Samacheer Kalvi Class 11 Business Maths Solutions Chapter 4 Trigonometry Exercise 4.2

 

Question 1. Find the values of the following:
(i) cosec 15°
(ii) sin (-105°)
(iii) cot 75°
Answer:
(i) To find cosec 15°, we first find sin 15°. We can write sin 15° as sin(45° - 30°).
Using the formula \( \sin(A - B) = \sin A \cos B - \cos A \sin B \):
\( \sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) \)
\( = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \)
\( = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \)
\( = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \)
\( = \frac{\sqrt{3}-1}{2\sqrt{2}} \)
Now, \( \text{cosec } 15^{\circ} = \frac{1}{\sin 15^{\circ}} \)
\( = \frac{1}{\frac{\sqrt{3}-1}{2\sqrt{2}}} \)
\( = \frac{2\sqrt{2}}{\sqrt{3}-1} \)
To simplify, we multiply the numerator and denominator by the conjugate of the denominator, \( \sqrt{3}+1 \).
\( = \frac{2\sqrt{2}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \)
\( = \frac{2\sqrt{2}(\sqrt{3}+1)}{(\sqrt{3})^2 - 1^2} \)
\( = \frac{2\sqrt{2}(\sqrt{3}+1)}{3 - 1} \)
\( = \frac{2\sqrt{2}(\sqrt{3}+1)}{2} \)
\( = \sqrt{2}(\sqrt{3}+1) \)
\( = \sqrt{6}+\sqrt{2} \)

(ii) To find sin (-105°), we can use the property \( \sin(-\theta) = -\sin \theta \).
\( \sin(-105^{\circ}) = -\sin(105^{\circ}) \)
Now, we find \( \sin(105^{\circ}) \). We can write it as \( \sin(60^{\circ} + 45^{\circ}) \).
Using the formula \( \sin(A + B) = \sin A \cos B + \cos A \sin B \):
\( \sin(105^{\circ}) = \sin(60^{\circ} + 45^{\circ}) \)
\( = \sin 60^{\circ} \cos 45^{\circ} + \cos 60^{\circ} \sin 45^{\circ} \)
\( = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \)
\( = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \)
\( = \frac{\sqrt{3}+1}{2\sqrt{2}} \)
So, \( \sin(-105^{\circ}) = -\frac{\sqrt{3}+1}{2\sqrt{2}} \)
We can also rationalize the denominator:
\( = -\frac{\sqrt{3}+1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \)
\( = -\frac{\sqrt{2}(\sqrt{3}+1)}{2 \times 2} \)
\( = -\frac{\sqrt{6}+\sqrt{2}}{4} \)

(iii) To find cot 75°, we first find tan 75°. We can write tan 75° as tan(30° + 45°).
Using the formula \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \):
\( \tan 75^{\circ} = \tan(30^{\circ} + 45^{\circ}) \)
\( = \frac{\tan 30^{\circ} + \tan 45^{\circ}}{1 - \tan 30^{\circ} \tan 45^{\circ}} \)
\( = \frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}} \cdot 1} \)
\( = \frac{\frac{1+\sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} \)
\( = \frac{1+\sqrt{3}}{\sqrt{3}-1} \)
Now, \( \cot 75^{\circ} = \frac{1}{\tan 75^{\circ}} \)
\( = \frac{1}{\frac{1+\sqrt{3}}{\sqrt{3}-1}} \)
\( = \frac{\sqrt{3}-1}{\sqrt{3}+1} \)
To simplify, we multiply the numerator and denominator by the conjugate of the denominator, \( \sqrt{3}-1 \).
\( = \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \)
\( = \frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2 - 1^2} \)
\( = \frac{3 - 2\sqrt{3} + 1}{3 - 1} \)
\( = \frac{4 - 2\sqrt{3}}{2} \)
\( = 2 - \sqrt{3} \)
In simple words: We break down the angles into common angles like 30°, 45°, and 60° because we know their trigonometric values. Then we use addition or subtraction formulas for sine, cosine, or tangent to find the value for the given angle. Finally, we simplify the fractions and rationalize the denominators if needed.

🎯 Exam Tip: Remember the standard trigonometric values for 0°, 30°, 45°, 60°, and 90° as they are frequently used in these types of problems. Also, practice rationalizing denominators with square roots to ensure your final answer is in the simplest form.

 

Question 2. Find the values of the following:
(i) sin 76° cos 16° – cos 76° sin 16°
(ii) \( \sin \frac{\pi}{4} \cos \frac{\pi}{12}+\cos \frac{\pi}{4} \sin \frac{\pi}{12} \)
(iii) cos 70° cos 10° – sin 70° sin 10°
(iv) cos² 15° – sin² 15°
Answer:
(i) We need to find the value of sin 76° cos 16° – cos 76° sin 16°.
This expression matches the formula for \( \sin(A - B) \), which is \( \sin A \cos B - \cos A \sin B \).
Here, \( A = 76^{\circ} \) and \( B = 16^{\circ} \).
So, \( \sin 76^{\circ} \cos 16^{\circ} - \cos 76^{\circ} \sin 16^{\circ} = \sin(76^{\circ} - 16^{\circ}) \)
\( = \sin(60^{\circ}) \)
We know that \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \).

(ii) We need to find the value of \( \sin \frac{\pi}{4} \cos \frac{\pi}{12}+\cos \frac{\pi}{4} \sin \frac{\pi}{12} \).
This expression matches the formula for \( \sin(A + B) \), which is \( \sin A \cos B + \cos A \sin B \).
Here, \( A = \frac{\pi}{4} \) and \( B = \frac{\pi}{12} \).
So, \( \sin \frac{\pi}{4} \cos \frac{\pi}{12}+\cos \frac{\pi}{4} \sin \frac{\pi}{12} = \sin(\frac{\pi}{4} + \frac{\pi}{12}) \)
First, add the angles inside the sine function:
\( \frac{\pi}{4} + \frac{\pi}{12} = \frac{3\pi}{12} + \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3} \)
Therefore, the expression becomes \( \sin(\frac{\pi}{3}) \).
We know that \( \sin(\frac{\pi}{3}) \) (or \( \sin 60^{\circ} \)) is \( \frac{\sqrt{3}}{2} \).

