CBSE Class 10 Mathematics Introduction to Trigonometry VBQs Set 04

Question. From the figure, write the values of \( \sin A, \cos A, \tan A, \csc A, \sec A, \cot A \).
Answer: \( \sin A = \frac{5}{13}, \cos A = \frac{12}{13}, \tan A = \frac{5}{12}, \csc A = \frac{13}{5}, \sec A = \frac{13}{12}, \cot A = \frac{12}{5} \)

Question. In \( \Delta ABC \), \( \angle B = 90^\circ \), If \( AB = 4 \) cm, \( BC = 3 \) cm and \( AC = 5 \) cm, write the values of \( \sin A, \cos A \) and \( \cot A \).
Answer: \( \sin A = \frac{3}{5}, \cos A = \frac{4}{5}, \cot A = \frac{4}{3} \)

Question. If \( \sin \theta = \frac{\sqrt{3}}{2} \), find the values of other t-ratios.
Answer: \( \cos \theta = \frac{1}{2}, \tan \theta = \sqrt{3}, \csc \theta = \frac{2}{\sqrt{3}}, \sec \theta = 2, \cot \theta = \frac{1}{\sqrt{3}} \)

Question. If \( \tan A = \frac{7}{24} \), find the value of \( \sin A + \cos A \).
Answer: \( \frac{31}{25} \)

Question. If \( \csc A = \frac{7}{4} \), prove that \( 1 + \tan^2 A = \sec^2 A \).
Answer: 1

Question. If \( \sec A = \frac{5}{4} \), prove that \( \cos^2 A = 1 - \sin^2 A \).
Answer: \( \frac{7}{18} \)

Question. If \( \tan A = \frac{2}{3} \), prove that \( \sin^2 A + \cos^2 A = 1 \).
Answer: \( \frac{5}{4} \)

Question. If \( \tan A = \frac{1}{3} \), prove that \( \csc^2 A = 1 + \cot^2 A \).
Answer: 3

Question. If \( \cot B = \frac{12}{5} \), show that \( \tan^2 B - \sin^2 B = \sin^2 B \tan^2 B \).
Answer: \( \sin \theta = \frac{2mn}{m^2 + n^2}, \cos \theta = \frac{m^2 - n^2}{m^2 + n^2}, \csc \theta = \frac{m^2 + n^2}{2mn}, \sec \theta = \frac{m^2 + n^2}{m^2 - n^2}, \cot \theta = \frac{m^2 - n^2}{2mn} \)

Question. If \( \cos A = \frac{1}{2} \), find the value of \( \frac{2 \sec A}{1 + \tan^2 A} \).
Answer: 1

Question. If \( 3 \tan \theta = 4 \), find the value of \( \frac{4 \sin \theta - 3 \cos \theta}{3 \sin \theta + 2 \cos \theta} \).
Answer: \( \frac{7}{18} \)

Question. If \( \cot A = 2 \), find the value of \( \frac{4 \cos A + 2 \sin A}{5 \cos A - 2 \sin A} \).
Answer: \( \frac{5}{4} \)

Question. If \( \csc \theta = \frac{13}{12} \), find the value of \( \frac{2 \sin \theta - 3 \cos \theta}{4 \sin \theta - 9 \cos \theta} \).
Answer: 3

Question. If \( \sec A = \frac{5}{4} \), prove that: \( \frac{3 \sin A - 4 \sin^3 A}{4 \cos^3 A - 3 \cos A} = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} \).
Answer: Proof established by substituting trigonometric ratios based on \( \sec A = \frac{5}{4} \).

Question. If \( \sin A = \frac{1}{3} \), show that: \( \cos A \csc A + \tan A \sec A = \frac{16\sqrt{2} + 3}{8} \).
Answer: LHS = \( \cos A \csc A + \tan A \sec A \). Substituting \( \sin A = \frac{1}{3} \), we get \( \cos A = \frac{2\sqrt{2}}{3} \). \( \frac{2\sqrt{2}}{3} \cdot 3 + \frac{1}{2\sqrt{2}} \cdot \frac{3}{2\sqrt{2}} = 2\sqrt{2} + \frac{3}{8} = \frac{16\sqrt{2} + 3}{8} = \) RHS.

