Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.4

Question 1. If \( \sin x = \frac{15}{17} \) and \( \cos y = \frac{12}{13} \), \( 0 < x < \frac{\pi}{2} \), \( 0 < y < \frac{\pi}{2} \), find the value of
(i) \( \sin(x + y) \)
(ii) \( \cos(x - y) \)
(iii) \( \tan(x + y) \)
Answer:
Given \( \sin x = \frac{15}{17} \) and \( 0 < x < \frac{\pi}{2} \). Since x is in the first quadrant, \( \cos x \) is positive.
We know that \( \cos^2 x = 1 - \sin^2 x \).
So, \( \cos^2 x = 1 - \left(\frac{15}{17}\right)^{2} \)
\( \cos^2 x = 1 - \frac{225}{289} \)
\( \cos^2 x = \frac{289 - 225}{289} \)
\( \cos^2 x = \frac{64}{289} \)
\( \cos x = \sqrt{\frac{64}{289}} = \frac{8}{17} \) (Since x is in the first quadrant, \( \cos x \) is positive).
Also given \( \cos y = \frac{12}{13} \) and \( 0 < y < \frac{\pi}{2} \). Since y is in the first quadrant, \( \sin y \) is positive.
We know that \( \sin^2 y = 1 - \cos^2 y \).
So, \( \sin^2 y = 1 - \left(\frac{12}{13}\right)^{2} \)
\( \sin^2 y = 1 - \frac{144}{169} \)
\( \sin^2 y = \frac{169 - 144}{169} \)
\( \sin^2 y = \frac{25}{169} \)
\( \sin y = \sqrt{\frac{25}{169}} = \frac{5}{13} \) (Since y is in the first quadrant, \( \sin y \) is positive).
Now we can find the values of the expressions:

(i) To find \( \sin(x + y) \):
We use the formula \( \sin(x + y) = \sin x \cos y + \cos x \sin y \).
Substitute the values:
\( \sin(x + y) = \left(\frac{15}{17}\right) \left(\frac{12}{13}\right) + \left(\frac{8}{17}\right) \left(\frac{5}{13}\right) \)
\( \sin(x + y) = \frac{180}{221} + \frac{40}{221} \)
\( \sin(x + y) = \frac{180 + 40}{221} = \frac{220}{221} \)

(ii) To find \( \cos(x - y) \):
We use the formula \( \cos(x - y) = \cos x \cos y + \sin x \sin y \).
Substitute the values:
\( \cos(x - y) = \left(\frac{8}{17}\right) \left(\frac{12}{13}\right) + \left(\frac{15}{17}\right) \left(\frac{5}{13}\right) \)
\( \cos(x - y) = \frac{96}{221} + \frac{75}{221} \)
\( \cos(x - y) = \frac{96 + 75}{221} = \frac{171}{221} \)

(iii) To find \( \tan(x + y) \):
First, find \( \tan x \) and \( \tan y \).
\( \tan x = \frac{\sin x}{\cos x} = \frac{15/17}{8/17} = \frac{15}{8} \)
\( \tan y = \frac{\sin y}{\cos y} = \frac{5/13}{12/13} = \frac{5}{12} \)
Now, use the formula \( \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \).
Substitute the values:
\( \tan(x + y) = \frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8} \cdot \frac{5}{12}} \)
\( \tan(x + y) = \frac{\frac{45 + 10}{24}}{1 - \frac{75}{96}} \)
\( \tan(x + y) = \frac{\frac{55}{24}}{\frac{96 - 75}{96}} \)
\( \tan(x + y) = \frac{55}{24} \cdot \frac{96}{21} \)
\( \tan(x + y) = \frac{55 \cdot 4}{21} = \frac{220}{21} \)
In simple words: First, we find the missing side values using the Pythagorean identity because the angles are in the first quadrant. Then, we use special trigonometry formulas to combine these values to get the sine, cosine, and tangent of the sum or difference of the angles.

🎯 Exam Tip: Always check the quadrant of the angles to determine the correct sign (+ or -) for sine, cosine, and tangent values. A common mistake is to overlook the quadrant information, which can lead to incorrect signs in the final answer.

Question 2. If \( \sin A = \frac{3}{5} \) and \( \cos B = \frac{9}{41} \), \( 0 < A < \frac{\pi}{2} \), \( 0 < B < \frac{\pi}{2} \), find the value of
(i) \( \sin(A + B) \)
(ii) \( \cos(A - B) \)
Answer:
Given \( \sin A = \frac{3}{5} \) and \( 0 < A < \frac{\pi}{2} \). Since A is in the first quadrant, \( \cos A \) is positive.
We know that \( \cos^2 A = 1 - \sin^2 A \).
So, \( \cos^2 A = 1 - \left(\frac{3}{5}\right)^{2} \)
\( \cos^2 A = 1 - \frac{9}{25} \)
\( \cos^2 A = \frac{25 - 9}{25} \)
\( \cos^2 A = \frac{16}{25} \)
\( \cos A = \sqrt{\frac{16}{25}} = \frac{4}{5} \) (Since A is in the first quadrant, \( \cos A \) is positive).
Also given \( \cos B = \frac{9}{41} \) and \( 0 < B < \frac{\pi}{2} \). Since B is in the first quadrant, \( \sin B \) is positive.
We know that \( \sin^2 B = 1 - \cos^2 B \).
So, \( \sin^2 B = 1 - \left(\frac{9}{41}\right)^{2} \)
\( \sin^2 B = 1 - \frac{81}{1681} \)
\( \sin^2 B = \frac{1681 - 81}{1681} \)
\( \sin^2 B = \frac{1600}{1681} \)
\( \sin B = \sqrt{\frac{1600}{1681}} = \frac{40}{41} \) (Since B is in the first quadrant, \( \sin B \) is positive).

(i) To find \( \sin(A + B) \):
We use the formula \( \sin(A + B) = \sin A \cos B + \cos A \sin B \).
Substitute the values:
\( \sin(A + B) = \left(\frac{3}{5}\right) \left(\frac{9}{41}\right) + \left(\frac{4}{5}\right) \left(\frac{40}{41}\right) \)
\( \sin(A + B) = \frac{27}{205} + \frac{160}{205} \)
\( \sin(A + B) = \frac{27 + 160}{205} = \frac{187}{205} \)

(ii) To find \( \cos(A - B) \):
We use the formula \( \cos(A - B) = \cos A \cos B + \sin A \sin B \).
Substitute the values:
\( \cos(A - B) = \left(\frac{4}{5}\right) \left(\frac{9}{41}\right) + \left(\frac{3}{5}\right) \left(\frac{40}{41}\right) \)
\( \cos(A - B) = \frac{36}{205} + \frac{120}{205} \)
\( \cos(A - B) = \frac{36 + 120}{205} = \frac{156}{205} \)
In simple words: First, we find the missing trigonometric ratios for angles A and B using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), making sure to choose the correct sign based on the quadrant. Then, we apply the compound angle formulas for sine and cosine to solve the problem.

🎯 Exam Tip: Remember to always clarify the quadrant of the given angles. This step is crucial for determining whether sine, cosine, or tangent values should be positive or negative when calculating the missing trigonometric ratios.

Question 3. Find \( \cos(x - y) \), given that \( \cos x = -\frac{4}{5} \) with \( \pi < x < \frac{3 \pi}{2} \) and \( \sin y = -\frac{24}{25} \) with \( \pi < y < \frac{3 \pi}{2} \).
Answer:
Given \( \cos x = -\frac{4}{5} \) and \( \pi < x < \frac{3 \pi}{2} \). Since x is in the third quadrant, \( \sin x \) is negative.
We know that \( \cos^2 x + \sin^2 x = 1 \).
So, \( \left(-\frac{4}{5}\right)^{2} + \sin^2 x = 1 \)
\( \frac{16}{25} + \sin^2 x = 1 \)
\( \sin^2 x = 1 - \frac{16}{25} \)
\( \sin^2 x = \frac{25 - 16}{25} \)
\( \sin^2 x = \frac{9}{25} \)
\( \sin x = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \) (Since x is in the third quadrant, \( \sin x \) is negative).
Also given \( \sin y = -\frac{24}{25} \) with \( \pi < y < \frac{3 \pi}{2} \). Since y is in the third quadrant, \( \cos y \) is negative.
We know that \( \sin^2 y + \cos^2 y = 1 \).
So, \( \left(-\frac{24}{25}\right)^{2} + \cos^2 y = 1 \)
\( \frac{576}{625} + \cos^2 y = 1 \)
\( \cos^2 y = 1 - \frac{576}{625} \)
\( \cos^2 y = \frac{625 - 576}{625} \)
\( \cos^2 y = \frac{49}{625} \)
\( \cos y = -\sqrt{\frac{49}{625}} = -\frac{7}{25} \) (Since y is in the third quadrant, \( \cos y \) is negative).

