CBSE Class 10 Mathematics Introduction to Trigonometry VBQs Set 05

Evaluate each of the following:

Question. \( \frac{\tan 45^\circ}{\sin 30^\circ + \cos 60^\circ} \)
Answer: \( \frac{1}{\frac{1}{2} + \frac{1}{2}} = 1 \)

Question. \( \frac{\tan 45^\circ}{\text{cosec } 30^\circ} + \frac{\sec 60^\circ}{\cot 45^\circ} - \frac{2 \sin 90^\circ}{\cos 0^\circ} \)
Answer: \( \frac{1}{2} + \frac{2}{1} - \frac{2(1)}{1} = \frac{1}{2} \)

Question. \( 3 \sin^2 30^\circ + 2 \tan^2 60^\circ - 5 \cos^2 45^\circ \)
Answer: \( 3(\frac{1}{2})^2 + 2(\sqrt{3})^2 - 5(\frac{1}{\sqrt{2}})^2 = \frac{3}{4} + 6 - \frac{5}{2} = \frac{3 + 24 - 10}{4} = \frac{17}{4} \)

Question. \( \frac{4}{\cot^2 30^\circ} + \frac{1}{\sin^2 30^\circ} - 2 \cos^2 45^\circ - \sin^2 0^\circ \)
Answer: \( \frac{4}{(\sqrt{3})^2} + \frac{1}{(\frac{1}{2})^2} - 2(\frac{1}{\sqrt{2}})^2 - 0 = \frac{4}{3} + 4 - 1 = \frac{13}{3} \)

Question. \( \cos^2 60^\circ \tan^2 30^\circ + \sin 30^\circ \cos 0^\circ \sin 60^\circ \tan 45^\circ \)
Answer: \( (\frac{1}{2})^2(\frac{1}{\sqrt{3}})^2 + (\frac{1}{2})(1)(\frac{\sqrt{3}}{2})(1) = \frac{1}{12} + \frac{\sqrt{3}}{4} = \frac{1 + 3\sqrt{3}}{12} \)

Prove the following:

Question. \( \frac{\cos 30^\circ + \sin 60^\circ}{1 + \cos 60^\circ + \sin 30^\circ} = \frac{\sqrt{3}}{2} \)
Answer: LHS = \( \frac{\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}}{1 + \frac{1}{2} + \frac{1}{2}} = \frac{\sqrt{3}}{2} \) = RHS. Hence Proved.

Question. \( 2(\cos^4 60^\circ + \sin^4 30^\circ) - (\tan^2 60^\circ + \cot^2 45^\circ) + 3 \sec^2 30^\circ = \frac{1}{4} \)
Answer: LHS = \( 2\left((\frac{1}{2})^4 + (\frac{1}{2})^4\right) - ((\sqrt{3})^2 + 1^2) + 3(\frac{2}{\sqrt{3}})^2 \)
\( = 2\left(\frac{1}{16} + \frac{1}{16}\right) - (3 + 1) + 3\left(\frac{4}{3}\right) = 2\left(\frac{1}{8}\right) - 4 + 4 = \frac{1}{4} \) = RHS. Hence Proved.

Question. \( 2(\cos^2 45^\circ + \tan^2 60^\circ) - 6 (\sin^2 45^\circ - \tan^2 30^\circ) = 6 \)
Answer: LHS = \( 2\left((\frac{1}{\sqrt{2}})^2 + (\sqrt{3})^2\right) - 6 \left((\frac{1}{\sqrt{2}})^2 - (\frac{1}{\sqrt{3}})^2\right) \)
\( = 2\left(\frac{1}{2} + 3\right) - 6 \left(\frac{1}{2} - \frac{1}{3}\right) = 2\left(\frac{7}{2}\right) - 6\left(\frac{1}{6}\right) = 7 - 1 = 6 \) = RHS. Hence Proved.