Question. If \( \tan A = \sqrt{2} - 1 \), show that \( \sin A \cos A = \frac{\sqrt{2}}{4} \).
Answer: Proof established by using the identity \( \sin A \cos A = \frac{\tan A}{\sec^2 A} = \frac{\tan A}{1 + \tan^2 A} \).

Question. If \( \tan \theta = \frac{2mn}{m^2 - n^2} \), find the values of other t-ratios of \( \theta \).
Answer: \( \sin \theta = \frac{2mn}{m^2 + n^2}, \cos \theta = \frac{m^2 - n^2}{m^2 + n^2}, \csc \theta = \frac{m^2 + n^2}{2mn}, \sec \theta = \frac{m^2 + n^2}{m^2 - n^2}, \cot \theta = \frac{m^2 - n^2}{2mn} \)

Question. If \( \tan \theta = \frac{p}{q} \), show that \( \frac{p \sin \theta - q \cos \theta}{p \sin \theta + q \cos \theta} = \left( \frac{p^2 - q^2}{p^2 + q^2} \right) \).
Answer: Dividing numerator and denominator by \( \cos \theta \), we get \( \frac{p \tan \theta - q}{p \tan \theta + q} = \frac{p(p/q) - q}{p(p/q) + q} = \frac{p^2 - q^2}{p^2 + q^2} \).

Question. If \( \tan \theta = \frac{m}{n} \), show that \( \frac{m \sin \theta - n \cos \theta}{m \sin \theta + n \cos \theta} = \frac{m^2 - n^2}{m^2 + n^2} \).
Answer: Dividing numerator and denominator by \( \cos \theta \), we get \( \frac{m \tan \theta - n}{m \tan \theta + n} = \frac{m(m/n) - n}{m(m/n) + n} = \frac{m^2 - n^2}{m^2 + n^2} \).

Question. In \( \Delta ABC \), right angled \( \angle A \) at C and
(i) Is \( \cos A = \cos B \) ?
(ii) Is \( \tan A = \tan B \) ?
What about the other trigonometric ratios for \( \angle A \) and \( \angle B \) ? Will they be equal ?
Answer: (i) Yes, (ii) Yes

Question. Evaluate: (i) \( \cos 30^\circ \cos 45^\circ - \sin 30^\circ \sin 45^\circ \) (ii) \( \tan 30^\circ \csc 60^\circ + \tan 60^\circ \sec 30^\circ \)
Answer: (i) \( \cos 30^\circ \cos 45^\circ - \sin 30^\circ \sin 45^\circ \)
\( = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} - \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} \times \sqrt{2}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} \)
(ii) \( \tan 30^\circ \csc 60^\circ + \tan 60^\circ \sec 30^\circ \)
\( = \frac{1}{\sqrt{3}} \times \frac{2}{\sqrt{3}} + \sqrt{3} \times \frac{2}{\sqrt{3}} = \frac{2}{3} + 2 = \frac{8}{3} = 2\frac{2}{3} \)

Question. Evaluate: \( \tan 60^\circ \csc^2 45^\circ + \sec^2 60^\circ \tan 45^\circ \)
Answer: \( \tan 60^\circ \csc^2 45^\circ + \sec^2 60^\circ \tan 45^\circ \) ... (1)
On substituting the values of various t-ratios in (1), we get
Given expression \( = \sqrt{3} \times (\sqrt{2})^2 + (2)^2 \times 1 = 2\sqrt{3} + 4 = 4 + 2\sqrt{3} \)