Now, to find \( \cos(x - y) \):
We use the formula \( \cos(x - y) = \cos x \cos y + \sin x \sin y \).
Substitute the values:
\( \cos(x - y) = \left(-\frac{4}{5}\right) \left(-\frac{7}{25}\right) + \left(-\frac{3}{5}\right) \left(-\frac{24}{25}\right) \)
\( \cos(x - y) = \frac{28}{125} + \frac{72}{125} \)
\( \cos(x - y) = \frac{28 + 72}{125} = \frac{100}{125} = \frac{4}{5} \)
In simple words: We find the missing sine and cosine values for x and y, remembering that both angles are in the third quadrant where both sine and cosine are negative. Then, we use the cosine difference formula to combine these values and get the final answer.

🎯 Exam Tip: Pay extra attention to the quadrant given for each angle. In the third quadrant, both sine and cosine are negative, which is crucial for correctly calculating their values using the Pythagorean identity.

Question 4. Find \( \sin(x - y) \), given that \( \sin x = \frac{8}{17} \) with \( 0 < x < \frac{\pi}{2} \), and \( \cos y = -\frac{24}{25} \) with \( \pi < y < \frac{3 \pi}{2} \).
Answer:
Given \( \sin x = \frac{8}{17} \) and \( 0 < x < \frac{\pi}{2} \). Since x is in the first quadrant, \( \cos x \) is positive.
We know that \( \sin^2 x + \cos^2 x = 1 \).
So, \( \left(\frac{8}{17}\right)^{2} + \cos^2 x = 1 \)
\( \frac{64}{289} + \cos^2 x = 1 \)
\( \cos^2 x = 1 - \frac{64}{289} \)
\( \cos^2 x = \frac{289 - 64}{289} \)
\( \cos^2 x = \frac{225}{289} \)
\( \cos x = \sqrt{\frac{225}{289}} = \frac{15}{17} \) (Since x is in the first quadrant, \( \cos x \) is positive).
Also given \( \cos y = -\frac{24}{25} \) with \( \pi < y < \frac{3 \pi}{2} \). Since y is in the third quadrant, \( \sin y \) is negative.
We know that \( \sin^2 y + \cos^2 y = 1 \).
So, \( \sin^2 y + \left(-\frac{24}{25}\right)^{2} = 1 \)
\( \sin^2 y + \frac{576}{625} = 1 \)
\( \sin^2 y = 1 - \frac{576}{625} \)
\( \sin^2 y = \frac{625 - 576}{625} \)
\( \sin^2 y = \frac{49}{625} \)
\( \sin y = -\sqrt{\frac{49}{625}} = -\frac{7}{25} \) (Since y is in the third quadrant, \( \sin y \) is negative).

Now, to find \( \sin(x - y) \):
We use the formula \( \sin(x - y) = \sin x \cos y - \cos x \sin y \).
Substitute the values:
\( \sin(x - y) = \left(\frac{8}{17}\right) \left(-\frac{24}{25}\right) - \left(\frac{15}{17}\right) \left(-\frac{7}{25}\right) \)
\( \sin(x - y) = -\frac{192}{425} - \left(-\frac{105}{425}\right) \)
\( \sin(x - y) = -\frac{192}{425} + \frac{105}{425} \)
\( \sin(x - y) = \frac{-192 + 105}{425} = -\frac{87}{425} \)
In simple words: First, we find the missing cosine for x (positive in Quadrant I) and the missing sine for y (negative in Quadrant III). Then, we plug these values into the sine difference formula to calculate the final answer.

🎯 Exam Tip: It is easy to make sign errors in these calculations. Always double-check the quadrant of each angle to ensure that you use the correct positive or negative sign for sine and cosine values.

Question 5. Find the value of
(i) \( \cos 105^\circ \)
(ii) \( \sin 105^\circ \)
(iii) \( \tan \frac{7 \pi}{12} \)
Answer:
(i) To find \( \cos 105^\circ \):
We can write \( 105^\circ \) as \( 60^\circ + 45^\circ \) or \( 90^\circ + 15^\circ \). Let's use \( 60^\circ + 45^\circ \).
Using the formula \( \cos(A + B) = \cos A \cos B - \sin A \sin B \):
\( \cos 105^\circ = \cos(60^\circ + 45^\circ) \)
\( \cos 105^\circ = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ \)
\( \cos 105^\circ = \left(\frac{1}{2}\right) \left(\frac{1}{\sqrt{2}}\right) - \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{\sqrt{2}}\right) \)
\( \cos 105^\circ = \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} \)
\( \cos 105^\circ = \frac{1 - \sqrt{3}}{2\sqrt{2}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \):
\( \cos 105^\circ = \frac{(1 - \sqrt{3})\sqrt{2}}{2\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{2} - \sqrt{6}}{4} \)

(ii) To find \( \sin 105^\circ \):
We can write \( 105^\circ \) as \( 60^\circ + 45^\circ \).
Using the formula \( \sin(A + B) = \sin A \cos B + \cos A \sin B \):
\( \sin 105^\circ = \sin(60^\circ + 45^\circ) \)
\( \sin 105^\circ = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \)
\( \sin 105^\circ = \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{\sqrt{2}}\right) \)
\( \sin 105^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \)
\( \sin 105^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \):
\( \sin 105^\circ = \frac{(\sqrt{3} + 1)\sqrt{2}}{2\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4} \)

(iii) To find \( \tan \frac{7 \pi}{12} \):
First, convert radians to degrees: \( \frac{7 \pi}{12} = \frac{7 \times 180^\circ}{12} = 7 \times 15^\circ = 105^\circ \).
So, we need to find \( \tan 105^\circ \).
We can write \( 105^\circ \) as \( 60^\circ + 45^\circ \).
Using the formula \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \):
\( \tan 105^\circ = \tan(60^\circ + 45^\circ) \)
\( \tan 105^\circ = \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ} \)
\( \tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} \)
\( \tan 105^\circ = \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \)
To rationalize the denominator, multiply by the conjugate \( \frac{1 + \sqrt{3}}{1 + \sqrt{3}} \):
\( \tan 105^\circ = \frac{(1 + \sqrt{3})(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \)
\( \tan 105^\circ = \frac{1^2 + 2\sqrt{3} + (\sqrt{3})^2}{1^2 - (\sqrt{3})^2} \)
\( \tan 105^\circ = \frac{1 + 2\sqrt{3} + 3}{1 - 3} \)
\( \tan 105^\circ = \frac{4 + 2\sqrt{3}}{-2} \)
\( \tan 105^\circ = - (2 + \sqrt{3}) \)
In simple words: To find these values, we break down the angle \( 105^\circ \) into two common angles like \( 60^\circ \) and \( 45^\circ \). Then, we use the sum formulas for sine, cosine, and tangent and simplify the expressions to get the final exact values.

🎯 Exam Tip: When dealing with special angles like \( 105^\circ \) or \( \frac{7\pi}{12} \), always try to express them as a sum or difference of standard angles (like \( 30^\circ, 45^\circ, 60^\circ, 90^\circ \)) for which you know the trigonometric values. Remember to rationalize the denominator in your final answer.

Question 6. Prove that
(i) \( \cos (30^\circ + x) = \frac{\sqrt{3} \cos x-\sin x}{2} \)
(ii) \( \cos (\pi + \theta) = -\cos \theta \)
(iii) \( \sin (\pi + \theta) = -\sin \theta \)
Answer:
(i) To prove \( \cos (30^\circ + x) = \frac{\sqrt{3} \cos x-\sin x}{2} \):
We start with the Left Hand Side (LHS): \( \cos (30^\circ + x) \)
Using the compound angle formula \( \cos(A + B) = \cos A \cos B - \sin A \sin B \):
\( \cos (30^\circ + x) = \cos 30^\circ \cos x - \sin 30^\circ \sin x \)
We know \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) and \( \sin 30^\circ = \frac{1}{2} \).
Substitute these values:
\( \cos (30^\circ + x) = \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x \)
\( \cos (30^\circ + x) = \frac{\sqrt{3} \cos x - \sin x}{2} \)
This is equal to the Right Hand Side (RHS). Hence proved.

(ii) To prove \( \cos (\pi + \theta) = -\cos \theta \):
We start with the Left Hand Side (LHS): \( \cos (\pi + \theta) \)
Using the compound angle formula \( \cos(A + B) = \cos A \cos B - \sin A \sin B \):
\( \cos (\pi + \theta) = \cos \pi \cos \theta - \sin \pi \sin \theta \)
We know \( \cos \pi = -1 \) and \( \sin \pi = 0 \).
Substitute these values:
\( \cos (\pi + \theta) = (-1) \cos \theta - (0) \sin \theta \)
\( \cos (\pi + \theta) = -\cos \theta - 0 \)
\( \cos (\pi + \theta) = -\cos \theta \)
This is equal to the Right Hand Side (RHS). Hence proved.