Question. Using the formula, \( \cos A = \sqrt{\frac{1 + \cos 2A}{2}} \), find the value of \( \cos 30^\circ \); given that \( \cos 60^\circ = \frac{1}{2} \).
Answer: \( \cos 30^\circ = \sqrt{\frac{1 + \cos 60^\circ}{2}} = \sqrt{\frac{1 + 1/2}{2}} = \sqrt{\frac{3/4}{1}} = \frac{\sqrt{3}}{2} \)

Question. Using the formula, \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \), find the value of \( \tan 60^\circ \); given that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
Answer: \( \tan 60^\circ = \frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = \frac{2(1/\sqrt{3})}{1 - (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{2/3} = \sqrt{3} \)

Question. If \(\sin (A - B) = \frac{1}{2}\) and \(\cos (A + B) = 0\)
Answer: \(\angle A = 60^\circ, \angle B = 30^\circ\)

Find the value of x in the following:

Question. If \(\cos (40^\circ + x) = \sin 30^\circ\)
Answer: \(x = 20^\circ\)

Question. If \(\frac{x \sin^2 30^\circ \cos^2 60^\circ}{4 \cos^2 45^\circ} = \frac{3 \sin^2 45^\circ + 2 \cos^2 30^\circ}{\sin^2 90^\circ - 4 \cos^2 45^\circ}\)
Answer: \(-96\)

Question. If \(\sin (A + B + C) = 1, \tan (A - B) = \frac{1}{\sqrt{3}}\) and \(\sec (A + C) = 2\), find \(A, B\) and \(C\) when they are acute.
Answer: \(\angle A = 60^\circ, \angle B = 30^\circ, \angle C = 0^\circ\)

SOLVED EXAMPLES

Evaluate each of the following without using trigonometric tables:

Question. \(\frac{3 \sin 62^\circ}{\cos 28^\circ} - \frac{\sec 42^\circ}{\text{cosec } 48^\circ}\)
Answer: Solution. \(\frac{3 \sin 62^\circ}{\cos 28^\circ} - \frac{\sec 42^\circ}{\text{cosec } 48^\circ} = \frac{3 \sin(90^\circ - 28^\circ)}{\cos 28^\circ} - \frac{\text{cosec}(90^\circ - 48^\circ)}{\text{cosec } 48^\circ} = \frac{3 \cos 28^\circ}{\cos 28^\circ} - \frac{\text{cosec } 48^\circ}{\text{cosec } 48^\circ} = 3 - 1 = 2 \)

Question. \(\frac{\cot 54^\circ}{\tan 36^\circ} + \frac{\tan 20^\circ}{\cot 70^\circ} - 2\)
Answer: Solution. \(\frac{\cot 54^\circ}{\tan 36^\circ} + \frac{\tan 20^\circ}{\cot 70^\circ} - 2 = \frac{\cot(90^\circ - 36^\circ)}{\tan 36^\circ} + \frac{\tan(90^\circ - 70^\circ)}{\cot 70^\circ} - 2 = \frac{\tan 36^\circ}{\tan 36^\circ} + \frac{\cot 70^\circ}{\cot 70^\circ} - 2 = 1 + 1 - 2 = 0 \)

Question. \(\frac{\tan 35^\circ}{\cot 55^\circ} + \frac{\cot 78^\circ}{\tan 12^\circ} - 1\)
Answer: Solution. \(\frac{\tan 35^\circ}{\cot 55^\circ} + \frac{\cot 78^\circ}{\tan 12^\circ} - 1 = \frac{\tan (90^\circ - 55^\circ)}{\cot 55^\circ} + \frac{\cot (90^\circ - 12^\circ)}{\tan 12^\circ} - 1 = \frac{\cot 55^\circ}{\cot 55^\circ} + \frac{\tan 12^\circ}{\tan 12^\circ} - 1 = 1 + 1 - 1 = 1 \)