Question. Evaluate:
(i) \( \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ \)
(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) \( \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ \)
On substituting the values of various t-ratios in (1), we get
Given expression \( = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1 \)
(ii) \( 2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ = 2 \times (1)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 - \left( \frac{\sqrt{3}}{2} \right)^2 = 2 + \frac{3}{4} - \frac{3}{4} = 2 \)
(iii) \( \frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ} = \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2} = \frac{\frac{1}{\sqrt{2}}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2 + 2\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{2} \times 2(\sqrt{3} + 1)} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \)
\( = \frac{\sqrt{3}(\sqrt{3}-1)}{\sqrt{2} \times 2 \times (3-1)} = \frac{3-\sqrt{3}}{4\sqrt{2}} = \frac{3\sqrt{2}-\sqrt{6}}{8} \)
(iv) \( \frac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ} = \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1} = \frac{\frac{\sqrt{3} + 2\sqrt{3} - 4}{2\sqrt{3}}}{\frac{4 + \sqrt{3} + 2\sqrt{3}}{2\sqrt{3}}} = \frac{3\sqrt{3}-4}{3\sqrt{3}+4} \)
\( = \frac{3\sqrt{3}-4}{3\sqrt{3}+4} \times \frac{3\sqrt{3}-4}{3\sqrt{3}-4} = \frac{27 + 16 - 24\sqrt{3}}{27 - 16} = \frac{43 - 24\sqrt{3}}{11} \)
(v) \( \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} = \frac{5(\frac{1}{2})^2 + 4(\frac{2}{\sqrt{3}})^2 - (1)^2}{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} \)
\( = \frac{\frac{5}{4} + \frac{16}{3} - 1}{\frac{1}{4} + \frac{3}{4}} = \frac{\frac{15 + 64 - 12}{12}}{\frac{4}{4}} = \frac{\frac{67}{12}}{1} = \frac{67}{12} \)

Question. 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) \( \sin 30^\circ \)

Answer: (a) \( \sin 60^\circ \)
Justification: \( \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \frac{2 \times \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2} = \sin 60^\circ \)

Question. (ii) \( \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \)
(a) \( \tan 90^\circ \)
(b) 1
(c) \( \sin 45^\circ \)
(d) 0

Answer: (d) 0
Justification: \( \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \)

Question. (iii) \( \sin 2A = 2 \sin A \) is true when \( A = \)
(a) \( 0^\circ \)
(b) \( 30^\circ \)
(c) \( 45^\circ \)
(d) \( 60^\circ \)

Answer: (a) \( 0^\circ \)
Justification: When \( A = 0^\circ \), \( \sin 2A = \sin 0^\circ = 0 \) and \( 2 \sin A = 2 \sin 0^\circ = 2 \times 0 = 0 \). or \( \sin 2A = 2 \sin A \), when \( A = 0^\circ \).

Question. (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: (c) \( \tan 60^\circ \)
Justification: \( \frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = \frac{2 \times \frac{1}{\sqrt{3}}}{1 - (\frac{1}{\sqrt{3}})^2} = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} = \tan 60^\circ \)

Question. If \( \tan(A + B) = \sqrt{3} \) and \( \tan(A - B) = \frac{1}{\sqrt{3}}; 0^\circ < A + B \le 90^\circ, A \ge B \), find \( A \) and \( B \).
Answer: We have, \( \tan(A + B) = \sqrt{3} \)

\( \implies \) \( \tan(A + B) = \tan 60^\circ \)

\( \implies \) \( A + B = 60^\circ \) ... (1)
Also, \( \tan(A - B) = \frac{1}{\sqrt{3}} \)

\( \implies \) \( \tan(A - B) = \tan 30^\circ \)

\( \implies \) \( A - B = 30^\circ \) ... (2)
Solving (1) and (2), we get \( A = 45^\circ \) and \( B = 15^\circ \).