(iii) We need to find the value of cos 70° cos 10° – sin 70° sin 10°.
This expression matches the formula for \( \cos(A + B) \), which is \( \cos A \cos B - \sin A \sin B \).
Here, \( A = 70^{\circ} \) and \( B = 10^{\circ} \).
So, \( \cos 70^{\circ} \cos 10^{\circ} - \sin 70^{\circ} \sin 10^{\circ} = \cos(70^{\circ} + 10^{\circ}) \)
\( = \cos(80^{\circ}) \).

(iv) We need to find the value of cos² 15° – sin² 15°.
This expression matches the formula for \( \cos 2A \), which is \( \cos^2 A - \sin^2 A \).
Here, \( A = 15^{\circ} \).
So, \( \cos^2 15^{\circ} - \sin^2 15^{\circ} = \cos(2 \times 15^{\circ}) \)
\( = \cos(30^{\circ}) \)
We know that \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \).
In simple words: For these problems, we look for patterns that match standard trigonometric addition or double-angle formulas. Once we identify the correct formula, we just plug in the angles and calculate the value. This helps to simplify complex expressions into simpler forms.

🎯 Exam Tip: Memorize the key trigonometric identities, especially for sum, difference, and double angles. Converting degrees to radians or vice-versa might be helpful in some cases, but the core is recognizing the identity.

 

Question 3. If sin A = \( \frac{3}{5} \), 0 < A < \( \frac{\pi}{2} \) and cos B = \( \frac{-12}{13} \), \( \pi < B < \frac{3\pi}{2} \) find the values of the following:
(i) cos(A + B)
(ii) sin(A - B)
(iii) tan(A - B)
Answer:
Given: \( \sin A = \frac{3}{5} \), where A is in the first quadrant (0 < A < \( \frac{\pi}{2} \)).
Given: \( \cos B = \frac{-12}{13} \), where B is in the third quadrant (\( \pi < B < \frac{3\pi}{2} \)).

For angle A (First Quadrant):
\( \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{5} \)
Using Pythagoras theorem, \( \text{Adjacent} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \).
So, \( \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5} \) (positive in Q1)
And \( \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4} \) (positive in Q1)

For angle B (Third Quadrant):
\( \cos B = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{-12}{13} \). (The negative sign indicates Q3)
Using Pythagoras theorem, \( \text{Opposite} = \sqrt{13^2 - (-12)^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \).
In the third quadrant, sine is negative, so \( \sin B = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{-5}{13} \).
In the third quadrant, tangent is positive, so \( \tan B = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{-5}{-12} = \frac{5}{12} \).

(i) Value of cos(A + B):
Using the formula \( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
Substitute the values:
\( \cos(A + B) = \left(\frac{4}{5}\right) \left(\frac{-12}{13}\right) - \left(\frac{3}{5}\right) \left(\frac{-5}{13}\right) \)
\( = \frac{-48}{65} - \frac{-15}{65} \)
\( = \frac{-48 + 15}{65} \)
\( = \frac{-33}{65} \)

(ii) Value of sin(A - B):
Using the formula \( \sin(A - B) = \sin A \cos B - \cos A \sin B \)
Substitute the values:
\( \sin(A - B) = \left(\frac{3}{5}\right) \left(\frac{-12}{13}\right) - \left(\frac{4}{5}\right) \left(\frac{-5}{13}\right) \)
\( = \frac{-36}{65} - \frac{-20}{65} \)
\( = \frac{-36 + 20}{65} \)
\( = \frac{-16}{65} \)

(iii) Value of tan(A - B):
Using the formula \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)
Substitute the values:
\( \tan(A - B) = \frac{\frac{3}{4} - \frac{5}{12}}{1 + \left(\frac{3}{4}\right) \left(\frac{5}{12}\right)} \)
First, simplify the numerator:
\( \frac{3}{4} - \frac{5}{12} = \frac{9}{12} - \frac{5}{12} = \frac{4}{12} = \frac{1}{3} \)
Next, simplify the denominator:
\( 1 + \frac{15}{48} = 1 + \frac{5}{16} = \frac{16}{16} + \frac{5}{16} = \frac{21}{16} \)
So, \( \tan(A - B) = \frac{\frac{1}{3}}{\frac{21}{16}} \)
\( = \frac{1}{3} \times \frac{16}{21} \)
\( = \frac{16}{63} \)
In simple words: First, we use the given sine and cosine values to find all six trigonometric ratios for angles A and B, remembering to consider which quadrant each angle is in. Then, we use the specific addition and subtraction formulas for cosine, sine, and tangent to calculate the required values by plugging in the ratios we found. Drawing a quick triangle can help visualize the sides.

🎯 Exam Tip: Always pay close attention to the quadrant of the angles, as this determines the sign of the trigonometric ratios (positive or negative). A common mistake is to forget the signs in the second, third, or fourth quadrants.

 

Question 4. If cos A = \( \frac{13}{14} \) and cos B = \( \frac{1}{7} \) where A, B are acute angles prove that A – B = \( \frac{\pi}{3} \)
Answer:
Given: \( \cos A = \frac{13}{14} \) and \( \cos B = \frac{1}{7} \). Both A and B are acute angles, meaning they are in the first quadrant.
We want to prove that \( A - B = \frac{\pi}{3} \). We can use the formula for \( \cos(A - B) \).
\( \cos(A - B) = \cos A \cos B + \sin A \sin B \)
First, we need to find \( \sin A \) and \( \sin B \). Since A and B are acute, their sine values will be positive.
For \( \sin A \):
\( \sin A = \sqrt{1 - \cos^2 A} \)
\( = \sqrt{1 - \left(\frac{13}{14}\right)^2} \)
\( = \sqrt{1 - \frac{169}{196}} \)
\( = \sqrt{\frac{196 - 169}{196}} \)
\( = \sqrt{\frac{27}{196}} \)
\( = \frac{\sqrt{27}}{\sqrt{196}} = \frac{3\sqrt{3}}{14} \)
So, \( \sin A = \frac{3\sqrt{3}}{14} \).