Question. \(\frac{\cot 40^\circ}{\tan 50^\circ} - \frac{1}{2}\left(\frac{\cos 35^\circ}{\sin 55^\circ}\right)\)
Answer: Solution. \(\frac{\cot 40^\circ}{\tan 50^\circ} - \frac{1}{2}\left(\frac{\cos 35^\circ}{\sin 55^\circ}\right) = \frac{\cot(90^\circ - 50^\circ)}{\tan 50^\circ} - \frac{1}{2}\frac{\cos (90^\circ - 55^\circ)}{\sin 55^\circ} = \frac{\tan 50^\circ}{\tan 50^\circ} - \frac{1 \sin 55^\circ}{2 \sin 55^\circ} = 1 - \frac{1}{2} = \frac{1}{2} \)

Question. \(\frac{\tan 50^\circ + \sec 50^\circ}{\cot 40^\circ + \text{cosec } 40^\circ} + \cos 40^\circ \cdot \text{cosec } 50^\circ\)
Answer: Solution. \(\frac{\tan 50^\circ + \sec 50^\circ}{\cot 40^\circ + \text{cosec } 40^\circ} + \cos 40^\circ \cdot \text{cosec } 50^\circ = \frac{\tan(90^\circ - 40^\circ) + \sec(90^\circ - 40^\circ)}{\cot 40^\circ + \text{cosec } 40^\circ} + \cos 40^\circ \text{cosec } (90^\circ - 40^\circ) \)
\( = \left[ \frac{\cot 40^\circ + \text{cosec } 40^\circ}{\cot 40^\circ + \text{cosec } 40^\circ} \right] + \cos 40^\circ \cdot \sec 40^\circ \)
Using: \(\tan (90^\circ - \theta) = \cot \theta\), \(\sec (90^\circ - \theta) = \text{cosec } \theta\), \(\text{cosec } (90^\circ - \theta) = \sec \theta\)
\( = 1 + 1 = 2 \)

Evaluate, without using trigonometric tables:

Question. \(\frac{\cos 80^\circ}{\sin 10^\circ} + \cos 59^\circ \text{cosec } 31^\circ\)
Answer: Solution. \(\frac{\cos 80^\circ}{\sin 10^\circ} + \cos 59^\circ \text{cosec } 31^\circ = \frac{\cos(90^\circ - 10^\circ)}{\sin 10^\circ} + \cos(90^\circ - 31^\circ) \cdot \text{cosec } 31^\circ\)
\( = \frac{\sin 10^\circ}{\sin 10^\circ} + \sin 31^\circ \times \frac{1}{\sin 31^\circ} = 1 + 1 = 2 \)

Question. \(\frac{\cos 75^\circ}{\sin 15^\circ} + \frac{\sin 12^\circ}{\cos 78^\circ} - \frac{\cos 18^\circ}{\sin 72^\circ}\)
Answer: Solution. \(\frac{\cos 75^\circ}{\sin 15^\circ} + \frac{\sin 12^\circ}{\cos 78^\circ} - \frac{\cos 18^\circ}{\sin 72^\circ} = \frac{\cos(90^\circ - 15^\circ)}{\sin 15^\circ} + \frac{\sin(90^\circ - 78^\circ)}{\cos 78^\circ} - \frac{\cos(90^\circ - 72^\circ)}{\sin 72^\circ}\)
\( = \frac{\sin 15^\circ}{\sin 15^\circ} + \frac{\cos 78^\circ}{\cos 78^\circ} - \frac{\sin 72^\circ}{\sin 72^\circ} = 1 + 1 - 1 = 1 \)