Question. 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 \( \cos \theta \) increase 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. Because When \( A = 60^\circ \) and \( B = 30^\circ \). Then, \( \sin(A + B) = \sin(60^\circ + 30^\circ) = \sin 90^\circ = 1 \) and, \( \sin A + \sin B = \sin 60^\circ + \sin 30^\circ = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3}+1}{2} \). So, \( \sin(A + B) \ne \sin A + \sin B \).
(ii) True. Because, it is clear from the table that the value of \( \sin \theta \) increases as \( \theta \) increases from \( 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ \).
(iii) False. Because it is clear from the table that the value of \( \cos \theta \) decreases as \( \theta \) increases.
(iv) False. Because it is only true for \( \theta = 45^\circ \). \( \sin 45^\circ = \frac{1}{\sqrt{2}} = \cos 45^\circ \).
(v) True. Because \( \tan 0^\circ = 0 \) and \( \cot 0^\circ = \frac{1}{\tan 0^\circ} = \frac{1}{0} \), i.e., not defined.

Question. Show that \( \csc^2 60^\circ \sec^2 30^\circ \cos 0^\circ \sin 45^\circ \cot^2 60^\circ \tan^2 60^\circ = \frac{8\sqrt{2}}{9} \)
Answer: L.H.S. \( = \csc^2 60^\circ \sec^2 30^\circ \cos 0^\circ \sin 45^\circ \cot^2 60^\circ \tan^2 60^\circ \)
\( = \left( \frac{2}{\sqrt{3}} \right)^2 \left( \frac{2}{\sqrt{3}} \right)^2 (1) \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{3}} \right)^2 (\sqrt{3})^2 = \frac{4}{3} \times \frac{4}{3} \times 1 \times \frac{1}{\sqrt{2}} \times \frac{1}{3} \times \frac{3}{1} = \frac{16}{9} \times \frac{1}{\sqrt{2}} = \frac{16}{9\sqrt{2}} = \frac{8\sqrt{2}}{9} = \) R.H.S.

Question. Show that \( \frac{1 - \sin 60^\circ}{\cos 60^\circ} = 2 - \sqrt{3} \)
Answer: L.H.S. \( = \frac{1 - \sin 60^\circ}{\cos 60^\circ} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\frac{2 - \sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3} = \) R.H.S.

Question. If \( \cos x = \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ \), find \( x \).
Answer: \( \cos x = \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ = \frac{1}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \)

\( \implies \) \( \cos x = \frac{\sqrt{3}}{2} = \cos 30^\circ \)

\( \implies \) \( x = 30^\circ \)

Question. Determine the value of \( x \), such that \( 2 \csc^2 30^\circ + x \sin^2 60^\circ - \frac{3}{4} \tan^2 30^\circ = 10 \)
Answer: \( 2 \csc^2 30^\circ + x \sin^2 60^\circ - \frac{3}{4} \tan^2 30^\circ = 10 \)

\( \implies \) \( 2(2)^2 + x \left( \frac{\sqrt{3}}{2} \right)^2 - \frac{3}{4} \left( \frac{1}{\sqrt{3}} \right)^2 = 10 \)

\( \implies \) \( 8 + \frac{3x}{4} - \frac{3}{4} \times \frac{1}{3} = 10 \)

\( \implies \) \( 8 + \frac{3x}{4} - \frac{1}{4} = 10 \)

\( \implies \) \( \frac{3x}{4} = 10 + \frac{1}{4} - 8 \)

\( \implies \) \( \frac{3x}{4} = 2 + \frac{1}{4} = \frac{9}{4} \)

\( \implies \) \( \frac{3x}{4} = \frac{9}{4} \)

\( \implies \) \( 12x = 36 \)

\( \implies \) \( x = 3 \)

Question. If \( \tan x = \sin 45^\circ \cos 45^\circ + \sin 30^\circ \), determine \( x \).
Answer: \( \tan x = \sin 45^\circ \cos 45^\circ + \sin 30^\circ = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1 \)
\( \tan x = 1 = \tan 45^\circ \)
\( x = 45^\circ \)