For \( \sin B \):
\( \sin B = \sqrt{1 - \cos^2 B} \)
\( = \sqrt{1 - \left(\frac{1}{7}\right)^2} \)
\( = \sqrt{1 - \frac{1}{49}} \)
\( = \sqrt{\frac{49 - 1}{49}} \)
\( = \sqrt{\frac{48}{49}} \)
\( = \frac{\sqrt{48}}{\sqrt{49}} = \frac{4\sqrt{3}}{7} \)
So, \( \sin B = \frac{4\sqrt{3}}{7} \).

Now substitute these values into the \( \cos(A - B) \) formula:
\( \cos(A - B) = \left(\frac{13}{14}\right) \left(\frac{1}{7}\right) + \left(\frac{3\sqrt{3}}{14}\right) \left(\frac{4\sqrt{3}}{7}\right) \)
\( = \frac{13}{98} + \frac{12 \times 3}{98} \)
\( = \frac{13}{98} + \frac{36}{98} \)
\( = \frac{13 + 36}{98} \)
\( = \frac{49}{98} \)
\( = \frac{1}{2} \)
We know that \( \cos 60^{\circ} = \frac{1}{2} \). Since A and B are acute, A - B will be between -90° and 90°. For a positive cosine value, A - B must be in the first quadrant.
Therefore, \( A - B = 60^{\circ} \).
Converting 60° to radians, \( 60^{\circ} = \frac{\pi}{3} \text{ radians} \).
Hence, \( A - B = \frac{\pi}{3} \).
In simple words: We used the given cosine values and the fact that angles A and B are acute to find their sine values. Then, we applied the cosine difference formula, \( \cos(A-B) \), and after calculating, we found that \( \cos(A-B) \) equals \( \frac{1}{2} \). Since we know that \( \cos 60^{\circ} \) is \( \frac{1}{2} \), this proves that A - B must be 60 degrees, which is \( \frac{\pi}{3} \) in radians.

🎯 Exam Tip: When given values for cosine or sine and information about the quadrant (like "acute angle"), immediately calculate the other required trigonometric ratios using \( \sin^2 \theta + \cos^2 \theta = 1 \). This prepares all necessary parts for using sum/difference formulas efficiently.

 

Question 5. Prove that 2 tan 80° = tan 85° – tan 5°.
Answer:
We need to prove that \( 2 \tan 80^{\circ} = \tan 85^{\circ} - \tan 5^{\circ} \).
Let's start with the right-hand side (RHS) of the equation:
RHS = \( \tan 85^{\circ} - \tan 5^{\circ} \)
We know that \( 80^{\circ} = 85^{\circ} - 5^{\circ} \).
Using the formula for \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \), we can write:
\( \tan(85^{\circ} - 5^{\circ}) = \frac{\tan 85^{\circ} - \tan 5^{\circ}}{1 + \tan 85^{\circ} \tan 5^{\circ}} \)
So, \( \tan 80^{\circ} = \frac{\tan 85^{\circ} - \tan 5^{\circ}}{1 + \tan 85^{\circ} \tan 5^{\circ}} \)
From this, we can write:
\( \tan 85^{\circ} - \tan 5^{\circ} = \tan 80^{\circ} (1 + \tan 85^{\circ} \tan 5^{\circ}) \)
We also know that \( \tan 85^{\circ} = \tan(90^{\circ} - 5^{\circ}) = \cot 5^{\circ} \).
Substitute \( \tan 85^{\circ} = \cot 5^{\circ} \) into the equation:
\( \tan 85^{\circ} - \tan 5^{\circ} = \tan 80^{\circ} (1 + \cot 5^{\circ} \tan 5^{\circ}) \)
Since \( \cot 5^{\circ} \tan 5^{\circ} = 1 \), the equation becomes:
\( \tan 85^{\circ} - \tan 5^{\circ} = \tan 80^{\circ} (1 + 1) \)
\( \tan 85^{\circ} - \tan 5^{\circ} = \tan 80^{\circ} (2) \)
\( \tan 85^{\circ} - \tan 5^{\circ} = 2 \tan 80^{\circ} \)
This proves the statement. The properties of complementary angles are very useful here.
In simple words: We start with the difference of the two tangent terms on the right side. We recognize that 85 degrees minus 5 degrees equals 80 degrees, which relates to the left side. By using the tangent subtraction formula and the fact that tan(90° - x) is cot(x), we can simplify the expression until it matches the left side, proving the statement.

🎯 Exam Tip: When proving identities involving differences or sums of tangent functions, consider if one angle can be expressed as a complementary angle of another. This often simplifies the expression to \( \tan x \cdot \cot x = 1 \), which is a key step in such proofs.

 

Question 6. If cot \( \alpha = \frac{1}{2} \), sec \( \beta = \frac{-5}{3} \), where \( \pi < \alpha < \frac{3\pi}{2} \) and \( \frac{\pi}{2} < \beta < \pi \), find the value of tan(\( \alpha + \beta \)). State the quadrant in which \( \alpha + \beta \) terminates.
Answer:
We need to find \( \tan(\alpha + \beta) \). The formula for this is \( \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \).
First, find \( \tan \alpha \) and \( \tan \beta \).

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

For angle \( \beta \):
Given \( \sec \beta = \frac{-5}{3} \). Since \( \cos \beta = \frac{1}{\sec \beta} \), we have \( \cos \beta = \frac{-3}{5} \).
The angle \( \beta \) is in the second quadrant (\( \frac{\pi}{2} < \beta < \pi \)). In the second quadrant, cosine is negative, which matches our value.
To find \( \tan \beta \), we can use a right triangle or \( \sin^2 \beta + \cos^2 \beta = 1 \).
Let's use a triangle. If \( \cos \beta = \frac{-3}{5} \), the adjacent side is 3 and the hypotenuse is 5 (ignoring the negative sign for now, just for side lengths).
The opposite side would be \( \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \).
In the second quadrant, sine is positive and tangent is negative.
So, \( \sin \beta = \frac{4}{5} \) and \( \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{4}{5}}{\frac{-3}{5}} = \frac{-4}{3} \).