Question. \(\frac{\sin 50^\circ}{\cos 40^\circ} + \frac{\text{cosec } 40^\circ}{\sec 50^\circ} - 4 \cos 50^\circ \text{cosec } 40^\circ\)
Answer: Solution. \(\frac{\sin 50^\circ}{\cos 40^\circ} + \frac{\text{cosec } 40^\circ}{\sec 50^\circ} - 4 \cos 50^\circ \text{cosec } 40^\circ = \frac{\sin (90^\circ - 40^\circ)}{\cos 40^\circ} + \frac{\text{cosec } (90^\circ - 50^\circ)}{\sec 50^\circ} - 4 \cos 50^\circ \text{cosec}(90^\circ - 50^\circ) \)
\( = \frac{\cos 40^\circ}{\cos 40^\circ} + \frac{\sec 50^\circ}{\sec 50^\circ} - 4 \cos 50^\circ \cdot \sec 50^\circ \)
[Using: \(\sin (90^\circ - \theta) = \cos \theta\); \(\text{cosec } (90^\circ - \theta) = \sec \theta\)]
\( = 1 + 1 - 4 \cos 50^\circ \cdot \frac{1}{\cos 50^\circ} = 1 + 1 - 4 = -2 \)

Question. \(\frac{\cos 35^\circ}{\sin 55^\circ} + \frac{\sin 11^\circ}{\cos 79^\circ} - \cos 28^\circ \text{cosec } 62^\circ\)
Answer: Solution. \(\frac{\cos 35^\circ}{\sin 55^\circ} + \frac{\sin 11^\circ}{\cos 79^\circ} - \cos 28^\circ \text{cosec } 62^\circ = \frac{\cos(90^\circ - 55^\circ)}{\sin 55^\circ} + \frac{\sin(90^\circ - 79^\circ)}{\cos 79^\circ} - \cos (90^\circ - 62^\circ) \text{cosec } 62^\circ\)
\( = \frac{\sin 55^\circ}{\sin 55^\circ} + \frac{\cos 79^\circ}{\cos 79^\circ} - \sin 62^\circ \times \frac{1}{\sin 62^\circ} = 1 + 1 - 1 = 1 \)

Evaluate each of the following, without using the trigonometric tables:

Question. \(2 \frac{\tan 53^\circ}{\cot 37^\circ} - \frac{\cot 80^\circ}{\tan 10^\circ}\)
Answer: Solution. \(2 \frac{\tan 53^\circ}{\cot 37^\circ} - \frac{\cot 80^\circ}{\tan 10^\circ} = 2 \frac{\tan(90^\circ - 37^\circ)}{\cot 37^\circ} - \frac{\cot(90^\circ - 10^\circ)}{\tan 10^\circ} = 2 \frac{\cot 37^\circ}{\cot 37^\circ} - \frac{\tan 10^\circ}{\tan 10^\circ} = 2 - 1 = 1 \)

Question. \(\frac{\sec 70^\circ}{\text{cosec } 20^\circ} + \frac{\sin 59^\circ}{\cos 31^\circ}\)
Answer: Solution. \(\frac{\sec 70^\circ}{\text{cosec } 20^\circ} + \frac{\sin 59^\circ}{\cos 31^\circ} = \frac{\sec (90^\circ - 20^\circ)}{\text{cosec } 20^\circ} + \frac{\sin (90^\circ - 31^\circ)}{\cos 31^\circ} = \frac{\text{cosec } 20^\circ}{\text{cosec } 20^\circ} + \frac{\cos 31^\circ}{\cos 31^\circ} = 1 + 1 = 2 \)

Question. \(\frac{\cos 70^\circ}{\sin 20^\circ} + \frac{\cos 59^\circ}{\sin 31^\circ} - 8 \sin^2 30^\circ\)
Answer: Solution. \(\frac{\cos 70^\circ}{\sin 20^\circ} + \frac{\cos 59^\circ}{\sin 31^\circ} - 8 \sin^2 30^\circ = \frac{\cos(90^\circ - 20^\circ)}{\sin 20^\circ} + \frac{\cos(90^\circ - 31^\circ)}{\sin 31^\circ} - 8\left(\frac{1}{2}\right)^2\)
\( = \frac{\sin 20^\circ}{\sin 20^\circ} + \frac{\sin 31^\circ}{\sin 31^\circ} - 8 \times \frac{1}{4} = 1 + 1 - 2 = 0 \)

