CBSE Class 10 Mathematics Trigonometry Notes Set 15

Trigonometric Angles

Trigonometric Ratios of 30° and 60°

Consider an equilateral triangle ABC with each side of length 2a. Since each angle of an equilateral triangle is \( 60^\circ \), each angle of \( \triangle ABC \) is \( 60^\circ \). Let AD be perpendicular from A on BC. Since the triangle is equilateral, AD is the bisector of \( \angle A \) and D is the mid-point of BC. Therefore BD = DC = a and \( \angle BAD = 30^\circ \).

Thus, in \( \triangle ABD \), \( \angle D \) is a right angle, hypotenuse AB = 2a and BD = a. By the Pythagoras theorem, \( AB^2 = AD^2 + BD^2 \Rightarrow 4a^2 = AD^2 + a^2 \Rightarrow AD = \sqrt{3}\,a \).

Trigonometric ratios of 30°. In right triangle ADB: Base = AD = \( \sqrt{3}a \), Perpendicular = BD = a, Hypotenuse = AB = 2a and \( \angle DAB = 30^\circ \).

\[ \sin 30^\circ = \frac{BD}{AB} = \frac{1}{2}, \quad \cos 30^\circ = \frac{AD}{AB} = \frac{\sqrt{3}}{2}, \quad \tan 30^\circ = \frac{BD}{AD} = \frac{1}{\sqrt{3}} \]

\[ \operatorname{cosec} 30^\circ = 2, \quad \sec 30^\circ = \frac{2}{\sqrt{3}}, \quad \cot 30^\circ = \sqrt{3} \]

Trigonometric ratios of 60°. In right triangle ADB: Base = BD = a, Perpendicular = AD = \( \sqrt{3}a \), Hypotenuse = AB = 2a and \( \angle ABD = 60^\circ \).

\[ \sin 60^\circ = \frac{AD}{AB} = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{BD}{AB} = \frac{1}{2}, \quad \tan 60^\circ = \frac{AD}{BD} = \sqrt{3} \]

\[ \operatorname{cosec} 60^\circ = \frac{2}{\sqrt{3}}, \quad \sec 60^\circ = 2, \quad \cot 60^\circ = \frac{1}{\sqrt{3}} \]

Trigonometric Ratios of 45°

Consider a right triangle ABC with right angle at B such that \( \angle A = 45^\circ \). Then \( \angle C = 180^\circ - 45^\circ - 90^\circ = 45^\circ \), so \( \angle A = \angle C \) and AB = BC. Let AB = BC = a. By the Pythagoras theorem, \( AC^2 = a^2 + a^2 = 2a^2 \Rightarrow AC = \sqrt{2}a \).

Thus, in \( \triangle ABC \): \( \angle A = 45^\circ \), Base = AB = a, Perpendicular = BC = a and Hypotenuse = AC = \( \sqrt{2}a \).

\[ \sin 45^\circ = \frac{BC}{AC} = \frac{1}{\sqrt{2}}, \quad \cos 45^\circ = \frac{AB}{AC} = \frac{1}{\sqrt{2}}, \quad \tan 45^\circ = \frac{BC}{AB} = 1 \]

\[ \operatorname{cosec} 45^\circ = \sqrt{2}, \quad \sec 45^\circ = \sqrt{2}, \quad \cot 45^\circ = 1 \]

Trigonometric Table

Solved Examples

Example 4. Given that \( \cos(A - B) = \cos A\cos B + \sin A\sin B \), find the value of \( \cos 15^\circ \).
Answer:
Put A = 45° and B = 30°: \( \cos(45^\circ - 30^\circ) = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ \).
\( \cos 15^\circ = \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2} = \dfrac{\sqrt{3} + 1}{2\sqrt{2}} \).

Check Your Progress

Check Your Progress 1. Evaluate \( \sin^2 45^\circ \cdot \cos^2 45^\circ + \sin^2 60^\circ \cdot \cos^2 60^\circ \).
Answer:
\( \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{3}{4} \times \dfrac{1}{4} = \dfrac{1}{4} + \dfrac{3}{16} = \) 7/16.

Check Your Progress 2. Find the value of x from the equation \( x\tan 45^\circ \sec 60^\circ = \cot^2 30^\circ \).
Answer:
\( x \times 1 \times 2 = (\sqrt{3})^2 = 3 \Rightarrow \) x = 3/2.