Now substitute \( \tan \alpha = 2 \) and \( \tan \beta = \frac{-4}{3} \) into the \( \tan(\alpha + \beta) \) formula:
\( \tan(\alpha + \beta) = \frac{2 + \left(\frac{-4}{3}\right)}{1 - (2) \left(\frac{-4}{3}\right)} \)
\( = \frac{\frac{6}{3} - \frac{4}{3}}{1 + \frac{8}{3}} \)
\( = \frac{\frac{2}{3}}{\frac{3}{3} + \frac{8}{3}} \)
\( = \frac{\frac{2}{3}}{\frac{11}{3}} \)
\( = \frac{2}{3} \times \frac{3}{11} \)
\( = \frac{2}{11} \)

The value of \( \tan(\alpha + \beta) \) is \( \frac{2}{11} \), which is positive.
To determine the quadrant of \( \alpha + \beta \):
We know \( \pi < \alpha < \frac{3\pi}{2} \) (i.e., \( 180^{\circ} < \alpha < 270^{\circ} \)).
We know \( \frac{\pi}{2} < \beta < \pi \) (i.e., \( 90^{\circ} < \beta < 180^{\circ} \)).
Adding the ranges for \( \alpha \) and \( \beta \):
\( \pi + \frac{\pi}{2} < \alpha + \beta < \frac{3\pi}{2} + \pi \)
\( \frac{3\pi}{2} < \alpha + \beta < \frac{5\pi}{2} \)
This means \( 270^{\circ} < \alpha + \beta < 450^{\circ} \).
Since \( \tan(\alpha + \beta) \) is positive, \( \alpha + \beta \) must be in either the first or third quadrant. Our range \( (270^{\circ}, 450^{\circ}) \) includes the fourth quadrant \( (270^{\circ}, 360^{\circ}) \) and the first quadrant \( (360^{\circ}, 450^{\circ}) \).
Because \( \tan(\alpha + \beta) \) is positive, it must lie in the first quadrant, specifically between \( 360^{\circ} \) and \( 450^{\circ} \), which is equivalent to the first quadrant for its principal value.
So, \( \alpha + \beta \) terminates in the first quadrant.
In simple words: First, we use the given cot \( \alpha \) and sec \( \beta \) values, along with their quadrant information, to find \( \tan \alpha \) and \( \tan \beta \). Then, we use the tangent addition formula to calculate \( \tan(\alpha + \beta) \). Since the result is positive, and the sum of the ranges for \( \alpha \) and \( \beta \) shows that \( \alpha + \beta \) could be in the first or fourth equivalent quadrant, we conclude that \( \alpha + \beta \) is in the first quadrant.

🎯 Exam Tip: Always sketch a quick right-angled triangle for the angle and use the quadrant information to correctly assign positive or negative signs to the trigonometric ratios. This is a crucial step before applying any addition or subtraction formulas.

 

Question 7. If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan \( 22\frac{1}{2}^{\circ} \).
Answer:
Given: \( A + B = 45^{\circ} \).
Take the tangent of both sides:
\( \tan(A + B) = \tan 45^{\circ} \)
Use the tangent addition formula \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). We know \( \tan 45^{\circ} = 1 \).
So, \( \frac{\tan A + \tan B}{1 - \tan A \tan B} = 1 \)
Multiply both sides by \( (1 - \tan A \tan B) \):
\( \tan A + \tan B = 1 - \tan A \tan B \)
Rearrange the terms to group all tangent terms on one side:
\( \tan A + \tan B + \tan A \tan B = 1 \)
To get the desired form, add 1 to both sides:
\( 1 + \tan A + \tan B + \tan A \tan B = 1 + 1 \)
\( 1 + \tan A + \tan B (1 + \tan A) = 2 \)
Factor out \( (1 + \tan A) \):
\( (1 + \tan A) (1 + \tan B) = 2 \). This proves the first part.

Now, to deduce the value of \( \tan 22\frac{1}{2}^{\circ} \):
Let \( A = B = 22\frac{1}{2}^{\circ} \).
If \( A = B = 22\frac{1}{2}^{\circ} \), then \( A + B = 22\frac{1}{2}^{\circ} + 22\frac{1}{2}^{\circ} = 45^{\circ} \). This fits the condition.
Substitute \( A = 22\frac{1}{2}^{\circ} \) and \( B = 22\frac{1}{2}^{\circ} \) into the proved identity \( (1 + \tan A) (1 + \tan B) = 2 \):
\( (1 + \tan 22\frac{1}{2}^{\circ}) (1 + \tan 22\frac{1}{2}^{\circ}) = 2 \)
\( (1 + \tan 22\frac{1}{2}^{\circ})^2 = 2 \)
Take the square root of both sides:
\( 1 + \tan 22\frac{1}{2}^{\circ} = \pm \sqrt{2} \)
Solve for \( \tan 22\frac{1}{2}^{\circ} \):
\( \tan 22\frac{1}{2}^{\circ} = -1 \pm \sqrt{2} \)
Since \( 22\frac{1}{2}^{\circ} \) is an acute angle (less than 90°), \( \tan 22\frac{1}{2}^{\circ} \) must be positive.
\( \sqrt{2} \approx 1.414 \).
If we choose \( -1 - \sqrt{2} \), the result is negative. If we choose \( -1 + \sqrt{2} \), the result is positive.
So, \( \tan 22\frac{1}{2}^{\circ} = \sqrt{2} - 1 \).
In simple words: First, we use the given condition \( A+B=45^{\circ} \) and the tangent addition formula. By simplifying the equation, we prove the identity \( (1+\tan A)(1+\tan B)=2 \). Then, to find \( \tan 22\frac{1}{2}^{\circ} \), we set A and B equal to \( 22\frac{1}{2}^{\circ} \) in the identity, solve for tan, and pick the positive value since \( 22\frac{1}{2}^{\circ} \) is an acute angle.

🎯 Exam Tip: When solving for an angle's trigonometric value, always consider the quadrant it lies in. This helps to determine the correct sign (positive or negative) of the result, especially when dealing with square roots.

