GSEB Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.2

Question 1.
(i) \( \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ \)
(ii) \( 2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ \)
(iii) \( \frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ} \)
(iv) \( \frac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ} \)
(v) \( \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} \)
Answer:
(i) We know the standard values for trigonometric ratios:
\( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
\( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \sin 30^\circ = \frac{1}{2} \)
\( \cos 60^\circ = \frac{1}{2} \)
Substitute these values into the expression:
\( \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ \)
\( = \left( \frac{\sqrt{3}}{2} \right) \times \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{2} \right) \times \left( \frac{1}{2} \right) \)
\( = \frac{3}{4} + \frac{1}{4} \)
\( = \frac{3+1}{4} \)
\( = \frac{4}{4} \)
\( = 1 \)
(ii) We use the standard trigonometric values:
\( \tan 45^\circ = 1 \)
\( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
Substitute these values into the expression:
\( 2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ \)
\( = 2(1)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 - \left( \frac{\sqrt{3}}{2} \right)^2 \)
\( = 2(1) + \frac{3}{4} - \frac{3}{4} \)
\( = 2 + 0 \)
\( = 2 \)
(iii) We use the standard trigonometric values:
\( \cos 45^\circ = \frac{1}{\sqrt{2}} \)
\( \sec 30^\circ = \frac{2}{\sqrt{3}} \)
\( \csc 30^\circ = 2 \)
Substitute these values into the expression:
\( \frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ} \)
\( = \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2} \)
\( = \frac{\frac{1}{\sqrt{2}}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}} \)
\( = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2 + 2\sqrt{3}} \)
\( = \frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} \)
To rationalize the denominator, multiply by the conjugate \( ( \sqrt{3}-1 ) \):
\( = \frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \)
\( = \frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{2}( ( \sqrt{3})^2 - 1^2 )} \)
\( = \frac{3 - \sqrt{3}}{2\sqrt{2}(3 - 1)} \)
\( = \frac{3 - \sqrt{3}}{2\sqrt{2}(2)} \)
\( = \frac{3 - \sqrt{3}}{4\sqrt{2}} \)
To further rationalize, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \):
\( = \frac{(3 - \sqrt{3})\sqrt{2}}{4\sqrt{2} \times \sqrt{2}} \)
\( = \frac{3\sqrt{2} - \sqrt{6}}{4 \times 2} \)
\( = \frac{3\sqrt{2} - \sqrt{6}}{8} \)
(iv) We use the standard trigonometric values:
\( \sin 30^\circ = \frac{1}{2} \)
\( \tan 45^\circ = 1 \)
\( \csc 60^\circ = \frac{2}{\sqrt{3}} \)
\( \sec 30^\circ = \frac{2}{\sqrt{3}} \)
\( \cos 60^\circ = \frac{1}{2} \)
\( \cot 45^\circ = 1 \)
Substitute these values into the expression:
\( \frac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ} \)
\( = \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1} \)
\( = \frac{\frac{1+2}{2} - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1+2}{2}} \)
\( = \frac{\frac{3}{2} - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{3}{2}} \)
Find a common denominator for the numerator and denominator separately:
Numerator: \( \frac{3\sqrt{3} - 4}{2\sqrt{3}} \)
Denominator: \( \frac{4 + 3\sqrt{3}}{2\sqrt{3}} \)
Substitute back into the main expression:
\( = \frac{\frac{3\sqrt{3} - 4}{2\sqrt{3}}}{\frac{4 + 3\sqrt{3}}{2\sqrt{3}}} \)
\( = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \)
To rationalize the denominator, multiply by the conjugate \( ( 3\sqrt{3} - 4 ) \):
\( = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \times \frac{3\sqrt{3} - 4}{3\sqrt{3} - 4} \)
\( = \frac{(3\sqrt{3} - 4)^2}{(3\sqrt{3})^2 - 4^2} \)
\( = \frac{(3\sqrt{3})^2 - 2(3\sqrt{3})(4) + 4^2}{9 \times 3 - 16} \)
\( = \frac{27 - 24\sqrt{3} + 16}{27 - 16} \)
\( = \frac{43 - 24\sqrt{3}}{11} \)
(v) We use the standard trigonometric values:
\( \cos 60^\circ = \frac{1}{2} \)
\( \sec 30^\circ = \frac{2}{\sqrt{3}} \)
\( \tan 45^\circ = 1 \)
\( \sin 30^\circ = \frac{1}{2} \)
Substitute these values into the expression:
\( \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} \)
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \), so the denominator \( \sin^2 30^\circ + \cos^2 30^\circ = 1 \).
So the expression simplifies to just the numerator:
\( 5 \left( \frac{1}{2} \right)^2 + 4 \left( \frac{2}{\sqrt{3}} \right)^2 - (1)^2 \)
\( = 5 \left( \frac{1}{4} \right) + 4 \left( \frac{4}{3} \right) - 1 \)
\( = \frac{5}{4} + \frac{16}{3} - 1 \)
To combine these, find a common denominator, which is 12:
\( = \frac{5 \times 3}{4 \times 3} + \frac{16 \times 4}{3 \times 4} - \frac{1 \times 12}{1 \times 12} \)
\( = \frac{15}{12} + \frac{64}{12} - \frac{12}{12} \)
\( = \frac{15 + 64 - 12}{12} \)
\( = \frac{79 - 12}{12} \)
\( = \frac{67}{12} \)
In simple words: For each part, we replace the angles with their known sine, cosine, tangent, secant, and cosecant values. Then, we do the math step by step, like squaring and adding, to find the final number. Some steps require rationalizing the denominator to simplify the fraction.