(iii) To prove \( \sin (\pi + \theta) = -\sin \theta \):
We start with the Left Hand Side (LHS): \( \sin (\pi + \theta) \)
Using the compound angle formula \( \sin(A + B) = \sin A \cos B + \cos A \sin B \):
\( \sin (\pi + \theta) = \sin \pi \cos \theta + \cos \pi \sin \theta \)
We know \( \sin \pi = 0 \) and \( \cos \pi = -1 \).
Substitute these values:
\( \sin (\pi + \theta) = (0) \cos \theta + (-1) \sin \theta \)
\( \sin (\pi + \theta) = 0 - \sin \theta \)
\( \sin (\pi + \theta) = -\sin \theta \)
This is equal to the Right Hand Side (RHS). Hence proved.
In simple words: For these proofs, we use the basic trigonometric sum formulas for cosine and sine. We replace the known angle values like \( 30^\circ \) and \( \pi \) with their exact trigonometric values and then simplify the expression to show that it matches the other side of the equation.

🎯 Exam Tip: When proving trigonometric identities, always start with one side (usually the more complex one) and apply identities and known values to transform it into the other side. Memorize the values of standard angles and compound angle formulas.

Question 7. Find a quadratic equation whose roots are \( \sin 15^\circ \) and \( \cos 15^\circ \).
Answer:
First, we need to find the values of \( \sin 15^\circ \) and \( \cos 15^\circ \).
We can write \( 15^\circ \) as \( 45^\circ - 30^\circ \).

To find \( \sin 15^\circ \):
Using the formula \( \sin(A - B) = \sin A \cos B - \cos A \sin B \):
\( \sin 15^\circ = \sin(45^\circ - 30^\circ) \)
\( \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \)
\( \sin 15^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) \)
\( \sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \)
\( \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \) --- (1)

To find \( \cos 15^\circ \):
Using the formula \( \cos(A - B) = \cos A \cos B + \sin A \sin B \):
\( \cos 15^\circ = \cos(45^\circ - 30^\circ) \)
\( \cos 15^\circ = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \)
\( \cos 15^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) \)
\( \cos 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \)
\( \cos 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \) --- (2)

The general form of a quadratic equation is \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \).
Let the roots be \( \alpha = \sin 15^\circ \) and \( \beta = \cos 15^\circ \).

Sum of roots \( (\alpha + \beta) = \sin 15^\circ + \cos 15^\circ \):
\( \alpha + \beta = \frac{\sqrt{3} - 1}{2\sqrt{2}} + \frac{\sqrt{3} + 1}{2\sqrt{2}} \)
\( \alpha + \beta = \frac{\sqrt{3} - 1 + \sqrt{3} + 1}{2\sqrt{2}} \)
\( \alpha + \beta = \frac{2\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}} \)
Rationalize: \( \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{2} \)

Product of roots \( (\alpha \beta) = \sin 15^\circ \cos 15^\circ \):
\( \alpha \beta = \left(\frac{\sqrt{3} - 1}{2\sqrt{2}}\right) \left(\frac{\sqrt{3} + 1}{2\sqrt{2}}\right) \)
\( \alpha \beta = \frac{(\sqrt{3})^2 - 1^2}{(2\sqrt{2})^2} \)
\( \alpha \beta = \frac{3 - 1}{4 \cdot 2} = \frac{2}{8} = \frac{1}{4} \)

Substitute the sum and product of roots into the quadratic equation form:
\( x^2 - \left(\frac{\sqrt{6}}{2}\right)x + \frac{1}{4} = 0 \)
To remove fractions, multiply the entire equation by 4:
\( 4x^2 - 4\left(\frac{\sqrt{6}}{2}\right)x + 4\left(\frac{1}{4}\right) = 0 \)
\( 4x^2 - 2\sqrt{6}x + 1 = 0 \)
This is the required quadratic equation.
In simple words: First, we find the exact values for sine and cosine of \( 15^\circ \) by breaking it down as \( 45^\circ - 30^\circ \). Then, we use these values as the roots to form a quadratic equation, which involves calculating their sum and product.

🎯 Exam Tip: Remember that a quadratic equation with roots \( \alpha \) and \( \beta \) is given by \( x^2 - (\alpha + \beta)x + \alpha\beta = 0 \). Always rationalize denominators in intermediate steps to simplify calculations for the sum and product of roots.

Question 8. Expand \( \cos(A + B + C) \). Hence prove that \( \cos A \cos B \cos C = \sin A \sin B \cos C + \sin B \sin C \cos A + \sin C \sin A \cos B \) if \( A + B + C = \frac{\pi}{2} \).
Answer:
To expand \( \cos(A + B + C) \):
Let \( X = A + B \) and \( Y = C \). Then \( \cos(A + B + C) = \cos(X + Y) \).
Using the formula \( \cos(X + Y) = \cos X \cos Y - \sin X \sin Y \):
\( \cos(A + B + C) = \cos(A + B) \cos C - \sin(A + B) \sin C \)
Now, expand \( \cos(A + B) \) and \( \sin(A + B) \):
\( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
\( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
Substitute these back into the expression for \( \cos(A + B + C) \):
\( \cos(A + B + C) = (\cos A \cos B - \sin A \sin B) \cos C - (\sin A \cos B + \cos A \sin B) \sin C \)
\( \cos(A + B + C) = \cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C \)

Now, to prove the second part: if \( A + B + C = \frac{\pi}{2} \).
If \( A + B + C = \frac{\pi}{2} \), then \( \cos(A + B + C) = \cos\left(\frac{\pi}{2}\right) = 0 \).
So, we have:
\( 0 = \cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C \)
Rearrange the terms to match the required proof:
\( \cos A \cos B \cos C = \sin A \sin B \cos C + \sin A \cos B \sin C + \cos A \sin B \sin C \)
This can be written as:
\( \cos A \cos B \cos C = \sin A \sin B \cos C + \sin B \sin C \cos A + \sin C \sin A \cos B \)
Thus, the identity is proved.
In simple words: First, we use a step-by-step method to expand \( \cos(A + B + C) \) using the sum formulas. Then, for the second part, we use the fact that if the sum of three angles is \( 90^\circ \), their cosine is 0. We set our expanded formula to 0 and rearrange it to prove the given identity.

🎯 Exam Tip: When expanding expressions with more than two terms, group them (e.g., \( (A+B)+C \)) to apply compound angle formulas iteratively. For proofs involving conditions like \( A+B+C = \frac{\pi}{2} \), remember to use the value of the trigonometric function of the sum (e.g., \( \cos(\frac{\pi}{2}) = 0 \)).

Question 9. Prove that
(i) \( \sin (45^\circ + \theta) – \sin (45^\circ – \theta) = \sqrt{2} \sin \theta \)
(ii) \( \sin (30^\circ + \theta) + \cos (60^\circ + \theta) = \cos \theta \)
Answer:
(i) To prove \( \sin (45^\circ + \theta) – \sin (45^\circ – \theta) = \sqrt{2} \sin \theta \):
We start with the Left Hand Side (LHS): \( \sin (45^\circ + \theta) – \sin (45^\circ – \theta) \)
Using the compound angle formulas:
\( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
\( \sin(A - B) = \sin A \cos B - \cos A \sin B \)
So, \( \sin (45^\circ + \theta) = \sin 45^\circ \cos \theta + \cos 45^\circ \sin \theta \)
And \( \sin (45^\circ - \theta) = \sin 45^\circ \cos \theta - \cos 45^\circ \sin \theta \)
Substitute these into the LHS:
LHS \( = (\sin 45^\circ \cos \theta + \cos 45^\circ \sin \theta) - (\sin 45^\circ \cos \theta - \cos 45^\circ \sin \theta) \)
LHS \( = \sin 45^\circ \cos \theta + \cos 45^\circ \sin \theta - \sin 45^\circ \cos \theta + \cos 45^\circ \sin \theta \)
LHS \( = 2 \cos 45^\circ \sin \theta \)
We know \( \cos 45^\circ = \frac{1}{\sqrt{2}} \).
LHS \( = 2 \cdot \frac{1}{\sqrt{2}} \sin \theta \)
LHS \( = \frac{2}{\sqrt{2}} \sin \theta \)
To simplify \( \frac{2}{\sqrt{2}} \), multiply numerator and denominator by \( \sqrt{2} \):
LHS \( = \frac{2\sqrt{2}}{2} \sin \theta = \sqrt{2} \sin \theta \)
This is equal to the Right Hand Side (RHS). Hence proved.