 

Question 8. Prove that
(i) sin(A + 60°) + sin(A – 60°) = sin A.
(ii) tan 4A tan 3A tan A + tan 3A + tan A – tan 4A = 0
Answer:
(i) Prove that \( \sin(A + 60^{\circ}) + \sin(A - 60^{\circ}) = \sin A \).
Start with the Left Hand Side (LHS):
LHS = \( \sin(A + 60^{\circ}) + \sin(A - 60^{\circ}) \)
Use the sum and difference formulas for sine:
\( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
\( \sin(A - B) = \sin A \cos B - \cos A \sin B \)
Substitute \( B = 60^{\circ} \):
\( \sin(A + 60^{\circ}) = \sin A \cos 60^{\circ} + \cos A \sin 60^{\circ} \)
\( \sin(A - 60^{\circ}) = \sin A \cos 60^{\circ} - \cos A \sin 60^{\circ} \)
Add these two expressions:
\( \sin(A + 60^{\circ}) + \sin(A - 60^{\circ}) = (\sin A \cos 60^{\circ} + \cos A \sin 60^{\circ}) + (\sin A \cos 60^{\circ} - \cos A \sin 60^{\circ}) \)
The \( \cos A \sin 60^{\circ} \) terms cancel out:
\( = 2 \sin A \cos 60^{\circ} \)
We know that \( \cos 60^{\circ} = \frac{1}{2} \).
\( = 2 \sin A \left(\frac{1}{2}\right) \)
\( = \sin A \)
This equals the Right Hand Side (RHS). So, the identity is proven.

(ii) Prove that \( \tan 4A \tan 3A \tan A + \tan 3A + \tan A - \tan 4A = 0 \).
We know that \( 4A = 3A + A \).
Take the tangent of both sides:
\( \tan(4A) = \tan(3A + A) \)
Use the tangent addition formula \( \tan(X + Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y} \):
\( \tan 4A = \frac{\tan 3A + \tan A}{1 - \tan 3A \tan A} \)
Multiply both sides by \( (1 - \tan 3A \tan A) \):
\( \tan 4A (1 - \tan 3A \tan A) = \tan 3A + \tan A \)
Distribute \( \tan 4A \) on the left side:
\( \tan 4A - \tan 4A \tan 3A \tan A = \tan 3A + \tan A \)
Rearrange the terms to set the equation to zero:
\( \tan 4A - \tan 4A \tan 3A \tan A - \tan 3A - \tan A = 0 \)
Multiply the entire equation by -1 to match the desired form:
\( -\tan 4A + \tan 4A \tan 3A \tan A + \tan 3A + \tan A = 0 \)
Rearranging the terms in the order given in the question:
\( \tan 4A \tan 3A \tan A + \tan 3A + \tan A - \tan 4A = 0 \). So, the identity is proven.
In simple words: For the first part, we use the sum and difference formulas for sine. When we add the two expanded expressions, some terms cancel out, leaving us with \( 2 \sin A \cos 60^{\circ} \), which simplifies to \( \sin A \). For the second part, we start by noting that \( 4A \) is the sum of \( 3A \) and \( A \). We apply the tangent addition formula, then rearrange the resulting equation to prove the given identity.

🎯 Exam Tip: When proving trigonometric identities, always start with the more complex side (usually the LHS) and simplify it step-by-step. Look for opportunities to apply fundamental identities or sum/difference formulas. For product-sum identities, remember that \( \cos 60^{\circ} \) is \( 1/2 \), which often helps in simplification.

 

Question 9.
(i) If tan \( \theta = 3 \) find tan \( 3\theta \)
(ii) If sin A = \( \frac{12}{13} \), find sin 3A.
Answer:
(i) We need to find \( \tan 3\theta \) given \( \tan \theta = 3 \).
Use the triple angle formula for tangent:
\( \tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \)
Substitute \( \tan \theta = 3 \):
\( \tan 3\theta = \frac{3(3) - (3)^3}{1 - 3(3)^2} \)
\( = \frac{9 - 27}{1 - 3(9)} \)
\( = \frac{-18}{1 - 27} \)
\( = \frac{-18}{-26} \)
\( = \frac{9}{13} \)

(ii) We need to find \( \sin 3A \) given \( \sin A = \frac{12}{13} \).
Use the triple angle formula for sine:
\( \sin 3A = 3 \sin A - 4 \sin^3 A \)
Substitute \( \sin A = \frac{12}{13} \):
\( \sin 3A = 3 \left(\frac{12}{13}\right) - 4 \left(\frac{12}{13}\right)^3 \)
\( = \frac{36}{13} - 4 \left(\frac{1728}{2197}\right) \)
\( = \frac{36}{13} - \frac{6912}{2197} \)
To subtract, find a common denominator, which is 2197 (since \( 13 \times 169 = 2197 \)).
\( = \frac{36 \times 169}{13 \times 169} - \frac{6912}{2197} \)
\( = \frac{6084}{2197} - \frac{6912}{2197} \)
\( = \frac{6084 - 6912}{2197} \)
\( = \frac{-828}{2197} \)
In simple words: For both parts, we directly use the specific triple angle formulas for tangent and sine. We substitute the given value of \( \tan \theta \) or \( \sin A \) into the respective formula and then perform the calculations to find the final answer. These formulas help to quickly find the trigonometric value of a triple angle from the single angle's value.

🎯 Exam Tip: Memorizing the triple angle formulas for sine, cosine, and tangent is essential for these types of questions. Be careful with calculations, especially when dealing with cubes and fractions, to avoid arithmetic errors.

 

Question 10. If sin A = \( \frac{3}{5} \), find the values of cos 3A and tan 3A.
Answer:
Given \( \sin A = \frac{3}{5} \). We need to find \( \cos 3A \) and \( \tan 3A \).
First, find \( \cos A \) and \( \tan A \). Assume A is an acute angle (in the first quadrant) since no quadrant is specified, and \( \sin A \) is positive.
Using \( \sin^2 A + \cos^2 A = 1 \):
\( \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \).
Since A is acute, \( \cos A \) is positive.
Now, \( \tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \).