Exam Tip: Memorizing the trigonometric ratios for common angles like 0°, 30°, 45°, 60°, and 90° is essential for quickly solving these types of problems.

Question 2. Choose the correct option and justify your choice:
(i) \( \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}} \)
(a) sin 60°
(b) cos 60°
(c) tan 60°
(d) sin 30°
(ii) \( \frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}} \)
(a) tan 90°
(b) 1
(c) sin 45°
(d) 0
(iii) sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°
(c) 45°
(d) 60°
(iv) \( \frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}} \)
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°
Answer:
(i) We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
Substitute this value into the expression:
\( \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}} = \frac{2 \left( \frac{1}{\sqrt{3}} \right)}{1 + \left( \frac{1}{\sqrt{3}} \right)^2} \)
\( = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} \)
\( = \frac{2}{\sqrt{3}} \times \frac{3}{4} \)
\( = \frac{6}{4\sqrt{3}} \)
\( = \frac{3}{2\sqrt{3}} \)
\( = \frac{3\sqrt{3}}{2\sqrt{3}\sqrt{3}} \)
\( = \frac{3\sqrt{3}}{2 \times 3} \)
\( = \frac{\sqrt{3}}{2} \)
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
Thus, the correct option is (a) sin 60°.
(ii) We know that \( \tan 45^\circ = 1 \).
Substitute this value into the expression:
\( \frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}} = \frac{1 - (1)^2}{1 + (1)^2} \)
\( = \frac{1 - 1}{1 + 1} \)
\( = \frac{0}{2} \)
\( = 0 \)
Thus, the correct option is (d) 0.
(iii) We need to find the value of A for which \( \sin 2A = 2 \sin A \) is true.
Let's check each option:
(a) If \( A = 0^\circ \):
LHS \( = \sin (2 \times 0^\circ) = \sin 0^\circ = 0 \)
RHS \( = 2 \sin 0^\circ = 2 \times 0 = 0 \)
Since LHS = RHS, \( \sin 2A = 2 \sin A \) is true for \( A = 0^\circ \).
(b) If \( A = 30^\circ \):
LHS \( = \sin (2 \times 30^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \)
RHS \( = 2 \sin 30^\circ = 2 \times \frac{1}{2} = 1 \)
LHS \( \neq \) RHS.
(c) If \( A = 45^\circ \):
LHS \( = \sin (2 \times 45^\circ) = \sin 90^\circ = 1 \)
RHS \( = 2 \sin 45^\circ = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \)
LHS \( \neq \) RHS.
(d) If \( A = 60^\circ \):
LHS \( = \sin (2 \times 60^\circ) = \sin 120^\circ = \sin (180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \)
RHS \( = 2 \sin 60^\circ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \)
LHS \( \neq \) RHS.
Thus, the correct option is (a) 0°.
(iv) We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
Substitute this value into the expression:
\( \frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}} = \frac{2 \left( \frac{1}{\sqrt{3}} \right)}{1 - \left( \frac{1}{\sqrt{3}} \right)^2} \)
\( = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \)
\( = \frac{2}{\sqrt{3}} \times \frac{3}{2} \)
\( = \frac{3}{\sqrt{3}} \)
\( = \frac{3\sqrt{3}}{\sqrt{3}\sqrt{3}} \)
\( = \frac{3\sqrt{3}}{3} \)
\( = \sqrt{3} \)
We know that \( \tan 60^\circ = \sqrt{3} \).
Thus, the correct option is (c) tan 60°.
In simple words: We check each question by putting in the known values for the angles and doing the math. For the first two, we calculate the exact number and see which answer option matches. For the third, we test each angle option to see when both sides of the equation are equal. For the last one, we calculate the value and find the matching trigonometric ratio.

Exam Tip: Remember the double angle identity \( \sin 2A = 2 \sin A \cos A \) and \( \tan 2A = \frac{2 \tan A}{1-\tan^2 A} \) and \( \sin 2A = \frac{2 \tan A}{1+\tan^2 A} \) which are useful shortcuts for these types of MCQs.