(ii) To prove \( \sin (30^\circ + \theta) + \cos (60^\circ + \theta) = \cos \theta \):
We start with the Left Hand Side (LHS): \( \sin (30^\circ + \theta) + \cos (60^\circ + \theta) \)
Using the compound angle formulas:
\( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
\( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
First term: \( \sin (30^\circ + \theta) = \sin 30^\circ \cos \theta + \cos 30^\circ \sin \theta \)
\( = \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta \)
Second term: \( \cos (60^\circ + \theta) = \cos 60^\circ \cos \theta - \sin 60^\circ \sin \theta \)
\( = \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \)
Now add both terms for LHS:
LHS \( = \left(\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta\right) + \left(\frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta\right) \)
LHS \( = \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \)
LHS \( = \left(\frac{1}{2} \cos \theta + \frac{1}{2} \cos \theta\right) + \left(\frac{\sqrt{3}}{2} \sin \theta - \frac{\sqrt{3}}{2} \sin \theta\right) \)
LHS \( = \cos \theta + 0 \)
LHS \( = \cos \theta \)
This is equal to the Right Hand Side (RHS). Hence proved.
In simple words: For both proofs, we expand the expressions using the sine and cosine sum/difference formulas. We then substitute the known values for \( 30^\circ, 45^\circ, \) and \( 60^\circ \) and simplify the terms to show that the left side equals the right side.

🎯 Exam Tip: Remember to use the standard trigonometric values for \( 30^\circ, 45^\circ, \) and \( 60^\circ \) accurately. Practice with the sum and difference formulas for sine and cosine to avoid common expansion errors.

Question 10. If \( a \cos (x + y) = b \cos(x – y) \), show that \( (a + b) \tan x = (a – b) \cot y \).
Answer:
Given the equation: \( a \cos (x + y) = b \cos(x - y) \)
Expand \( \cos(x + y) \) and \( \cos(x - y) \) using the compound angle formulas:
\( \cos(x + y) = \cos x \cos y - \sin x \sin y \)
\( \cos(x - y) = \cos x \cos y + \sin x \sin y \)
Substitute these expansions into the given equation:
\( a(\cos x \cos y - \sin x \sin y) = b(\cos x \cos y + \sin x \sin y) \)
Distribute a and b:
\( a \cos x \cos y - a \sin x \sin y = b \cos x \cos y + b \sin x \sin y \)
Group terms with 'a' and 'b' on opposite sides by moving all terms containing \( \sin x \sin y \) to one side and terms containing \( \cos x \cos y \) to the other side:
\( a \cos x \cos y - b \cos x \cos y = a \sin x \sin y + b \sin x \sin y \)
Factor out common terms:
\( (a - b) \cos x \cos y = (a + b) \sin x \sin y \)
To obtain \( \tan x \) and \( \cot y \), divide both sides by \( \cos x \cos y \) and \( \sin x \sin y \). Specifically, to get \( \frac{\sin x}{\cos x} \) (for \( \tan x \)) and \( \frac{\cos y}{\sin y} \) (for \( \cot y \)), we can divide by \( \cos x \sin y \).
Divide both sides by \( \cos x \sin y \):
\( (a - b) \frac{\cos x \cos y}{\cos x \sin y} = (a + b) \frac{\sin x \sin y}{\cos x \sin y} \)
Simplify the fractions:
\( (a - b) \frac{\cos y}{\sin y} = (a + b) \frac{\sin x}{\cos x} \)
Substitute \( \frac{\cos y}{\sin y} = \cot y \) and \( \frac{\sin x}{\cos x} = \tan x \):
\( (a - b) \cot y = (a + b) \tan x \)
This is the required identity. Hence proved.
In simple words: We start by expanding the given equation using the sum and difference formulas for cosine. Then, we move similar terms to opposite sides of the equation. Finally, we divide by appropriate trigonometric terms to transform them into tangent and cotangent, showing the desired result.

🎯 Exam Tip: When proving identities involving tangent and cotangent, a good strategy is to expand all compound angle formulas first. Then, strategically divide both sides of the equation by products of sine and cosine terms to generate the required tangent and cotangent expressions.

Question 12. Prove that \( \sin 75^\circ - \sin 15^\circ = \cos 105^\circ + \cos 15^\circ \).
Answer: We will start with the Right Hand Side (RHS) of the equation and work to simplify it to match the Left Hand Side (LHS).
RHS \( = \cos 105^\circ + \cos 15^\circ \)
We can rewrite \( \cos 105^\circ \) as \( \cos (90^\circ + 15^\circ) \) and \( \cos 15^\circ \) as \( \cos (90^\circ - 75^\circ) \).
Using the identities \( \cos (90^\circ + \theta) = -\sin \theta \) and \( \cos (90^\circ - \theta) = \sin \theta \):
RHS \( = -\sin 15^\circ + \sin 75^\circ \)
RHS \( = \sin 75^\circ - \sin 15^\circ \)
This is equal to the Left Hand Side (LHS). This shows the identity is true.
In simple words: To prove this, we change the right side of the equation using angle rules. We turn \( \cos 105^\circ \) into \( -\sin 15^\circ \) and \( \cos 15^\circ \) into \( \sin 75^\circ \). After this, both sides match.

🎯 Exam Tip: When proving identities, always try to express angles in terms of \( 90^\circ \), \( 180^\circ \), or \( 270^\circ \) to use trigonometric reduction formulas.

Question 13. Show that \( \tan 75^\circ + \cot 75^\circ = 4 \).
Answer: First, let's find the value of \( \tan 75^\circ \). We can write \( \tan 75^\circ \) as \( \tan (45^\circ + 30^\circ) \).
Using the formula \( \tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \):
\( \tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \)
We know \( \tan 45^\circ = 1 \) and \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
\( \tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)
To simplify, multiply the numerator and denominator by the conjugate \( (\sqrt{3} + 1) \):
\( \tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \)

Next, let's find \( \cot 75^\circ \). We know \( \cot \theta = \frac{1}{\tan \theta} \).
\( \cot 75^\circ = \frac{1}{\tan 75^\circ} = \frac{1}{2 + \sqrt{3}} \)
To simplify, multiply the numerator and denominator by the conjugate \( (2 - \sqrt{3}) \):
\( \cot 75^\circ = \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3} \)

Now, add \( \tan 75^\circ \) and \( \cot 75^\circ \):
\( \tan 75^\circ + \cot 75^\circ = (2 + \sqrt{3}) + (2 - \sqrt{3}) \)
\( = 2 + \sqrt{3} + 2 - \sqrt{3} \)
\( = 4 \)
Thus, it is shown that \( \tan 75^\circ + \cot 75^\circ = 4 \). This method relies on simplifying fractions with square roots.
In simple words: First, we find the exact value of \( \tan 75^\circ \) using the sum formula and simplifying the root. Then we find \( \cot 75^\circ \) by taking 1 divided by \( \tan 75^\circ \) and simplifying again. When we add these two values, the square root parts cancel out, leaving just 4.

🎯 Exam Tip: Remember to rationalize the denominator by multiplying by the conjugate when dealing with expressions involving square roots in fractions.

Question 14. Prove that \( \cos(A + B) \cos C - \cos(B + C) \cos A = \sin B \sin (C - A) \).
Answer: Let's start with the Left Hand Side (LHS) of the equation.
LHS \( = \cos(A + B) \cos C - \cos(B + C) \cos A \)
Using the cosine sum formula \( \cos(X+Y) = \cos X \cos Y - \sin X \sin Y \):
\( \cos(A + B) = (\cos A \cos B - \sin A \sin B) \)
\( \cos(B + C) = (\cos B \cos C - \sin B \sin C) \)

Substitute these into the LHS:
LHS \( = (\cos A \cos B - \sin A \sin B) \cos C - (\cos B \cos C - \sin B \sin C) \cos A \)
Now, expand the terms:
LHS \( = \cos A \cos B \cos C - \sin A \sin B \cos C - (\cos B \cos C \cos A - \sin B \sin C \cos A) \)
LHS \( = \cos A \cos B \cos C - \sin A \sin B \cos C - \cos A \cos B \cos C + \sin B \sin C \cos A \)
The terms \( \cos A \cos B \cos C \) and \( -\cos A \cos B \cos C \) cancel each other out.
LHS \( = -\sin A \sin B \cos C + \sin B \sin C \cos A \)
Factor out \( \sin B \):
LHS \( = \sin B (\sin C \cos A - \sin A \cos C) \)
This expression \( (\sin C \cos A - \sin A \cos C) \) is the expansion of \( \sin (C - A) \).
LHS \( = \sin B \sin (C - A) \)
This matches the Right Hand Side (RHS), completing the proof. The proof demonstrates the application of compound angle formulas.
In simple words: We expand both \( \cos(A+B) \) and \( \cos(B+C) \) using a standard math rule. Then, we multiply everything out and simplify. Many terms cancel, and the remaining parts combine to form \( \sin B \) multiplied by \( \sin(C-A) \), which is what we needed to show.

🎯 Exam Tip: When dealing with complex trigonometric identities, always expand the compound angles first, then look for common terms to factor or cancel.