Now, use the triple angle formulas:

For \( \cos 3A \):
\( \cos 3A = 4 \cos^3 A - 3 \cos A \)
Substitute \( \cos A = \frac{4}{5} \):
\( \cos 3A = 4 \left(\frac{4}{5}\right)^3 - 3 \left(\frac{4}{5}\right) \)
\( = 4 \left(\frac{64}{125}\right) - \frac{12}{5} \)
\( = \frac{256}{125} - \frac{12 \times 25}{5 \times 25} \)
\( = \frac{256}{125} - \frac{300}{125} \)
\( = \frac{256 - 300}{125} \)
\( = \frac{-44}{125} \)

For \( \tan 3A \):
\( \tan 3A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} \)
Substitute \( \tan A = \frac{3}{4} \):
\( \tan 3A = \frac{3\left(\frac{3}{4}\right) - \left(\frac{3}{4}\right)^3}{1 - 3\left(\frac{3}{4}\right)^2} \)
\( = \frac{\frac{9}{4} - \frac{27}{64}}{1 - 3\left(\frac{9}{16}\right)} \)
\( = \frac{\frac{9 \times 16}{4 \times 16} - \frac{27}{64}}{1 - \frac{27}{16}} \)
\( = \frac{\frac{144}{64} - \frac{27}{64}}{\frac{16}{16} - \frac{27}{16}} \)
\( = \frac{\frac{144 - 27}{64}}{\frac{16 - 27}{16}} \)
\( = \frac{\frac{117}{64}}{\frac{-11}{16}} \)
\( = \frac{117}{64} \times \frac{16}{-11} \)
\( = \frac{117}{4 \times (-11)} \)
\( = \frac{117}{-44} = -\frac{117}{44} \)
In simple words: First, we use the given \( \sin A \) value to find \( \cos A \) and \( \tan A \) by drawing a right triangle and applying the Pythagorean theorem. Then, we use the specific triple angle formulas for cosine and tangent, substituting the found values, to calculate \( \cos 3A \) and \( \tan 3A \).

🎯 Exam Tip: Always calculate all necessary single angle trigonometric ratios (sine, cosine, tangent) first from the given information before plugging them into triple angle formulas. This helps prevent errors and ensures all required values are available.

 

Question 11. Prove that \( \frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}+\frac{\sin (A-B)}{\cos A \cos B}=0 \)
Answer:
We need to prove that \( \frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}+\frac{\sin (A-B)}{\cos A \cos B}=0 \).
Let's simplify each term separately.

Consider the first term: \( \frac{\sin (B-C)}{\cos B \cos C} \)
Using the formula \( \sin(X-Y) = \sin X \cos Y - \cos X \sin Y \):
\( = \frac{\sin B \cos C - \cos B \sin C}{\cos B \cos C} \)
Divide each part of the numerator by the denominator:
\( = \frac{\sin B \cos C}{\cos B \cos C} - \frac{\cos B \sin C}{\cos B \cos C} \)
\( = \tan B - \tan C \) ......(1)

Similarly, for the second term: \( \frac{\sin (C-A)}{\cos C \cos A} \)
\( = \frac{\sin C \cos A - \cos C \sin A}{\cos C \cos A} \)
\( = \frac{\sin C \cos A}{\cos C \cos A} - \frac{\cos C \sin A}{\cos C \cos A} \)
\( = \tan C - \tan A \) ......(2)

And for the third term: \( \frac{\sin (A-B)}{\cos A \cos B} \)
\( = \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B} \)
\( = \frac{\sin A \cos B}{\cos A \cos B} - \frac{\cos A \sin B}{\cos A \cos B} \)
\( = \tan A - \tan B \) ......(3)

Now, add the results from (1), (2), and (3):
\( (\tan B - \tan C) + (\tan C - \tan A) + (\tan A - \tan B) \)
\( = \tan B - \tan C + \tan C - \tan A + \tan A - \tan B \)
All terms cancel each other out:
\( = 0 \)
Thus, \( \frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}+\frac{\sin (A-B)}{\cos A \cos B}=0 \). The identity is proven.
In simple words: We take each fraction separately and expand the sine term in the numerator using the difference formula. Then, we split each fraction into two parts. After simplifying, each fraction turns into a difference of two tangent terms. When we add all three simplified expressions together, all the tangent terms cancel each other out, resulting in zero.

🎯 Exam Tip: When proving an identity with multiple fractional terms, it's often effective to simplify each term individually first. Look for ways to break down complex expressions into simpler trigonometric ratios like tangent, which can reveal cancellations.

 

Question 12. If tan A – tan B = x and cot B – cot A = y prove that cot(A - B) = \( \frac{1}{x}+\frac{1}{y} \).
Answer:
Given: \( \tan A - \tan B = x \) and \( \cot B - \cot A = y \).
We need to prove that \( \cot(A - B) = \frac{1}{x} + \frac{1}{y} \).

Let's start by expressing \( \frac{1}{x} + \frac{1}{y} \) in terms of tangents and cotangents.
\( \frac{1}{x} = \frac{1}{\tan A - \tan B} \)
\( \frac{1}{y} = \frac{1}{\cot B - \cot A} \)

We know that \( \cot \theta = \frac{1}{\tan \theta} \). So, rewrite \( \frac{1}{y} \) using tangents:
\( \frac{1}{y} = \frac{1}{\frac{1}{\tan B} - \frac{1}{\tan A}} \)
\( = \frac{1}{\frac{\tan A - \tan B}{\tan A \tan B}} \)
\( = \frac{\tan A \tan B}{\tan A - \tan B} \)

Now, add \( \frac{1}{x} \) and \( \frac{1}{y} \):
\( \frac{1}{x} + \frac{1}{y} = \frac{1}{\tan A - \tan B} + \frac{\tan A \tan B}{\tan A - \tan B} \)
Since both terms have the same denominator, we can combine them:
\( = \frac{1 + \tan A \tan B}{\tan A - \tan B} \)

Recall the formula for \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \).
The expression we have is the reciprocal of \( \tan(A - B) \).
So, \( \frac{1 + \tan A \tan B}{\tan A - \tan B} = \frac{1}{\tan(A - B)} = \cot(A - B) \).
Thus, we have proven that \( \cot(A - B) = \frac{1}{x} + \frac{1}{y} \).
In simple words: We start by converting the given expressions for x and y into terms of tangent. Then, we find \( \frac{1}{x} \) and \( \frac{1}{y} \). Adding these two fractions together gives us a new fraction that looks exactly like the reciprocal of the tangent difference formula. Since \( \frac{1}{\tan(\text{angle})} \) is \( \cot(\text{angle}) \), this proves the required identity.