Question. \(\frac{\cos 70^\circ}{\sin 20^\circ} + \frac{\cos 55^\circ \text{cosec } 35^\circ}{\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ}\)
Answer: Solution. \(\frac{\cos 70^\circ}{\sin 20^\circ} + \frac{\cos 55^\circ \text{cosec } 35^\circ}{\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ} \)
\( = \frac{\cos (90^\circ - 20^\circ)}{\sin 20^\circ} + \frac{\cos (90^\circ - 35^\circ) \times \text{cosec } 35^\circ}{\tan 5^\circ \tan 25^\circ \times 1 \times \tan (90^\circ - 25^\circ) \times \tan (90^\circ - 5^\circ)} \)
\( = \frac{\sin 20^\circ}{\sin 20^\circ} + \frac{\sin 35^\circ \times \frac{1}{\sin 35^\circ}}{\tan 5^\circ \tan 25^\circ \cdot \cot 25^\circ \cdot \cot 5^\circ} = 1 + \frac{1}{\tan 5^\circ \tan 25^\circ \cdot \frac{1}{\tan 25^\circ} \cdot \frac{1}{\tan 5^\circ}} \)
\( = 1 + 1 = 2 \) (Using \(\tan \theta \cdot \cot \theta = 1\))

Without using trigonometric tables, find the value of each of the following:

Question. \(\tan 15^\circ \tan 20^\circ \tan 70^\circ \tan 75^\circ\)
Answer: Solution. \(\tan 15^\circ \tan 20^\circ \tan 70^\circ \tan 75^\circ = \tan 15^\circ \tan 20^\circ \tan (90^\circ - 20^\circ) \tan (90^\circ - 15^\circ) = \tan 15^\circ \tan 20^\circ \cot 20^\circ \cot 15^\circ \)
\( = \tan 15^\circ \tan 20^\circ \left(\frac{1}{\tan 20^\circ}\right) \left(\frac{1}{\tan 15^\circ}\right) = 1 \)

Question. \(\tan 5^\circ \tan 25^\circ \tan 30^\circ \tan 65^\circ \tan 85^\circ\)
Answer: Solution. \(\tan 5^\circ \tan 25^\circ \tan 30^\circ \tan 65^\circ \tan 85^\circ = \tan 5^\circ \tan 25^\circ \tan 30^\circ \tan (90^\circ - 25^\circ) \tan (90^\circ - 5^\circ) = \tan 5^\circ \tan 25^\circ \tan 30^\circ \cot 25^\circ \cot 5^\circ \)
\( = \tan 5^\circ \tan 25^\circ \tan 30^\circ \frac{1}{\tan 25^\circ} \frac{1}{\tan 5^\circ} = \tan 30^\circ = \frac{1}{\sqrt{3}} \) [Using \(\tan 30^\circ = \frac{1}{\sqrt{3}}\)]

Question. \(\cot 12^\circ \cot 38^\circ \cot 52^\circ \cot 60^\circ \cot 78^\circ\)
Answer: Solution. \(\cot 12^\circ \cot 38^\circ \cot 52^\circ \cot 60^\circ \cot 78^\circ = \cot 12^\circ \cot 38^\circ \cot [90^\circ - 38^\circ] \cot 60^\circ \cot [90^\circ - 12^\circ] \)
\( = \cot 12^\circ \cot 38^\circ \tan 38^\circ \cot 60^\circ \tan 12^\circ \) [Using \(\cot (90^\circ - \theta) = \tan \theta\)]
\( = \frac{1}{\tan 12^\circ} \frac{1}{\tan 38^\circ} \cdot \tan 38^\circ \cdot \tan 12^\circ \cdot \cot 60^\circ = \cot 60^\circ = \frac{1}{\sqrt{3}} \) [Using \(\cot 60^\circ = \frac{1}{\sqrt{3}}\)]