Question 15. Prove that \( \sin (n + 1)\theta \cdot \sin(n - 1)\theta + \cos(n + 1)\theta \cdot \cos(n - 1)\theta = \cos 2\theta \), where \( n \in Z \).
Answer: Let's take the Left Hand Side (LHS) of the equation.
LHS \( = \sin (n + 1)\theta \cdot \sin(n - 1)\theta + \cos(n + 1)\theta \cdot \cos(n - 1)\theta \)
This expression looks like the cosine difference formula: \( \cos(A - B) = \cos A \cos B + \sin A \sin B \).
Let \( A = (n + 1)\theta \) and \( B = (n - 1)\theta \).
Then the LHS becomes \( \cos(A - B) \).
LHS \( = \cos[(n + 1)\theta - (n - 1)\theta] \)
LHS \( = \cos[n\theta + \theta - (n\theta - \theta)] \)
LHS \( = \cos[n\theta + \theta - n\theta + \theta] \)
LHS \( = \cos[ (n\theta - n\theta) + (\theta + \theta) ] \)
LHS \( = \cos[0 + 2\theta] \)
LHS \( = \cos 2\theta \)
This matches the Right Hand Side (RHS), completing the proof. This identity is a good example of applying sum and difference angle formulas.
In simple words: We see that the left side of the equation looks like a famous rule for cosine of (A minus B). If we say A is \( (n+1)\theta \) and B is \( (n-1)\theta \), then the expression becomes \( \cos(A-B) \). When we subtract B from A, all the 'n' parts cancel out, leaving just \( 2\theta \). So the left side becomes \( \cos 2\theta \), which matches the right side.

🎯 Exam Tip: Recognize common trigonometric identities like \( \cos(A-B) \) to simplify complex expressions quickly. This saves time and reduces calculation errors.

Question 16. If \( x \cos \theta = y \cos\left(\theta+\frac{2 \pi}{3}\right) = z \cos \left(\theta+\frac{4 \pi}{3}\right) \), find the value of \( xy + yz + zx \).
Answer: Let \( x \cos \theta = y \cos\left(\theta+\frac{2 \pi}{3}\right) = z \cos \left(\theta+\frac{4 \pi}{3}\right) = \lambda \) (where \( \lambda \) is a constant).
From this, we can write expressions for \( \cos \theta \), \( \cos\left(\theta+\frac{2 \pi}{3}\right) \), and \( \cos \left(\theta+\frac{4 \pi}{3}\right) \):
\( \cos \theta = \frac{\lambda}{x} \)
\( \cos\left(\theta+\frac{2 \pi}{3}\right) = \frac{\lambda}{y} \)
\( \cos \left(\theta+\frac{4 \pi}{3}\right) = \frac{\lambda}{z} \)

We know the identity: \( \cos \theta + \cos\left(\theta+\frac{2 \pi}{3}\right) + \cos \left(\theta+\frac{4 \pi}{3}\right) = 0 \). This identity is useful for sums of cosines with specific phase shifts.

Substitute the expressions for \( \frac{\lambda}{x}, \frac{\lambda}{y}, \frac{\lambda}{z} \):
\( \frac{\lambda}{x} + \frac{\lambda}{y} + \frac{\lambda}{z} = 0 \)
Factor out \( \lambda \):
\( \lambda \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) = 0 \)
Since \( \lambda \) cannot be zero (as \( x, y, z \) would then be undefined), the term in the parenthesis must be zero.
\( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \)
To combine these fractions, find a common denominator, which is \( xyz \):
\( \frac{yz + xz + xy}{xyz} = 0 \)
For this fraction to be zero, the numerator must be zero (assuming \( xyz \neq 0 \)).
\( xy + yz + zx = 0 \)
Thus, the value of \( xy + yz + zx \) is 0. This problem showcases how constant ratios can simplify trigonometric sums.
In simple words: We set all three parts of the equation equal to a constant, called lambda. This lets us write \( \cos \theta \), \( \cos(\theta+\frac{2\pi}{3}) \), and \( \cos(\theta+\frac{4\pi}{3}) \) as \( \lambda \) divided by \( x, y, z \). We know that these three cosine values always add up to zero. So, \( \frac{\lambda}{x} + \frac{\lambda}{y} + \frac{\lambda}{z} = 0 \). When we simplify this sum of fractions, we get \( \frac{xy + yz + zx}{xyz} = 0 \). This means that \( xy + yz + zx \) must be zero.

🎯 Exam Tip: When given an expression where several terms are equal, setting them to a constant (like \( \lambda \)) often simplifies the problem. Remember the identity \( \cos \theta + \cos(\theta+120^\circ) + \cos(\theta+240^\circ) = 0 \).

Question 17. Prove that:
(i) \( \sin(A + B) \cdot \sin(A - B) = \sin^2 A - \sin^2 B \)
(ii) \( \cos(A + B) \cdot \cos(A - B) = \cos^2 A - \sin^2 B = \cos^2 B - \sin^2 A \)
(iii) \( \sin^2 (A + B) - \sin^2(A - B) = \sin 2A \sin 2B \)
(iv) \( \cos 8\theta \cdot \cos 2\theta = \cos^2 5\theta - \sin^2 3\theta \)
Answer:
(i) Let's prove \( \sin(A + B) \cdot \sin(A - B) = \sin^2 A - \sin^2 B \).
Start with LHS: \( \sin(A + B) \cdot \sin(A - B) \)
Using the sine sum and difference formulas: \( \sin(X+Y) = \sin X \cos Y + \cos X \sin Y \) and \( \sin(X-Y) = \sin X \cos Y - \cos X \sin Y \)
LHS \( = (\sin A \cos B + \cos A \sin B) (\sin A \cos B - \cos A \sin B) \)
This is in the form \( (x + y)(x - y) = x^2 - y^2 \), where \( x = \sin A \cos B \) and \( y = \cos A \sin B \).
LHS \( = (\sin A \cos B)^2 - (\cos A \sin B)^2 \)
LHS \( = \sin^2 A \cos^2 B - \cos^2 A \sin^2 B \)
Now, express \( \cos^2 B \) as \( (1 - \sin^2 B) \) and \( \cos^2 A \) as \( (1 - \sin^2 A) \):
LHS \( = \sin^2 A (1 - \sin^2 B) - (1 - \sin^2 A) \sin^2 B \)
LHS \( = \sin^2 A - \sin^2 A \sin^2 B - (\sin^2 B - \sin^2 A \sin^2 B) \)
LHS \( = \sin^2 A - \sin^2 A \sin^2 B - \sin^2 B + \sin^2 A \sin^2 B \)
The terms \( -\sin^2 A \sin^2 B \) and \( +\sin^2 A \sin^2 B \) cancel out.
LHS \( = \sin^2 A - \sin^2 B \)
This matches the RHS, proving the identity. This shows how product formulas can simplify to squared terms.

(ii) Let's prove \( \cos(A + B) \cdot \cos(A - B) = \cos^2 A - \sin^2 B = \cos^2 B - \sin^2 A \).
Start with LHS: \( \cos(A + B) \cdot \cos(A - B) \)
Using the cosine sum and difference formulas: \( \cos(X+Y) = \cos X \cos Y - \sin X \sin Y \) and \( \cos(X-Y) = \cos X \cos Y + \sin X \sin Y \)
LHS \( = (\cos A \cos B - \sin A \sin B) (\cos A \cos B + \sin A \sin B) \)
This is in the form \( (x - y)(x + y) = x^2 - y^2 \), where \( x = \cos A \cos B \) and \( y = \sin A \sin B \).
LHS \( = (\cos A \cos B)^2 - (\sin A \sin B)^2 \)
LHS \( = \cos^2 A \cos^2 B - \sin^2 A \sin^2 B \)

To get the first part of RHS \( (\cos^2 A - \sin^2 B) \), replace \( \cos^2 B \) with \( (1 - \sin^2 B) \) and \( \sin^2 A \) with \( (1 - \cos^2 A) \).
LHS \( = \cos^2 A (1 - \sin^2 B) - (1 - \cos^2 A) \sin^2 B \)
LHS \( = \cos^2 A - \cos^2 A \sin^2 B - \sin^2 B + \cos^2 A \sin^2 B \)
The terms \( -\cos^2 A \sin^2 B \) and \( +\cos^2 A \sin^2 B \) cancel out.
LHS \( = \cos^2 A - \sin^2 B \)

To get the second part of RHS \( (\cos^2 B - \sin^2 A) \), go back to \( \cos^2 A \cos^2 B - \sin^2 A \sin^2 B \).
Replace \( \cos^2 A \) with \( (1 - \sin^2 A) \) and \( \sin^2 B \) with \( (1 - \cos^2 B) \).
LHS \( = (1 - \sin^2 A) \cos^2 B - \sin^2 A (1 - \cos^2 B) \)
LHS \( = \cos^2 B - \sin^2 A \cos^2 B - \sin^2 A + \sin^2 A \cos^2 B \)
The terms \( -\sin^2 A \cos^2 B \) and \( +\sin^2 A \cos^2 B \) cancel out.
LHS \( = \cos^2 B - \sin^2 A \)
Both parts of the RHS are proven. This identity is useful for simplifying products of cosines.