🎯 Exam Tip: When proving identities, especially those involving cotangent, it is often helpful to convert everything into tangent terms first. This simplifies the expressions and allows you to use familiar tangent formulas more easily.

 

Question 13. If sin \( \alpha + \sin \beta = a \) and cos \( \alpha + \cos \beta = b \), then prove that cos(\( \alpha – \beta \)) = \( \frac{a^{2}+b^{2}-2}{2} \)
Answer:
Given: \( \sin \alpha + \sin \beta = a \) and \( \cos \alpha + \cos \beta = b \).
We need to prove that \( \cos(\alpha - \beta) = \frac{a^2 + b^2 - 2}{2} \).

Let's square both given equations:
1. \( (\sin \alpha + \sin \beta)^2 = a^2 \)
\( \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta = a^2 \) ......(Equation 1)

2. \( (\cos \alpha + \cos \beta)^2 = b^2 \)
\( \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta = b^2 \) ......(Equation 2)

Now, add Equation 1 and Equation 2:
\( (\sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta) + (\cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta) = a^2 + b^2 \)
Rearrange the terms:
\( (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + 2 \sin \alpha \sin \beta + 2 \cos \alpha \cos \beta = a^2 + b^2 \)
Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( 1 + 1 + 2 (\sin \alpha \sin \beta + \cos \alpha \cos \beta) = a^2 + b^2 \)
\( 2 + 2 (\cos \alpha \cos \beta + \sin \alpha \sin \beta) = a^2 + b^2 \)
Recall the formula for \( \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \).
Substitute this into the equation:
\( 2 + 2 \cos(\alpha - \beta) = a^2 + b^2 \)
Now, solve for \( \cos(\alpha - \beta) \):
\( 2 \cos(\alpha - \beta) = a^2 + b^2 - 2 \)
\( \cos(\alpha - \beta) = \frac{a^2 + b^2 - 2}{2} \). This proves the identity.
In simple words: We start by squaring both of the given equations separately. Then, we add these two squared equations together. By grouping terms and using the basic trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we simplify the expression. Finally, we recognize the formula for \( \cos(\alpha - \beta) \) within the simplified expression and rearrange it to match the required proof.

🎯 Exam Tip: Squaring and adding equations is a common technique in trigonometry, especially when dealing with sums of sine and cosine terms that relate to sum/difference identities. Always look for opportunities to use \( \sin^2 \theta + \cos^2 \theta = 1 \) to simplify expressions.

 

Question 14. Find the value of tan\( \frac{\pi}{8} \).
Answer:
We need to find the value of \( \tan \frac{\pi}{8} \).
We know that \( \frac{\pi}{8} \) is equivalent to \( \frac{180^{\circ}}{8} = 22.5^{\circ} \).

**Method 1: Using \( \tan 2A \) formula**
Let \( A = \frac{\pi}{8} \). Then \( 2A = \frac{\pi}{4} \).
We know the formula \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \).
Substitute \( 2A = \frac{\pi}{4} \) and \( A = \frac{\pi}{8} \):
\( \tan \frac{\pi}{4} = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}} \)
We know \( \tan \frac{\pi}{4} = 1 \). Let \( t = \tan \frac{\pi}{8} \).
So, \( 1 = \frac{2t}{1 - t^2} \)
Cross-multiply:
\( 1 - t^2 = 2t \)
Rearrange into a quadratic equation:
\( t^2 + 2t - 1 = 0 \)
Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a=1, b=2, c=-1 \).
\( t = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} \)
\( t = \frac{-2 \pm \sqrt{4 + 4}}{2} \)
\( t = \frac{-2 \pm \sqrt{8}}{2} \)
\( t = \frac{-2 \pm 2\sqrt{2}}{2} \)
\( t = -1 \pm \sqrt{2} \)
Since \( \frac{\pi}{8} \) (or \( 22.5^{\circ} \)) is an acute angle, \( \tan \frac{\pi}{8} \) must be positive.
\( \sqrt{2} \approx 1.414 \). So, \( -1 + \sqrt{2} \) is positive, and \( -1 - \sqrt{2} \) is negative.
Therefore, \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \).

**Method 2: Using the half-angle formula for tangent**
We know the formula \( \tan^2 A = \frac{1 - \cos 2A}{1 + \cos 2A} \).
Let \( A = \frac{\pi}{8} \), so \( 2A = \frac{\pi}{4} \).
\( \tan^2 \frac{\pi}{8} = \frac{1 - \cos \frac{\pi}{4}}{1 + \cos \frac{\pi}{4}} \)
We know \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \).
\( \tan^2 \frac{\pi}{8} = \frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} \)
Multiply numerator and denominator by \( \sqrt{2} \):
\( = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \)
Rationalize the denominator by multiplying by \( \frac{\sqrt{2} - 1}{\sqrt{2} - 1} \):
\( = \frac{(\sqrt{2} - 1)^2}{(\sqrt{2})^2 - 1^2} \)
\( = \frac{2 - 2\sqrt{2} + 1}{2 - 1} \)
\( = \frac{3 - 2\sqrt{2}}{1} \)
So, \( \tan^2 \frac{\pi}{8} = 3 - 2\sqrt{2} \).
To find \( \tan \frac{\pi}{8} \), take the square root:
\( \tan \frac{\pi}{8} = \sqrt{3 - 2\sqrt{2}} \)
This can be simplified: \( 3 - 2\sqrt{2} = 2 - 2\sqrt{2} + 1 = (\sqrt{2})^2 - 2\sqrt{2} + 1^2 = (\sqrt{2} - 1)^2 \).
So, \( \tan \frac{\pi}{8} = \sqrt{(\sqrt{2} - 1)^2} = |\sqrt{2} - 1| \).
Since \( \sqrt{2} \approx 1.414 \), \( \sqrt{2} - 1 \) is positive. Therefore, \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \).