Check Your Progress 3. If A = 60° and B = 30°, verify that \( \sin A\cos B + \cos A\sin B = \sin(A + B) \).
Answer:
LHS \( = \dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{3}{4} + \dfrac{1}{4} = 1 \). RHS \( = \sin 90^\circ = 1 \). Verified.

Check Your Progress 4. If \( \theta \) is an acute angle and \( \sin\theta = \cos\theta \), find the value of \( 2\tan^2\theta + \sin^2\theta - 1 \).
Answer:
\( \sin\theta = \cos\theta \Rightarrow \theta = 45^\circ \). The value is \( 2(1) + \dfrac{1}{2} - 1 = \) 3/2.

Check Your Progress 5. Evaluate: (a) \( \sin^2 60^\circ + \cos^2 45^\circ \); (b) \( 3\cos^2 30^\circ + \tan^2 60^\circ \); (c) \( 4\sin^2 45^\circ + 3\tan^2 30^\circ + 8\sin 45^\circ\cos 45^\circ \); (d) \( 2\sin^2 30^\circ - 3\cos^2 60^\circ + \cot^2 30^\circ \); (e) \( \dfrac{\sin 30^\circ\cos 60^\circ}{\tan 45^\circ\cot 45^\circ} \); (f) \( \dfrac{\sin 60^\circ\tan 60^\circ}{\cos 30^\circ\cot 30^\circ} \); (g) \( \dfrac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} \); (h) \( 4(\sin^4 30^\circ + \cos^4 60^\circ) - 3(\sin^2 45^\circ - \sin 90^\circ) \).
Answer:
(a) \( \tfrac{3}{4} + \tfrac{1}{2} = \tfrac{5}{4} \). (b) \( 3 \times \tfrac{3}{4} + 3 = \tfrac{21}{4} \). (c) \( 4 \times \tfrac{1}{2} + 3 \times \tfrac{1}{3} + 8 \times \tfrac{1}{2} = 2 + 1 + 4 = 7 \). (d) \( \tfrac{2}{4} - \tfrac{3}{4} + 3 = \tfrac{11}{4} \).
(e) \( \dfrac{\frac{1}{2} \cdot \frac{1}{2}}{1 \cdot 1} = \tfrac{1}{4} \). (f) \( \dfrac{\frac{\sqrt{3}}{2} \cdot \sqrt{3}}{\frac{\sqrt{3}}{2} \cdot \sqrt{3}} = 1 \). (g) \( \dfrac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \tfrac{1}{2} \). (h) \( 4\left(\tfrac{1}{16} + \tfrac{1}{16}\right) - 3\left(\tfrac{1}{2} - 1\right) = \tfrac{1}{2} + \tfrac{3}{2} = 2 \).
Answers: (a) 5/4, (b) 21/4, (c) 7, (d) 11/4, (e) 1/4, (f) 1, (g) 1/2, (h) 2.

Complementary Angles

We know that two angles are said to be complementary if their sum is \( 90^\circ \). Thus, angles \( \theta \) and \( (90^\circ - \theta) \) are complementary. We shall now express the trigonometric ratios of the complementary angle \( (90^\circ - \theta) \) of a given angle \( \theta \) in terms of the trigonometric ratios of \( \theta \), where \( \theta \) is an acute angle.

Consider a right-angled triangle ABC, where \( \angle ABC = 90^\circ \), \( \angle ACB = \theta \), so that \( \angle BAC = 90^\circ - \theta \). Clearly \( \angle BAC \) and \( \angle ACB \) are complementary angles.

With reference to the angle \( \theta \), BC is the base, AB is the perpendicular and AC is the hypotenuse. Hence

\[ \sin\theta = \frac{AB}{AC}, \ \cos\theta = \frac{BC}{AC}, \ \tan\theta = \frac{AB}{BC}, \ \cot\theta = \frac{BC}{AB}, \ \sec\theta = \frac{AC}{BC}, \ \operatorname{cosec}\theta = \frac{AC}{AB} \]

Again, with reference to the angle \( \angle BAC = 90^\circ - \theta \), AB is the base, BC is the perpendicular and AC is the hypotenuse. Hence

\[ \sin(90^\circ - \theta) = \frac{BC}{AC} = \cos\theta, \qquad \cos(90^\circ - \theta) = \frac{AB}{AC} = \sin\theta \]

\[ \tan(90^\circ - \theta) = \frac{BC}{AB} = \cot\theta, \qquad \cot(90^\circ - \theta) = \frac{AB}{BC} = \tan\theta \]

\[ \sec(90^\circ - \theta) = \frac{AC}{AB} = \operatorname{cosec}\theta, \qquad \operatorname{cosec}(90^\circ - \theta) = \frac{AC}{BC} = \sec\theta \]