(iii) Let's prove \( \sin^2 (A + B) - \sin^2(A - B) = \sin 2A \sin 2B \).
Recall the identity from part (i): \( \sin^2 X - \sin^2 Y = \sin(X+Y) \sin(X-Y) \).
Here, let \( X = (A+B) \) and \( Y = (A-B) \).
LHS \( = \sin[(A+B) + (A-B)] \sin[(A+B) - (A-B)] \)
LHS \( = \sin[A+B+A-B] \sin[A+B-A+B] \)
LHS \( = \sin[2A] \sin[2B] \)
LHS \( = \sin 2A \sin 2B \)
This matches the RHS, proving the identity. This is a compact way to prove the identity using a previously derived result.

(iv) Let's prove \( \cos 8\theta \cdot \cos 2\theta = \cos^2 5\theta - \sin^2 3\theta \).
Start with RHS: \( \cos^2 5\theta - \sin^2 3\theta \).
Using the identity from part (ii) (the form \( \cos^2 X - \sin^2 Y = \cos(X+Y)\cos(X-Y) \)), let \( X = 5\theta \) and \( Y = 3\theta \).
RHS \( = \cos(5\theta + 3\theta) \cos(5\theta - 3\theta) \)
RHS \( = \cos(8\theta) \cos(2\theta) \)
RHS \( = \cos 8\theta \cdot \cos 2\theta \)
This matches the LHS, proving the identity. This solution uses a reverse application of the identities.
In simple words:
(i) We expand \( \sin(A+B) \) and \( \sin(A-B) \), then multiply them. We use the \( (x+y)(x-y) \) rule, and then replace \( \cos^2 \) with \( 1-\sin^2 \). After simplifying, we get \( \sin^2 A - \sin^2 B \).
(ii) We expand \( \cos(A+B) \) and \( \cos(A-B) \), then multiply them. We use the \( (x-y)(x+y) \) rule. Then, by swapping \( \cos^2 \) for \( 1-\sin^2 \) (or vice versa), we can get both \( \cos^2 A - \sin^2 B \) and \( \cos^2 B - \sin^2 A \).
(iii) We use a shortcut identity that says \( \sin^2 X - \sin^2 Y \) is equal to \( \sin(X+Y) \sin(X-Y) \). When we apply this with \( A+B \) and \( A-B \), the terms add and subtract to give \( \sin 2A \sin 2B \).
(iv) We work backward from the right side. Using the rule from part (ii) in reverse, \( \cos^2 X - \sin^2 Y \) equals \( \cos(X+Y)\cos(X-Y) \). We apply this with \( X=5\theta \) and \( Y=3\theta \), which leads directly to \( \cos 8\theta \cdot \cos 2\theta \).

🎯 Exam Tip: Mastering the sum/difference and product-to-sum trigonometric identities is crucial for solving such proof-based problems. Remember that \( \sin^2 X - \sin^2 Y = \sin(X+Y)\sin(X-Y) \) and \( \cos^2 X - \sin^2 Y = \cos(X+Y)\cos(X-Y) \).

Question 18. Show that \( \cos^2 A + \cos^2 B - 2 \cos A \cos B \cos (A + B) = \sin^2(A + B) \).
Answer: Let's start with the Left Hand Side (LHS) of the equation.
LHS \( = \cos^2 A + \cos^2 B - 2 \cos A \cos B \cos (A + B) \)
Using the cosine sum formula: \( \cos (A + B) = \cos A \cos B - \sin A \sin B \)
Substitute this into the LHS:
LHS \( = \cos^2 A + \cos^2 B - 2 \cos A \cos B (\cos A \cos B - \sin A \sin B) \)
Expand the term:
LHS \( = \cos^2 A + \cos^2 B - 2 \cos^2 A \cos^2 B + 2 \cos A \cos B \sin A \sin B \)
Rearrange the terms:
LHS \( = (\cos^2 A - \cos^2 A \cos^2 B) + (\cos^2 B - \cos^2 A \cos^2 B) + 2 \sin A \cos A \sin B \cos B \)
Factor out common terms:
LHS \( = \cos^2 A (1 - \cos^2 B) + \cos^2 B (1 - \cos^2 A) + 2 \sin A \cos A \sin B \cos B \)
We know \( 1 - \cos^2 B = \sin^2 B \) and \( 1 - \cos^2 A = \sin^2 A \).
LHS \( = \cos^2 A \sin^2 B + \cos^2 B \sin^2 A + 2 \sin A \cos A \sin B \cos B \)
This expression is in the form \( x^2 + y^2 + 2xy \), which is \( (x + y)^2 \).
Here, \( x = \sin A \cos B \) and \( y = \cos A \sin B \).
LHS \( = (\sin A \cos B + \cos A \sin B)^2 \)
The term inside the parenthesis is the expansion of \( \sin (A + B) \).
LHS \( = (\sin (A + B))^2 \)
LHS \( = \sin^2 (A + B) \)
This matches the Right Hand Side (RHS), completing the proof. This identity is a powerful simplification of the sum of squared cosines.
In simple words: We start with the left side and use the formula for \( \cos(A+B) \). After we put this into the equation and multiply everything out, we rearrange the terms. We then replace \( 1-\cos^2 B \) with \( \sin^2 B \) and \( 1-\cos^2 A \) with \( \sin^2 A \). The simplified expression perfectly matches the formula for \( (\sin A \cos B + \cos A \sin B)^2 \), which is just \( \sin^2(A+B) \).

🎯 Exam Tip: Look for opportunities to complete a square (like \( (x+y)^2 \)) or use fundamental identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) to simplify complex expressions.

Question 19. If \( \cos (\alpha - \beta) + \cos(\beta - \gamma) + \cos(\gamma - \alpha) = -\frac{3}{2} \), then prove that \( \cos \alpha + \cos \beta + \cos \gamma = 0 \) and \( \sin \alpha + \sin \beta + \sin \gamma = 0 \).
Answer: Given the equation: \( \cos (\alpha - \beta) + \cos(\beta - \gamma) + \cos(\gamma - \alpha) = -\frac{3}{2} \)
Using the cosine difference formula: \( \cos(X-Y) = \cos X \cos Y + \sin X \sin Y \)
Expand each term:
\( (\cos \alpha \cos \beta + \sin \alpha \sin \beta) + (\cos \beta \cos \gamma + \sin \beta \sin \gamma) + (\cos \gamma \cos \alpha + \sin \gamma \sin \alpha) = -\frac{3}{2} \)
Multiply the entire equation by 2:
\( 2 \cos \alpha \cos \beta + 2 \sin \alpha \sin \beta + 2 \cos \beta \cos \gamma + 2 \sin \beta \sin \gamma + 2 \cos \gamma \cos \alpha + 2 \sin \gamma \sin \alpha = -3 \)
Rearrange the equation and move the -3 to the left side:
\( 2 \cos \alpha \cos \beta + 2 \sin \alpha \sin \beta + 2 \cos \beta \cos \gamma + 2 \sin \beta \sin \gamma + 2 \cos \gamma \cos \alpha + 2 \sin \gamma \sin \alpha + 3 = 0 \)
We can rewrite 3 as \( 1+1+1 \). And we know that \( 1 = \cos^2 \theta + \sin^2 \theta \).
So, \( 3 = (\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + (\cos^2 \gamma + \sin^2 \gamma) \).
Substitute this into the equation:
\( 2 \cos \alpha \cos \beta + 2 \sin \alpha \sin \beta + 2 \cos \beta \cos \gamma + 2 \sin \beta \sin \gamma + 2 \cos \gamma \cos \alpha + 2 \sin \gamma \sin \alpha + (\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + (\cos^2 \gamma + \sin^2 \gamma) = 0 \)
Rearrange to group terms that form squares:
\( (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2 \cos \alpha \cos \beta + 2 \cos \beta \cos \gamma + 2 \cos \gamma \cos \alpha) \)
\( + (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma + 2 \sin \alpha \sin \beta + 2 \sin \beta \sin \gamma + 2 \sin \gamma \sin \alpha) = 0 \)
The first bracket is the expansion of \( (\cos \alpha + \cos \beta + \cos \gamma)^2 \).
The second bracket is the expansion of \( (\sin \alpha + \sin \beta + \sin \gamma)^2 \).
So, the equation becomes:
\( (\cos \alpha + \cos \beta + \cos \gamma)^2 + (\sin \alpha + \sin \beta + \sin \gamma)^2 = 0 \)
For the sum of two squares to be zero, each square must be zero (because squares are always non-negative).