Question. \(\tan 7^\circ \tan 23^\circ \tan 60^\circ \tan 67^\circ \tan 83^\circ\)
Answer: Solution. \(\tan 7^\circ \tan 23^\circ \tan 60^\circ \tan 67^\circ \tan 83^\circ = \tan 7^\circ \tan 23^\circ \cdot \sqrt{3} \cdot \tan (90^\circ - 23^\circ) \cdot \tan (90^\circ - 7^\circ) \) [Using \(\tan 60^\circ = \sqrt{3}\)]
\( = \tan 7^\circ \tan 23^\circ \cdot \sqrt{3} \cdot \cot 23^\circ \cdot \cot 7^\circ \) [Using \(\tan (90^\circ - \theta) = \cot \theta\)]
\( = \tan 7^\circ \tan 23^\circ \cdot \sqrt{3} \cdot \frac{1}{\tan 23^\circ} \cdot \frac{1}{\tan 7^\circ} = \sqrt{3} \)

Question. \(\tan 35^\circ \tan 40^\circ \tan 45^\circ \tan 50^\circ \tan 55^\circ\)
Answer: Solution. \(\tan 35^\circ \tan 40^\circ \tan 45^\circ \tan 50^\circ \tan 55^\circ = \tan (90^\circ - 55^\circ) \cdot \tan (90^\circ - 50^\circ) \cdot 1 \cdot \tan 50^\circ \cdot \tan 55^\circ \) [Using \(\tan 45^\circ = 1\)]
\( = \cot 55^\circ \cot 50^\circ \tan 50^\circ \tan 55^\circ \)
\( = \frac{1}{\tan 55^\circ} \times \frac{1}{\tan 50^\circ} \cdot \tan 50^\circ \cdot \tan 55^\circ = 1 \)

Find the value of each of the following without using trigonometric tables:

Question. \(\tan (55^\circ - \theta) - \cot(35^\circ + \theta)\)
Answer: Solution. \(\tan(55^\circ - \theta) - \cot(35^\circ + \theta) = \tan(55^\circ - \theta) - \cot [90^\circ - (55^\circ - \theta)] = \tan(55^\circ - \theta) - \tan(55^\circ - \theta) = 0 \)

Question. \(\text{cosec } (65^\circ + \theta) - \sec (25^\circ - \theta)\)
Answer: Solution. \(\text{cosec } (65^\circ + \theta) - \sec (25^\circ - \theta) = \text{cosec } (90^\circ - 25^\circ + \theta) - \sec (25^\circ - \theta) = \text{cosec } [90^\circ - (25^\circ - \theta)] - \sec (25^\circ - \theta) = \sec (25^\circ - \theta) - \sec (25^\circ - \theta) = 0 \)

Question. Evaluate: \(\text{cosec } (65^\circ + \theta) - \sec (25^\circ - \theta) - \tan(55^\circ - \theta) + \cot(35^\circ + \theta)\)
Answer: Solution. \(\text{cosec}(65^\circ + \theta) - \sec (25^\circ - \theta) - \tan(55^\circ - \theta) + \cot (35^\circ + \theta) \)
\( = \text{cosec}(90^\circ - 25^\circ + \theta) - \sec(25^\circ - \theta) - \tan(90^\circ - 35^\circ - \theta) + \cot(35^\circ + \theta) \)
\( = \text{cosec}[90^\circ - (25^\circ - \theta)] - \sec (25^\circ - \theta) - \tan [90^\circ - (35^\circ + \theta)] + \cot (35^\circ + \theta) \)
\( = \sec(25^\circ - \theta) - \sec(25^\circ - \theta) - \cot(35^\circ + \theta) + \cot(35^\circ + \theta) = 0 \)