Question 3. If \( \tan (A + B) = \sqrt{3} \) and \( \tan (A - B) = \frac{1}{\sqrt{3}} \), \( 0^\circ < A + B \leq 90^\circ \), \( A > B \), find A and B.
Answer: We are given two equations involving tangent functions:
1. \( \tan (A + B) = \sqrt{3} \)
2. \( \tan (A - B) = \frac{1}{\sqrt{3}} \)

From equation 1:
We know that \( \tan 60^\circ = \sqrt{3} \).
So, \( \tan (A + B) = \tan 60^\circ \)
\( \implies A + B = 60^\circ \) (Equation I)

From equation 2:
We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
So, \( \tan (A - B) = \tan 30^\circ \)
\( \implies A - B = 30^\circ \) (Equation II)

Now, we have a system of two linear equations:
I: \( A + B = 60^\circ \)
II: \( A - B = 30^\circ \)

Add Equation I and Equation II:
\( (A + B) + (A - B) = 60^\circ + 30^\circ \)
\( 2A = 90^\circ \)
\( A = \frac{90^\circ}{2} \)
\( \implies A = 45^\circ \)

Substitute the value of A into Equation I:
\( 45^\circ + B = 60^\circ \)
\( B = 60^\circ - 45^\circ \)
\( \implies B = 15^\circ \)

We check if the conditions are met:
\( 0^\circ < A + B \leq 90^\circ \implies 0^\circ < 45^\circ + 15^\circ \leq 90^\circ \implies 0^\circ < 60^\circ \leq 90^\circ \) (True)
\( A > B \implies 45^\circ > 15^\circ \) (True)

Therefore, \( A = 45^\circ \) and \( B = 15^\circ \).
In simple words: We get two easy equations by using what we know about tangent values for special angles. We then solve these two equations together to find the values for A and B.

Exam Tip: Always remember to check your calculated values of A and B against the given conditions (like \( A+B \leq 90^\circ \) and \( A>B \)) to make sure they are valid.

Question 4. State whether the following are true or false. Justify your answer.
(i) \( \sin (A + B) = \sin A + \sin B \).
(ii) The value of \( \sin \theta \) increases as \( \theta \) increases.
(iii) The value of \( \cos \theta \) increases as \( \theta \) increases.
(iv) \( \sin \theta = \cos \theta \) for all values of \( \theta \).
(v) \( \cot \theta \) is not defined for \( \theta = 0^\circ \).
Answer:
(i) False.
To justify, let's take specific values for A and B. Let \( A = 30^\circ \) and \( B = 60^\circ \).
LHS: \( \sin (A + B) = \sin (30^\circ + 60^\circ) = \sin 90^\circ = 1 \).
RHS: \( \sin A + \sin B = \sin 30^\circ + \sin 60^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \).
Since \( 1 \neq \frac{1 + \sqrt{3}}{2} \), LHS \( \neq \) RHS. Therefore, the statement is false. This shows that \( \sin (A + B) \) is not equal to \( \sin A + \sin B \).

(ii) True.
Let's examine the values of \( \sin \theta \) for \( \theta \) ranging from \( 0^\circ \) to \( 90^\circ \):
\( \sin 0^\circ = 0 \)
\( \sin 30^\circ = 0.5 \)
\( \sin 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 \)
\( \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \)
\( \sin 90^\circ = 1 \)
As \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \), the value of \( \sin \theta \) increases from 0 to 1.

(iii) False.
Let's examine the values of \( \cos \theta \) for \( \theta \) ranging from \( 0^\circ \) to \( 90^\circ \):
\( \cos 0^\circ = 1 \)
\( \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \)
\( \cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 \)
\( \cos 60^\circ = 0.5 \)
\( \cos 90^\circ = 0 \)
As \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \), the value of \( \cos \theta \) decreases from 1 to 0. Therefore, the statement is false.

(iv) False.
The statement \( \sin \theta = \cos \theta \) is only true for a specific value of \( \theta \), which is \( 45^\circ \).
For example, if we take \( \theta = 30^\circ \):
\( \sin 30^\circ = \frac{1}{2} \)
\( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
Since \( \frac{1}{2} \neq \frac{\sqrt{3}}{2} \), \( \sin 30^\circ \neq \cos 30^\circ \).
Therefore, \( \sin \theta = \cos \theta \) is not true for all values of \( \theta \).

(v) True.
The cotangent function is defined as \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).
For \( \theta = 0^\circ \):
\( \cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ} = \frac{1}{0} \).
Division by zero is undefined in mathematics. Therefore, \( \cot 0^\circ \) is not defined, making the statement true.
In simple words: We check each statement. For false ones, we give an example where it doesn't work. For true ones, we show why it's correct using known angle values and rules. We remember that sin generally goes up, cos generally goes down, and some values like cot at 0 degrees are not defined.

Exam Tip: When proving a "false" statement, a single counter-example is enough. For "true" statements, general reasoning or checking the trend of values for common angles (0°, 30°, 45°, 60°, 90°) is usually sufficient.