\( \implies (\cos \alpha + \cos \beta + \cos \gamma)^2 = 0 \)
\( \implies \cos \alpha + \cos \beta + \cos \gamma = 0 \)

\( \implies (\sin \alpha + \sin \beta + \sin \gamma)^2 = 0 \)
\( \implies \sin \alpha + \sin \beta + \sin \gamma = 0 \)
Both conditions are proven. This identity elegantly uses algebraic identities within trigonometry.
In simple words: We first expand the given equation using the rule for \( \cos(\text{angle} - \text{angle}) \). Then, we multiply everything by 2 and add 3 (which is \( \cos^2 \alpha + \sin^2 \alpha \) plus two similar terms) to both sides. By grouping the terms carefully, we can see that the equation becomes the sum of two squared expressions: \( (\cos \alpha + \cos \beta + \cos \gamma)^2 \) and \( (\sin \alpha + \sin \beta + \sin \gamma)^2 \). For the sum of two squares to be zero, each square must be zero, which means both \( \cos \alpha + \cos \beta + \cos \gamma \) and \( \sin \alpha + \sin \beta + \sin \gamma \) must be zero.

🎯 Exam Tip: When faced with an expression like \( A^2 + B^2 = 0 \), remember that this implies \( A=0 \) and \( B=0 \). Also, don't forget the identity \( \cos^2 \theta + \sin^2 \theta = 1 \) can be used creatively.

Question 20. Show that \( \tan \left(\frac{\pi}{4} + A\right) = \frac{1 + \tan A}{1 - \tan A} \) and \( \tan \left(\frac{\pi}{4} - A\right) = \frac{1 - \tan A}{1 + \tan A} \).
Answer:
(i) Let's prove \( \tan \left(\frac{\pi}{4} + A\right) = \frac{1 + \tan A}{1 - \tan A} \).
Using the tangent sum formula: \( \tan(X+Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y} \)
Let \( X = \frac{\pi}{4} \) and \( Y = A \). We know \( \tan \frac{\pi}{4} = \tan 45^\circ = 1 \).
LHS \( = \tan \left(\frac{\pi}{4} + A\right) = \frac{\tan \frac{\pi}{4} + \tan A}{1 - \tan \frac{\pi}{4} \tan A} \)
LHS \( = \frac{1 + \tan A}{1 - 1 \cdot \tan A} \)
LHS \( = \frac{1 + \tan A}{1 - \tan A} \)
This matches the RHS, proving the first identity. This shows a useful simplification for \( \tan(45^\circ + A) \).

(ii) Let's prove \( \tan \left(\frac{\pi}{4} - A\right) = \frac{1 - \tan A}{1 + \tan A} \).
Using the tangent difference formula: \( \tan(X-Y) = \frac{\tan X - \tan Y}{1 + \tan X \tan Y} \)
Let \( X = \frac{\pi}{4} \) and \( Y = A \). We know \( \tan \frac{\pi}{4} = \tan 45^\circ = 1 \).
LHS \( = \tan \left(\frac{\pi}{4} - A\right) = \frac{\tan \frac{\pi}{4} - \tan A}{1 + \tan \frac{\pi}{4} \tan A} \)
LHS \( = \frac{1 - \tan A}{1 + 1 \cdot \tan A} \)
LHS \( = \frac{1 - \tan A}{1 + \tan A} \)
This matches the RHS, proving the second identity. This demonstrates a similar simplification for \( \tan(45^\circ - A) \).
In simple words:
(i) For the first proof, we use the sum formula for tangent, \( \tan(X+Y) \). We replace X with \( \frac{\pi}{4} \) (which is \( 45^\circ \)) and Y with A. Since \( \tan 45^\circ \) is 1, the formula simplifies to \( \frac{1 + \tan A}{1 - \tan A} \).
(ii) For the second proof, we use the difference formula for tangent, \( \tan(X-Y) \). Again, we use X as \( \frac{\pi}{4} \) and Y as A. Because \( \tan 45^\circ \) is 1, the formula simplifies to \( \frac{1 - \tan A}{1 + \tan A} \).

🎯 Exam Tip: These two identities are very common and useful. It's often helpful to memorize them as direct applications of the tangent sum/difference formulas with \( 45^\circ \).

Question 21. Prove that \( \cot (A + B) = \frac{\cot A \cot B-1}{\cot A+\cot B} \).
Answer: Let's start with the Left Hand Side (LHS) of the equation.
LHS \( = \cot (A + B) \)
We know that \( \cot \theta = \frac{1}{\tan \theta} \). So, \( \cot (A + B) = \frac{1}{\tan (A + B)} \).
Using the tangent sum formula: \( \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
Substitute this into the expression for \( \cot (A + B) \):
LHS \( = \frac{1}{\frac{\tan A + \tan B}{1 - \tan A \tan B}} \)
LHS \( = \frac{1 - \tan A \tan B}{\tan A + \tan B} \)
Now, we want to express this in terms of cotangent. We know \( \tan \theta = \frac{1}{\cot \theta} \).
Replace all \( \tan \) terms with \( \frac{1}{\cot} \):
LHS \( = \frac{1 - \frac{1}{\cot A} \cdot \frac{1}{\cot B}}{\frac{1}{\cot A} + \frac{1}{\cot B}} \)
In the numerator, combine the terms: \( 1 - \frac{1}{\cot A \cot B} = \frac{\cot A \cot B - 1}{\cot A \cot B} \)
In the denominator, combine the terms: \( \frac{1}{\cot A} + \frac{1}{\cot B} = \frac{\cot B + \cot A}{\cot A \cot B} \)
Substitute these back into the LHS expression:
LHS \( = \frac{\frac{\cot A \cot B - 1}{\cot A \cot B}}{\frac{\cot B + \cot A}{\cot A \cot B}} \)
The \( \cot A \cot B \) in the denominator of both the main numerator and main denominator cancel out.
LHS \( = \frac{\cot A \cot B - 1}{\cot B + \cot A} \)
This matches the Right Hand Side (RHS), completing the proof. This derivation shows how to convert tangent formulas into cotangent formulas.
In simple words: We start by rewriting \( \cot (A+B) \) as \( \frac{1}{\tan (A+B)} \). Then we use the well-known formula for \( \tan (A+B) \). After that, we replace every \( \tan \) with \( \frac{1}{\cot} \) and simplify the complex fraction by finding common denominators. The \( \cot A \cot B \) terms cancel out from the top and bottom of the big fraction, leaving us with the desired formula for \( \cot (A+B) \).

🎯 Exam Tip: When asked to prove a cotangent identity, it's often easiest to start with the reciprocal of the corresponding tangent identity and then convert all tangent terms to cotangent terms.

Question 22. If \( \tan x = \frac{n}{n+1} \) and \( \tan y = \frac{1}{2n+1} \), find \( \tan (x + y) \).
Answer: We are given:
\( \tan x = \frac{n}{n+1} \)
\( \tan y = \frac{1}{2n+1} \)
We need to find \( \tan (x + y) \).
Using the tangent sum formula: \( \tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \)
Substitute the given values of \( \tan x \) and \( \tan y \) into the formula:
\( \tan (x + y) = \frac{\frac{n}{n+1} + \frac{1}{2n+1}}{1 - \frac{n}{n+1} \cdot \frac{1}{2n+1}} \)

First, calculate the numerator:
\( \frac{n}{n+1} + \frac{1}{2n+1} = \frac{n(2n+1) + 1(n+1)}{(n+1)(2n+1)} \)
\( = \frac{2n^2 + n + n + 1}{(n+1)(2n+1)} = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)} \)

Next, calculate the denominator:
\( 1 - \frac{n}{(n+1)(2n+1)} = \frac{(n+1)(2n+1) - n}{(n+1)(2n+1)} \)
\( = \frac{2n^2 + n + 2n + 1 - n}{(n+1)(2n+1)} = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)} \)

Now, substitute the simplified numerator and denominator back into the \( \tan (x + y) \) formula:
\( \tan (x + y) = \frac{\frac{2n^2 + 2n + 1}{(n+1)(2n+1)}}{\frac{2n^2 + 2n + 1}{(n+1)(2n+1)}} \)
Since the numerator and denominator are identical, the fraction simplifies to 1.
\( \tan (x + y) = 1 \)
This problem demonstrates how algebraic simplification is key in trigonometric problems. The answer is 1.
In simple words: We use the formula for \( \tan(x+y) \). We put the given values of \( \tan x \) and \( \tan y \) into the formula. Then, we carefully add the fractions in the top part (numerator) and subtract the multiplied fractions from 1 in the bottom part (denominator). After doing the math, we find that the top and bottom parts become exactly the same. When you divide a number by itself, the answer is always 1.

🎯 Exam Tip: Always simplify the numerator and denominator separately in such problems before combining them, to avoid mistakes and keep calculations clear.