**Method 3: Using the half-angle formula \( \tan A = \frac{\sin 2A}{1 + \cos 2A} \)**
Let \( A = \frac{\pi}{8} \). Then \( 2A = \frac{\pi}{4} \).
\( \tan \frac{\pi}{8} = \frac{\sin \frac{\pi}{4}}{1 + \cos \frac{\pi}{4}} \)
Substitute \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \):
\( = \frac{\frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} \)
Multiply numerator and denominator by \( \sqrt{2} \):
\( = \frac{1}{\sqrt{2} + 1} \)
Rationalize the denominator:
\( = \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} \)
\( = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2} \)
\( = \frac{\sqrt{2} - 1}{2 - 1} \)
\( = \sqrt{2} - 1 \)
All three methods yield the same result. The value of \( \tan \frac{\pi}{8} \) is \( \sqrt{2} - 1 \).
In simple words: To find the value of tan \( \frac{\pi}{8} \), we can use different trigonometric formulas that relate an angle to its half or double. We let \( \frac{\pi}{8} \) be 'A', so \( 2A \) becomes \( \frac{\pi}{4} \), whose tangent and cosine values are known. By plugging these into the chosen formula and solving, we get a quadratic equation for tan \( \frac{\pi}{8} \). Since \( \frac{\pi}{8} \) is a small angle, its tangent must be positive, which helps us pick the correct answer.

🎯 Exam Tip: When asked to find trigonometric values for half-angles like \( \frac{\pi}{8} \) or \( 22.5^{\circ} \), use half-angle formulas or relate them to known double angles. Always select the positive root if the angle is in the first quadrant, as tangent values are positive there.

 

Question 15. If tan \( \alpha = \frac{1}{7} \), sin \( \beta = \frac{1}{\sqrt{10}} \). Prove that \( \alpha + 2\beta = \frac{\pi}{4} \) where \( 0 < \alpha < \frac{\pi}{2} \) and \( 0 < \beta < \frac{\pi}{2} \).
Answer:
Given: \( \tan \alpha = \frac{1}{7} \), \( \sin \beta = \frac{1}{\sqrt{10}} \). Both \( \alpha \) and \( \beta \) are acute angles (in the first quadrant).
We need to prove that \( \alpha + 2\beta = \frac{\pi}{4} \). We can do this by finding \( \tan(\alpha + 2\beta) \) and showing it equals \( \tan \frac{\pi}{4} = 1 \).

First, find \( \tan \alpha \), which is given as \( \frac{1}{7} \).

Next, find \( \tan \beta \) from \( \sin \beta \). Since \( \beta \) is in the first quadrant, all trigonometric ratios are positive.
If \( \sin \beta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{\sqrt{10}} \), then using Pythagoras theorem, \( \text{Adjacent} = \sqrt{(\sqrt{10})^2 - 1^2} = \sqrt{10 - 1} = \sqrt{9} = 3 \).
So, \( \tan \beta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1}{3} \).

Next, find \( \tan 2\beta \):
Using the double angle formula \( \tan 2\beta = \frac{2 \tan \beta}{1 - \tan^2 \beta} \):
\( \tan 2\beta = \frac{2\left(\frac{1}{3}\right)}{1 - \left(\frac{1}{3}\right)^2} \)
\( = \frac{\frac{2}{3}}{1 - \frac{1}{9}} \)
\( = \frac{\frac{2}{3}}{\frac{9 - 1}{9}} \)
\( = \frac{\frac{2}{3}}{\frac{8}{9}} \)
\( = \frac{2}{3} \times \frac{9}{8} \)
\( = \frac{18}{24} = \frac{3}{4} \)
So, \( \tan 2\beta = \frac{3}{4} \).

Now, find \( \tan(\alpha + 2\beta) \):
Using the sum formula \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \), with \( A = \alpha \) and \( B = 2\beta \):
\( \tan(\alpha + 2\beta) = \frac{\tan \alpha + \tan 2\beta}{1 - \tan \alpha \tan 2\beta} \)
Substitute \( \tan \alpha = \frac{1}{7} \) and \( \tan 2\beta = \frac{3}{4} \):
\( = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \left(\frac{1}{7}\right) \left(\frac{3}{4}\right)} \)
For the numerator: \( \frac{1}{7} + \frac{3}{4} = \frac{4 + 21}{28} = \frac{25}{28} \)
For the denominator: \( 1 - \frac{3}{28} = \frac{28 - 3}{28} = \frac{25}{28} \)
So, \( \tan(\alpha + 2\beta) = \frac{\frac{25}{28}}{\frac{25}{28}} = 1 \).

We know that \( \tan \frac{\pi}{4} = 1 \).
Since \( \alpha \) and \( \beta \) are acute angles, \( 0 < \alpha < \frac{\pi}{2} \) and \( 0 < 2\beta < \pi \).
Therefore, \( 0 < \alpha + 2\beta < \frac{\pi}{2} + \pi = \frac{3\pi}{2} \).
Since \( \tan(\alpha + 2\beta) = 1 \) and \( \alpha + 2\beta \) is in the range \( (0, \frac{3\pi}{2}) \), the principal value must be \( \frac{\pi}{4} \).
Hence, \( \alpha + 2\beta = \frac{\pi}{4} \). The proof is complete.
In simple words: We are given \( \tan \alpha \) and \( \sin \beta \), and both angles are acute. First, we find \( \tan \beta \) from \( \sin \beta \) using a right triangle. Then, we calculate \( \tan 2\beta \) using the double angle formula. Finally, we use the sum formula for tangent, \( \tan(\alpha + 2\beta) \), and plug in the values. The result is 1, which is the tangent of \( \frac{\pi}{4} \), thus proving the identity.

🎯 Exam Tip: When proving identities like this, calculate the tangent of the combined angle \( (\alpha + 2\beta) \). If the result is a known tangent value (like 1 for \( \frac{\pi}{4} \)), the proof is complete. Always check the quadrant for the sum of angles to confirm the principal value.

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