Question. Without using tables, evaluate: \(\sin(50^\circ + \theta) - \cos(40^\circ - \theta) + \tan 1^\circ \tan 10^\circ \tan 20^\circ \tan 70^\circ \tan 80^\circ \tan 89^\circ\)
Answer: Solution. \(\sin(50^\circ + \theta) - \cos(40^\circ - \theta) + \tan 1^\circ \tan 10^\circ \tan 20^\circ \tan 70^\circ \tan 80^\circ \tan 89^\circ \)
\( = \sin (90^\circ - 40^\circ + \theta) - \cos(40^\circ - \theta) + \tan 1^\circ \tan 10^\circ \tan 20^\circ \tan(90^\circ - 20^\circ) \tan(90^\circ - 10^\circ) \tan (90^\circ - 1^\circ) \)
\( = \sin[(90^\circ - (40^\circ - \theta))] - \cos(40^\circ - \theta) + \tan 1^\circ \tan 10^\circ \tan 20^\circ \cot 20^\circ \cot 10^\circ \cot 1^\circ \) [Using \(\tan(90^\circ - \theta) = \cot \theta\)]
\( = \cos (40^\circ - \theta) - \cos (40^\circ - \theta) + \tan 1^\circ \tan 10^\circ \tan 20^\circ \frac{1}{\tan 20^\circ} \frac{1}{\tan 10^\circ} \frac{1}{\tan 1^\circ} = 0 + 1 \cdot 1 \cdot 1 = 1 \)
[Using \(\sin(90^\circ - \theta) = \cos \theta\)]

Question. If \(\tan 2A = \cot (A - 18^\circ)\), where \(2A\) is an acute angle, find the value of \(A\).
Answer: Solution. We have, \(\tan 2A = \cot (A - 18^\circ) \dots (1) \)
\( \implies \) \( \cot (90^\circ - 2A) = \cot (A - 18^\circ) \) [Using \(\cot (90^\circ - 2A) = \tan 2A\)]
\( \implies \) \( 90^\circ - 2A = A - 18^\circ \)
\( \implies \) \( 3A = 108^\circ \) [Using \(90^\circ - 2A\) and \(A - 18^\circ\) both are acute angles]
\( \implies \) \( \angle A = 36^\circ \)

Question. If \(\tan A = \cot B\), prove that \(A + B = 90^\circ\).
Answer: Solution. We have, \(\tan A = \cot B \)
\( \implies \) \( \tan A = \tan(90^\circ - B) \) [Using \(\tan (90^\circ - B) = \cot B\)]
\( \implies \) \( A = 90^\circ - B \)
\( \implies \) \( A + B = 90^\circ \). Proved.

Question. If \(\sec 4A = \text{cosec } (A - 20^\circ)\), where \(4A\) is an acute angle, find the value of \(A\).
Answer: Solution. We have, \(\sec 4A = \text{cosec } (A - 20^\circ) \)
\( \implies \) \( \text{cosec } (90^\circ - 4A) = \text{cosec } (A - 20^\circ) \) [Using \(\text{cosec } (90^\circ - \theta) = \sec \theta\)]
\( \implies \) \( 90^\circ - 4A = A - 20^\circ \)
\( \implies \) \( 5A = 90^\circ + 20^\circ \)
\( \implies \) \( 5A = 110^\circ \)
\( \implies \) \( \angle A = 22^\circ \)

Question. If \(A, B\) and \(C\) are the interior angles of a triangle \(ABC\), show that \(\sin \frac{B + C}{2} = \cos \frac{A}{2}\).
Answer: Solution. Since \(A, B,\) and \(C\) are the interior angles of a triangle \(ABC\)
\( \implies \) \( A + B + C = 180^\circ \)
\( \implies \) \( \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ \)
\( \implies \) \( \frac{B}{2} + \frac{C}{2} = 90^\circ - \frac{A}{2} \)
\( \implies \) \( \sin \left( \frac{B + C}{2} \right) = \sin \left( 90^\circ - \frac{A}{2} \right) \)
\( \implies \) \( \sin \frac{B + C}{2} = \cos \frac{A}{2} \). Proved.