Question 23. Prove that \( \tan \left(\frac{\pi}{4}+\theta\right) \tan \left(\frac{3 \pi}{4}+\theta\right) = -1 \).
Answer: Let's evaluate each tangent term separately.
First term: \( \tan \left(\frac{\pi}{4}+\theta\right) \)
Using the identity \( \tan \left(\frac{\pi}{4}+A\right) = \frac{1 + \tan A}{1 - \tan A} \) (from Question 20, or expand using sum formula):
\( \tan \left(\frac{\pi}{4}+\theta\right) = \frac{1 + \tan \theta}{1 - \tan \theta} \) (Equation 1)

Second term: \( \tan \left(\frac{3 \pi}{4}+\theta\right) \)
We can write \( \frac{3 \pi}{4} \) as \( \pi - \frac{\pi}{4} \).
So, \( \tan \left(\frac{3 \pi}{4}+\theta\right) = \tan \left(\pi - \frac{\pi}{4}+\theta\right) = \tan \left(\pi + \left(\theta - \frac{\pi}{4}\right)\right) \)
Since \( \tan (\pi + X) = \tan X \):
\( = \tan \left(\theta - \frac{\pi}{4}\right) \)
We know \( \tan(X-Y) = \frac{\tan X - \tan Y}{1 + \tan X \tan Y} \).
\( \tan \left(\theta - \frac{\pi}{4}\right) = \frac{\tan \theta - \tan \frac{\pi}{4}}{1 + \tan \theta \tan \frac{\pi}{4}} \)
Since \( \tan \frac{\pi}{4} = 1 \):
\( = \frac{\tan \theta - 1}{1 + \tan \theta} \)
We can factor out -1 from the numerator to make it look similar to Equation 1:
\( = \frac{- (1 - \tan \theta)}{1 + \tan \theta} \) (Equation 2)

Now, multiply Equation 1 and Equation 2:
LHS \( = \tan \left(\frac{\pi}{4}+\theta\right) \tan \left(\frac{3 \pi}{4}+\theta\right) \)
LHS \( = \left(\frac{1 + \tan \theta}{1 - \tan \theta}\right) \cdot \left(\frac{- (1 - \tan \theta)}{1 + \tan \theta}\right) \)
The term \( (1 + \tan \theta) \) in the numerator of the first fraction cancels with the denominator of the second.
The term \( (1 - \tan \theta) \) in the denominator of the first fraction cancels with the numerator of the second.
LHS \( = -1 \)
This matches the RHS, completing the proof. This identity uses the periodic properties of the tangent function.
In simple words: We first find a simple form for \( \tan(\frac{\pi}{4}+\theta) \) using a common identity. For the second term, \( \tan(\frac{3\pi}{4}+\theta) \), we rewrite \( \frac{3\pi}{4} \) as \( \pi - \frac{\pi}{4} \), which helps us simplify the expression using known tangent properties. After simplifying both parts, we multiply them. The parts cancel out perfectly, leaving us with \( -1 \).

🎯 Exam Tip: Remember that \( \tan(\pi + X) = \tan X \) and \( \tan(\pi - X) = -\tan X \). Using these properties is key to simplifying complex angle expressions efficiently.

Question 24. Find the value of \( \tan (\alpha + \beta) \), given that \( \cot \alpha = \frac{1}{2} \), \( \alpha \in \left(\pi, \frac{3 \pi}{2}\right) \) and \( \sec \beta = -\frac{5}{3} \), \( \beta \in \left(\frac{\pi}{2}, \pi\right) \).
Answer: We need to find \( \tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \). So we need to find \( \tan \alpha \) and \( \tan \beta \).

For \( \alpha \):
Given \( \cot \alpha = \frac{1}{2} \).
Since \( \tan \alpha = \frac{1}{\cot \alpha} \), we have \( \tan \alpha = \frac{1}{1/2} = 2 \).
The angle \( \alpha \) is in the interval \( \left(\pi, \frac{3 \pi}{2}\right) \), which is the third quadrant. In the third quadrant, \( \tan \alpha \) is positive, which matches our value of 2.

For \( \beta \):
Given \( \sec \beta = -\frac{5}{3} \).
We know the identity \( \sec^2 \beta - \tan^2 \beta = 1 \).
So, \( \tan^2 \beta = \sec^2 \beta - 1 \).
\( \tan^2 \beta = \left(-\frac{5}{3}\right)^2 - 1 = \frac{25}{9} - 1 = \frac{25 - 9}{9} = \frac{16}{9} \)
Therefore, \( \tan \beta = \pm \sqrt{\frac{16}{9}} = \pm \frac{4}{3} \).
The angle \( \beta \) is in the interval \( \left(\frac{\pi}{2}, \pi\right) \), which is the second quadrant. In the second quadrant, \( \tan \beta \) is negative.
So, we must choose \( \tan \beta = -\frac{4}{3} \).

Now, substitute \( \tan \alpha = 2 \) and \( \tan \beta = -\frac{4}{3} \) into the formula for \( \tan (\alpha + \beta) \):
\( \tan (\alpha + \beta) = \frac{2 + \left(-\frac{4}{3}\right)}{1 - 2 \cdot \left(-\frac{4}{3}\right)} \)
\( = \frac{2 - \frac{4}{3}}{1 + \frac{8}{3}} \)
Calculate the numerator: \( 2 - \frac{4}{3} = \frac{6}{3} - \frac{4}{3} = \frac{2}{3} \)
Calculate the denominator: \( 1 + \frac{8}{3} = \frac{3}{3} + \frac{8}{3} = \frac{11}{3} \)
So, \( \tan (\alpha + \beta) = \frac{\frac{2}{3}}{\frac{11}{3}} \)
\( = \frac{2}{3} \cdot \frac{3}{11} \)
\( = \frac{2}{11} \)
The value of \( \tan (\alpha + \beta) \) is \( \frac{2}{11} \). This calculation emphasizes the importance of correctly identifying the quadrant for trigonometric function signs.
In simple words: First, we find \( \tan \alpha \) from \( \cot \alpha \) and check its sign based on the given quadrant for \( \alpha \). Then, we find \( \tan \beta \) from \( \sec \beta \) using a squared identity, and we pick the correct sign (negative in this case) because \( \beta \) is in the second quadrant. Finally, we put these \( \tan \alpha \) and \( \tan \beta \) values into the \( \tan(A+B) \) formula and simplify the fractions to get the final answer.

🎯 Exam Tip: Always pay close attention to the given quadrant for each angle. The quadrant determines the sign of trigonometric functions and is crucial for correctly calculating values from identities.

Question 25. If \( \theta + \phi = \alpha \) and \( \tan \theta = k \tan \phi \), then prove that \( \sin (\theta - \phi) = \frac{k-1}{k+1} \sin \alpha \).
Answer: We are given:
1. \( \theta + \phi = \alpha \)
2. \( \tan \theta = k \tan \phi \)
We need to prove \( \sin (\theta - \phi) = \frac{k-1}{k+1} \sin \alpha \).

From (2), we can write \( \frac{\tan \theta}{\tan \phi} = k \).
Rewrite \( \tan \theta \) and \( \tan \phi \) in terms of sine and cosine:
\( \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \phi}{\cos \phi}} = k \)
\( \frac{\sin \theta \cos \phi}{\cos \theta \sin \phi} = k \)
Now, apply the componendo and dividendo rule (or simply subtract 1 from both sides and add 1 to both sides, then divide the results). This rule states if \( \frac{a}{b} = k \), then \( \frac{a-b}{a+b} = \frac{k-1}{k+1} \).
So, \( \frac{\sin \theta \cos \phi - \cos \theta \sin \phi}{\sin \theta \cos \phi + \cos \theta \sin \phi} = \frac{k-1}{k+1} \)
The numerator is the expansion of \( \sin (\theta - \phi) \).
The denominator is the expansion of \( \sin (\theta + \phi) \).
\( \frac{\sin (\theta - \phi)}{\sin (\theta + \phi)} = \frac{k-1}{k+1} \)
From (1), we know \( \theta + \phi = \alpha \). Substitute this into the equation:
\( \frac{\sin (\theta - \phi)}{\sin \alpha} = \frac{k-1}{k+1} \)
Multiply both sides by \( \sin \alpha \):
\( \sin (\theta - \phi) = \frac{k-1}{k+1} \sin \alpha \)
This proves the identity. This problem nicely uses the componendo and dividendo property with trigonometric functions.
In simple words: We are given two facts: that two angles add up to \( \alpha \), and that the tangent of one is 'k' times the tangent of the other. We rewrite the tangent relationship using sine and cosine. Then, we use a special math rule called componendo and dividendo, which helps to change ratios. Applying this rule transforms the equation into \( \frac{\sin(\theta-\phi)}{\sin(\theta+\phi)} = \frac{k-1}{k+1} \). Since we know \( \theta+\phi = \alpha \), we can replace it, and then simply move \( \sin \alpha \) to the other side to get the final answer.

🎯 Exam Tip: The componendo and dividendo rule (if \( \frac{a}{b} = c \implies \frac{a-b}{a+b} = \frac{c-1}{c+1} \)) is a powerful tool in such problems. Recognizing when to apply it can greatly simplify proofs.