Question. If \( \sin(A + B) = 1 \) and \( \cos(A - B) = \frac{\sqrt{3}}{2} \), then find \( A \) and \( B \).
Answer: \( \sin(A + B) = 1 \)

\( \implies \) \( \sin(A + B) = \sin 90^\circ \)

\( \implies \) \( A + B = 90^\circ \) ... (1)
And, \( \cos(A - B) = \frac{\sqrt{3}}{2} \)

\( \implies \) \( \cos(A - B) = \cos 30^\circ \)

\( \implies \) \( A - B = 30^\circ \) ... (2)
Adding (1) and (2), we get \( 2A = 120^\circ \)

\( \implies \) \( A = 60^\circ \)
Putting \( A = 60^\circ \) in (1), we get \( 60^\circ + B = 90^\circ \)

\( \implies \) \( B = 30^\circ \)
Hence, \( A = 60^\circ, B = 30^\circ \).

Question. Show that \( \frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ} = \tan 30^\circ \).
Answer: L.H.S. \( = \frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \frac{1}{\sqrt{3}}} = \frac{\frac{3-1}{\sqrt{3}}}{2} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \)
Also R.H.S. \( = \tan 30^\circ = \frac{1}{\sqrt{3}} \)
Hence, L.H.S. = R.H.S. Proved.

Question. If \( \sin \theta = \cos \theta \), find the value of \( 2 \tan^2 \theta + \sin^2 \theta - 1 \).
Answer: \( \sin \theta = \cos \theta \)

\( \implies \) \( \frac{\sin \theta}{\cos \theta} = 1 \)

\( \implies \) \( \tan \theta = 1 \text{ and } \tan \theta = \tan 45^\circ \)

\( \implies \) \( \theta = 45^\circ \)
\( 2 \tan^2 \theta + \sin^2 \theta - 1 = 2 \tan^2 45^\circ + \sin^2 45^\circ - 1 \)
\( = 2(1)^2 + \left( \frac{1}{\sqrt{2}} \right)^2 - 1 = 2 + \frac{1}{2} - 1 = \frac{3}{2} \)

Question. Find \( \sin 75^\circ \) by using the formula \( \sin(A + B) = \sin A \cos B + \cos A \sin B \).
Answer: \( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
Putting \( A = 45^\circ \) and \( B = 30^\circ \)
\( \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \)
\( = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)

Question. Find \( \sin 15^\circ \) by using the formula \( \sin(A - B) = \sin A \cos B - \cos A \sin B \).
Answer: \( \sin(A - B) = \sin A \cos B - \cos A \sin B \)
On replacing \( A \) by \( 45^\circ \) and \( B \) by \( 30^\circ \) in the above formula we get
\( \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \)
\( = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)

Question. Find \( \cos 15^\circ \) by using the formula \( \cos(A - B) = \cos A \cos B + \sin A \sin B \).
Answer: On substituting \( A = 45^\circ \) and \( B = 30^\circ \), we get
\( \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \)
\( = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)

Question. Find \( \cos 75^\circ \) by using the formula \( \cos(A + B) = \cos A \cos B - \sin A \sin B \).
Answer: \( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
On replacing \( A \) by \( 45^\circ \) and \( B \) by \( 30^\circ \), we get
\( \cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ \)
\( = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)

Question. Using the formula \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \), find the value of \( \tan 75^\circ \).
Answer: On substituting \( A = 45^\circ \) and \( B = 30^\circ \), we get
\( \tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)
\( = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 1 + 2\sqrt{3}}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \).

Question. Find the value of \( \tan 15^\circ \) by using the formula \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \).
Answer: Putting \( A = 45^\circ \) and \( B = 30^\circ \), we get
\( \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \)
\( \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1) \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \)
\( = \frac{3 + 1 - 2\sqrt{3}}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \).