CBSE Class 10 Mathematics Trigonometry Notes Set 12

Trigonometric Ratios of Complementary Angles

Two angles are said to be complementary if their sum is \( 90^\circ \). Thus \( \theta \) and \( (90^\circ - \theta) \) are complementary angles.

Consider \( \triangle ABC \) in which \( \angle B = 90^\circ \) and \( \angle A = \theta \). Then \( \angle C = 90^\circ - \theta \). For \( \angle C \), Base \( = BC \), Perpendicular \( = AB \) and Hypotenuse \( = AC \).

\( \sin(90^\circ - \theta) = \frac{AB}{AC} = \cos \theta \)
\( \cos(90^\circ - \theta) = \frac{BC}{AC} = \sin \theta \)
\( \tan(90^\circ - \theta) = \frac{AB}{BC} = \cot \theta \)
\( \operatorname{cosec}(90^\circ - \theta) = \frac{1}{\sin(90^\circ - \theta)} = \frac{1}{\cos \theta} = \sec \theta \)
\( \sec(90^\circ - \theta) = \frac{1}{\cos(90^\circ - \theta)} = \frac{1}{\sin \theta} = \operatorname{cosec} \theta \)
\( \cot(90^\circ - \theta) = \frac{1}{\tan(90^\circ - \theta)} = \frac{1}{\cot \theta} = \tan \theta \)

Hence: (i) \( \sin(90^\circ - \theta) = \cos \theta \) (ii) \( \cos(90^\circ - \theta) = \sin \theta \) (iii) \( \tan(90^\circ - \theta) = \cot \theta \) (iv) \( \operatorname{cosec}(90^\circ - \theta) = \sec \theta \) (v) \( \sec(90^\circ - \theta) = \operatorname{cosec} \theta \) (vi) \( \cot(90^\circ - \theta) = \tan \theta \).

Solved Examples

Example 1. Evaluate: \( \frac{\sec 39^\circ}{\operatorname{cosec} 51^\circ} + \frac{2}{\sqrt{3}} \tan 17^\circ \tan 38^\circ \tan 60^\circ \tan 52^\circ \tan 73^\circ \).
Answer:
\( \frac{\sec 39^\circ}{\operatorname{cosec}(90^\circ - 39^\circ)} + \frac{2}{\sqrt{3}} \tan 17^\circ \tan 38^\circ \cdot \sqrt{3} \cdot \tan(90^\circ - 38^\circ) \tan(90^\circ - 17^\circ) \)
\( = \frac{\sec 39^\circ}{\sec 39^\circ} + \frac{2}{\sqrt{3}} \times \sqrt{3} \times (\tan 17^\circ \cot 17^\circ)(\tan 38^\circ \cot 38^\circ) \) [using \( \operatorname{cosec}(90^\circ - \theta) = \sec \theta \) and \( \tan(90^\circ - \theta) = \cot \theta \)]
\( = 1 + 2 \times (1) \times (1) = 3 \) [since \( \tan \theta \cdot \cot \theta = 1 \)].

Example 2. Without using tables, show that \( \cos 35^\circ \cos 55^\circ - \sin 35^\circ \sin 55^\circ = 0 \).
Answer:
LHS \( = \cos 35^\circ \cos 55^\circ - \sin(90^\circ - 55^\circ) \sin(90^\circ - 35^\circ) \) [since \( \sin(90^\circ - \theta) = \cos \theta \)]
\( = \cos 35^\circ \cos 55^\circ - \cos 55^\circ \cos 35^\circ = 0 = \) RHS.

Example 3. Evaluate: (i) \( \frac{\sin 18^\circ}{\cos 72^\circ} \) (ii) \( \frac{\tan 26^\circ}{\cot 64^\circ} \) (iii) \( \cos 48^\circ - \sin 42^\circ \) (iv) \( \operatorname{cosec} 31^\circ - \sec 59^\circ \).
Answer:
(i) \( \frac{\sin 18^\circ}{\cos 72^\circ} = \frac{\sin(90^\circ - 72^\circ)}{\cos 72^\circ} = \frac{\cos 72^\circ}{\cos 72^\circ} = 1 \)
(ii) \( \frac{\tan 26^\circ}{\cot 64^\circ} = \frac{\tan(90^\circ - 64^\circ)}{\cot 64^\circ} = \frac{\cot 64^\circ}{\cot 64^\circ} = 1 \)
(iii) \( \cos 48^\circ - \sin 42^\circ = \cos(90^\circ - 42^\circ) - \sin 42^\circ = \sin 42^\circ - \sin 42^\circ = 0 \)
(iv) \( \operatorname{cosec} 31^\circ - \sec 59^\circ = \operatorname{cosec}(90^\circ - 59^\circ) - \sec 59^\circ = \sec 59^\circ - \sec 59^\circ = 0 \)

Example 4. Show that: (i) \( \tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ = 1 \) (ii) \( \cos 38^\circ \cos 52^\circ - \sin 38^\circ \sin 52^\circ = 0 \).
Answer:
(i) LHS \( = \tan 48^\circ \tan 23^\circ \tan(90^\circ - 48^\circ) \tan(90^\circ - 23^\circ) = \tan 48^\circ \tan 23^\circ \cot 48^\circ \cot 23^\circ = 1 = \) RHS.
(ii) \( \cos 38^\circ = \cos(90^\circ - 52^\circ) = \sin 52^\circ \) and \( \cos 52^\circ = \cos(90^\circ - 38^\circ) = \sin 38^\circ \). So LHS \( = \sin 52^\circ \sin 38^\circ - \sin 38^\circ \sin 52^\circ = 0 = \) RHS.

Example 5. If \( \tan 2A = \cot(A - 18^\circ) \), where \( 2A \) is an acute angle, find the value of A.
Answer:
\( \tan 2A = \cot(90^\circ - 2A) \), so \( \cot(90^\circ - 2A) = \cot(A - 18^\circ) \).
Equating the angles: \( 90^\circ - 2A = A - 18^\circ \Rightarrow 108^\circ = 3A \Rightarrow A = 36^\circ \).

Example 6. If \( \tan A = \cot B \), prove that \( A + B = 90^\circ \).
Answer:
\( \cot B = \tan(90^\circ - B) \), so \( \tan A = \tan(90^\circ - B) \).
Since both angles are acute, \( A = 90^\circ - B \), i.e. \( A + B = 90^\circ \). Hence proved.

Example 7. If \( \sec 4A = \operatorname{cosec}(A - 20^\circ) \), where \( 4A \) is an acute angle, find the value of A.
Answer:
\( \sec 4A = \operatorname{cosec}(90^\circ - 4A) \), so \( \operatorname{cosec}(90^\circ - 4A) = \operatorname{cosec}(A - 20^\circ) \).
Equating the angles: \( 90^\circ - 4A = A - 20^\circ \Rightarrow 110^\circ = 5A \Rightarrow A = 22^\circ \).

Example 8. If A, B and C are the interior angles of a triangle ABC, then show that \( \sin\left( \frac{B + C}{2} \right) = \cos \frac{A}{2} \).
Answer:
In a triangle, \( A + B + C = 180^\circ \Rightarrow B + C = 180^\circ - A \Rightarrow \frac{B + C}{2} = 90^\circ - \frac{A}{2} \).
\( \therefore \sin\left( \frac{B + C}{2} \right) = \sin\left( 90^\circ - \frac{A}{2} \right) = \cos \frac{A}{2} \). Hence proved.

Example 9. Express \( \sin 67^\circ + \cos 75^\circ \) in terms of trigonometric ratios of angles between \( 0^\circ \) and \( 45^\circ \).
Answer:
\( \sin 67^\circ = \sin(90^\circ - 23^\circ) = \cos 23^\circ \) and \( \cos 75^\circ = \cos(90^\circ - 15^\circ) = \sin 15^\circ \).
Therefore \( \sin 67^\circ + \cos 75^\circ = \cos 23^\circ + \sin 15^\circ \).