Solved Examples

Example 7. Find the value of \( \dfrac{\sin 59^\circ}{\cos 31^\circ} + \dfrac{\cos 20^\circ}{\sin 70^\circ} - 2\dfrac{\cos 31^\circ}{\sin 59^\circ} \).
Answer:
\( \sin 59^\circ = \sin(90^\circ - 31^\circ) = \cos 31^\circ \), \( \cos 20^\circ = \sin 70^\circ \) and \( \cos 31^\circ = \sin 59^\circ \).
The value is \( 1 + 1 - 2(1) = \) 0.

Example 8. Prove that \( \operatorname{cosec}(65^\circ + \theta) - \sec(25^\circ - \theta) - \tan(55^\circ - \theta) + \cot(35^\circ + \theta) = 0 \).
Answer:
\( \operatorname{cosec}(65^\circ + \theta) = \operatorname{cosec}\{90^\circ - (25^\circ - \theta)\} = \sec(25^\circ - \theta) \) ... (i)
\( \cot(35^\circ + \theta) = \cot\{90^\circ - (55^\circ - \theta)\} = \tan(55^\circ - \theta) \) ... (ii)
LHS \( = \sec(25^\circ - \theta) - \sec(25^\circ - \theta) - \tan(55^\circ - \theta) + \tan(55^\circ - \theta) = 0 = \) RHS. Hence proved.

Example 9. If \( \sec 3A = \operatorname{cosec}(A - 10^\circ) \), where 4A is an acute angle, find the value of A.
Answer:
\( \sec 3A = \operatorname{cosec}(90^\circ - 3A) \), so \( \operatorname{cosec}(90^\circ - 3A) = \operatorname{cosec}(A - 10^\circ) \).
\( 90^\circ - 3A = A - 10^\circ \Rightarrow 4A = 100^\circ \Rightarrow \) A = 25°.

Example 10. If A, B, C are the interior angles of a triangle ABC, then prove that \( \tan\left(\dfrac{A + B}{2}\right) = \cot\left(\dfrac{C}{2}\right) \).
Answer:
In \( \triangle ABC \), \( A + B + C = 180^\circ \Rightarrow A + B = 180^\circ - C \), so \( \dfrac{A + B}{2} = 90^\circ - \dfrac{C}{2} \).
\( \tan\left(\dfrac{A + B}{2}\right) = \tan\left(90^\circ - \dfrac{C}{2}\right) = \cot\left(\dfrac{C}{2}\right) \). Hence proved.

Check Your Progress

Check Your Progress 1. Simplify \( \dfrac{\cot 70^\circ}{\tan 20^\circ} + \dfrac{\cos 52^\circ}{\sin 38^\circ} \).
Answer:
\( \cot 70^\circ = \tan 20^\circ \) and \( \cos 52^\circ = \sin 38^\circ \), so the value is \( 1 + 1 = \) 2.

Check Your Progress 2. Prove that \( \cos 72^\circ \cos 18^\circ - \sin 72^\circ \sin 18^\circ = 0 \).
Answer:
\( \cos 18^\circ = \sin 72^\circ \) and \( \sin 18^\circ = \cos 72^\circ \). So the expression is \( \cos 72^\circ\sin 72^\circ - \sin 72^\circ\cos 72^\circ = 0 \).

Check Your Progress 3. If \( \sin A = \cos B \), what is A + B equal to?
Answer:
\( \cos B = \sin(90^\circ - B) \), so \( A = 90^\circ - B \) (acute angles) and \( A + B = \) 90°.

Check Your Progress 4. Evaluate: \( \cos(40^\circ + \theta) - \sin(50^\circ - \theta) \).
Answer:
\( \sin(50^\circ - \theta) = \cos(90^\circ - 50^\circ + \theta) = \cos(40^\circ + \theta) \). The value is 0.

Check Your Progress 5. Evaluate: \( \tan 13^\circ \cdot \tan 21^\circ \cdot \tan 69^\circ \cdot \tan 77^\circ \).
Answer:
\( \tan 77^\circ = \cot 13^\circ \) and \( \tan 69^\circ = \cot 21^\circ \), so the product is \( (\tan 13^\circ\cot 13^\circ)(\tan 21^\circ\cot 21^\circ) = \) 1.