.
\( \tan z + \tan x - \tan y = \tan y \)
\( \tan x + \tan z = 2 \tan y \)
Rearranging this, we get \( \tan x - \tan y = \tan y - \tan z \).
This means \( \tan x, \tan y, \text{ and } \tan z \) are in A.P.
Hence proved. This problem combines the concept of Arithmetic Progression with trigonometric identities.
In simple words: If numbers are in A.P., the middle number is the average of the first and last. We use this rule with the given sine expressions. Then, we expand and simplify the trigonometric terms. By dividing by \( \cos x \cos y \cos z \), we convert the sines and cosines to tangents and show that they also form an A.P.

🎯 Exam Tip: For A.P. problems, always start with the definition \( 2b = a+c \). When dealing with trigonometric expressions, be ready to expand compound angles and simplify by dividing by product of cosines to convert to tangents.

Question 11. If \( \operatorname{cosec} A + \sec A = \operatorname{cosec} B + \sec B \) prove that \( \cot\left(\frac{A+B}{2}\right) = \tan A \tan B \).
Answer:
We are given \( \operatorname{cosec} A + \sec A = \operatorname{cosec} B + \sec B \).
We need to prove that \( \cot\left(\frac{A+B}{2}\right) = \tan A \tan B \).
Let's rewrite the given equation in terms of sine and cosine:
\( \frac{1}{\sin A} + \frac{1}{\cos A} = \frac{1}{\sin B} + \frac{1}{\cos B} \)
Combine terms on each side:
\( \frac{\cos A + \sin A}{\sin A \cos A} = \frac{\cos B + \sin B}{\sin B \cos B} \)
Rearrange the terms to group sines and cosines:
\( (\cos A + \sin A) \sin B \cos B = (\cos B + \sin B) \sin A \cos A \)
\( \cos A \sin B \cos B + \sin A \sin B \cos B = \cos B \sin A \cos A + \sin B \sin A \cos A \)
Move terms to one side:
\( \cos A \sin B \cos B - \cos B \sin A \cos A = \sin B \sin A \cos A - \sin A \sin B \cos B \)
Factor out common terms:
\( \cos A \cos B (\sin B - \sin A) = \sin A \sin B (\cos A - \cos B) \)
Now, isolate the terms to form \( \tan A \tan B \) on one side and a fraction of differences on the other:
\( \frac{\sin B - \sin A}{\cos A - \cos B} = \frac{\sin A \sin B}{\cos A \cos B} \)
We know that \( \frac{\sin A \sin B}{\cos A \cos B} = \tan A \tan B \). So the RHS is proven.
Now, let's simplify the LHS using sum-to-product formulas:
LHS \( = \frac{\sin B - \sin A}{\cos A - \cos B} \)
Numerator: \( \sin B - \sin A = 2 \cos\left(\frac{B+A}{2}\right) \sin\left(\frac{B-A}{2}\right) \)
Denominator: \( \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \)
Substitute these into the LHS:
\( = \frac{2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{B-A}{2}\right)}{-2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)} \)
We know that \( \sin\left(\frac{B-A}{2}\right) = -\sin\left(\frac{A-B}{2}\right) \). So, the numerator can be written as \( -2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \).
\( = \frac{-2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)}{-2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)} \)
Cancel out \( -2 \sin\left(\frac{A-B}{2}\right) \):
\( = \frac{\cos\left(\frac{A+B}{2}\right)}{\sin\left(\frac{A+B}{2}\right)} \)
\( = \cot\left(\frac{A+B}{2}\right) \)
So, we have \( \cot\left(\frac{A+B}{2}\right) = \tan A \tan B \).
Hence proved. This demonstrates a relationship derived from initial conditions involving cosecant and secant.
In simple words: We start by changing cosecant and secant into sine and cosine. Then, we rearrange the terms and use sum-to-product formulas for sine and cosine differences. This helps simplify the expression on one side to cotangent of a half-angle and the other side to a product of tangents.

🎯 Exam Tip: When given an equation with cosecants and secants, always convert them to sines and cosines first. This allows you to apply standard sum-to-product or other identities more easily.