CBSE Class 10 Mathematics Trigonometry Assignment Set 23

Values of Trigonometric Ratios (Exercise II)

Question 1. Evaluate the following: (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) \( \frac{\cos 45^\circ}{\sec 30^\circ + \operatorname{cosec} 30^\circ} \) (iv) \( \frac{\sin 30^\circ + \tan 45^\circ - \operatorname{cosec} 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ} \) (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) \( \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1 \).
(ii) \( 2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ = 2(1)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 - \left( \frac{\sqrt{3}}{2} \right)^2 = 2 + 0 = 2 \).
(iii) \( \cos 45^\circ = \frac{1}{\sqrt{2}},\ \sec 30^\circ = \frac{2}{\sqrt{3}},\ \operatorname{cosec} 30^\circ = 2 \). So \( \frac{\cos 45^\circ}{\sec 30^\circ + \operatorname{cosec} 30^\circ} = \frac{\frac{1}{\sqrt{2}}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}} = \frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} \). Multiplying numerator and denominator by \( (\sqrt{3} - 1) \): \( \frac{\sqrt{3}(\sqrt{3} - 1)}{2\sqrt{2}(3 - 1)} = \frac{3 - \sqrt{3}}{4\sqrt{2}} \). Multiplying by \( \sqrt{2} \): \( \frac{3\sqrt{2} - \sqrt{6}}{8} \).
(iv) Numerator \( = \frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}} = \frac{3\sqrt{3} - 4}{2\sqrt{3}} \). Denominator \( = \frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{3}{2} = \frac{4 + 3\sqrt{3}}{2\sqrt{3}} \).
So the value \( = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \times \frac{3\sqrt{3} - 4}{3\sqrt{3} - 4} = \frac{27 - 24\sqrt{3} + 16}{27 - 16} = \frac{43 - 24\sqrt{3}}{11} \).
(v) \( \frac{5\left( \frac{1}{2} \right)^2 + 4\left( \frac{2}{\sqrt{3}} \right)^2 - 1^2}{\left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = \frac{\frac{5}{4} + \frac{16}{3} - 1}{\frac{1}{4} + \frac{3}{4}} = \frac{\frac{15 + 64 - 12}{12}}{1} = \frac{67}{12} \).
In simple words: Replace every ratio by its table value and simplify step by step; parts (i) and (ii) simplify to whole numbers.

Teacher's Note:
1. Write each standard value from the table before substituting.
2. Rationalise the denominator only at the last step.

Question 2. Choose the correct option and justify your choice. (i) \( \frac{2\tan 30^\circ}{1 + \tan^2 30^\circ} = \) (a) \( \sin 60^\circ \) (b) \( \cos 60^\circ \) (c) \( \tan 60^\circ \) (d) \( \sin 30^\circ \) (ii) \( \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \) (a) \( \tan 90^\circ \) (b) 1 (c) \( \sin 45^\circ \) (d) 0 (iii) \( \sin 2A = 2\sin A \) is true when \( A = \) (a) \( 0^\circ \) (b) \( 30^\circ \) (c) \( 45^\circ \) (d) \( 60^\circ \) (iv) \( \frac{2\tan 30^\circ}{1 - \tan^2 30^\circ} = \) (a) \( \cos 60^\circ \) (b) \( \sin 60^\circ \) (c) \( \tan 60^\circ \) (d) \( \sin 30^\circ \)
Answer:
(i) Option (a). \( \frac{2 \times \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2} = \sin 60^\circ \).
(ii) Option (d). \( \frac{1 - 1^2}{1 + 1^2} = \frac{0}{2} = 0 \).
(iii) Option (a). Using \( \sin 2A = 2\sin A \cos A \), the relation \( 2\sin A \cos A = 2\sin A \) gives \( \sin A(\cos A - 1) = 0 \), so \( \sin A = 0 \) or \( \cos A = 1 \). Both give \( A = 0^\circ \). Check: \( \sin 0^\circ = 0 = 2\sin 0^\circ \); at \( 30^\circ \) we get \( \sin 60^\circ = \frac{\sqrt{3}}{2} \neq 1 = 2\sin 30^\circ \), and at \( 45^\circ \) we get \( 1 \neq \sqrt{2} \).
(iv) Option (c). \( \frac{2 \times \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \sqrt{3} = \tan 60^\circ \).
In simple words: Put the table values in the left side and match the number with the option.

Teacher's Note:
1. Write each standard value from the table before substituting.
2. Rationalise the denominator only at the last step.

Question 3. If \( \tan(A + B) = \sqrt{3} \) and \( \tan(A - B) = \frac{1}{\sqrt{3}} \), \( 0^\circ < A + B \le 90^\circ \), \( A > B \), find A and B.
Answer:
\( \tan(A + B) = \sqrt{3} = \tan 60^\circ \Rightarrow A + B = 60^\circ \) ....(i)
\( \tan(A - B) = \frac{1}{\sqrt{3}} = \tan 30^\circ \Rightarrow A - B = 30^\circ \) ....(ii)
Adding (i) and (ii): \( 2A = 90^\circ \Rightarrow A = 45^\circ \). Putting in (i): \( 45^\circ + B = 60^\circ \Rightarrow B = 15^\circ \).
Therefore \( A = 45^\circ \) and \( B = 15^\circ \).
In simple words: Match each tangent value with its standard angle to get two simple linear equations.

Teacher's Note:
1. Write the standard angle for each given tangent value.
2. Add the two equations to find A, then substitute to find B.

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) \( \cot A \) is not defined for \( A = 0^\circ \).
Answer:
(i) False. Take \( A = 30^\circ,\ B = 60^\circ \). LHS \( = \sin 90^\circ = 1 \), RHS \( = \sin 30^\circ + \sin 60^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \neq 1 \).
(ii) True. \( \sin 0^\circ = 0,\ \sin 30^\circ = \frac{1}{2},\ \sin 45^\circ = \frac{1}{\sqrt{2}},\ \sin 60^\circ = \frac{\sqrt{3}}{2},\ \sin 90^\circ = 1 \); the values increase from 0 to 1 as \( \theta \) goes from \( 0^\circ \) to \( 90^\circ \).
(iii) False. \( \cos 0^\circ = 1,\ \cos 30^\circ = \frac{\sqrt{3}}{2},\ \cos 45^\circ = \frac{1}{\sqrt{2}},\ \cos 60^\circ = \frac{1}{2},\ \cos 90^\circ = 0 \); the values decrease as \( \theta \) increases.
(iv) False. \( \sin \theta = \cos \theta \) only at \( \theta = 45^\circ \); for example \( \sin 30^\circ = \frac{1}{2} \) but \( \cos 30^\circ = \frac{\sqrt{3}}{2} \).
(v) True. \( \cot A = \frac{\cos A}{\sin A} \). At \( A = 0^\circ \), \( \cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ} = \frac{1}{0} \), which is undefined.
In simple words: Use the table of standard values to test each statement; one counter-example proves a statement false.

Teacher's Note:
1. Choose a convenient pair of angles from the table for a counter-example.
2. Remember that division by zero is undefined.