Trigonometry | Part -2

Trigonometric Ratios of a Right-Angled Triangle (Acute Angle Relations)

In trigonometry, the ratios of the sides of a right-angled triangle with respect to its acute angles form the foundation of all trigonometric identities. For a right-angled triangle, these are known as the six trigonometric functions or ratios.

Definition of Trigonometric Ratios

Consider a right-angled triangle ABC, right-angled at C, with acute angle \( \theta \) at vertex A. Let:

• Opposite side = side opposite to angle \( \theta \)
• Adjacent side = side next to angle \( \theta \)
• Hypotenuse = the side opposite the right angle (longest side)

The six trigonometric ratios are defined as:

• \( \sin \theta = \frac{opposite}{hypotenuse}\)
• \( \cos \theta = \frac{adjacent}{hypotenuse}\)
• \( \tan \theta = \frac{opposite}{adjacent}\)
• \( \cot \theta = \frac{adjacent}{opposite}\)
• \( \sec \theta = \frac{hypotenuse}{adjacent}\)
• \( \csc \theta = \frac{hypotenuse}{opposite}\)

Quotient Identities (Relations among Trig Ratios)

These identities express trigonometric functions as quotients of each other:

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
• \( \cot \theta = \frac{\cos \theta}{\sin \theta} \)

Reciprocal Identities

• \( \sin \theta = \frac{1}{\csc \theta} \)
• \( \cos \theta = \frac{1}{\sec \theta} \)
• \( \tan \theta = \frac{1}{\cot \theta} \)

Square Relations (Pythagorean Identities)

• \( \sin^2 \theta + \cos^2 \theta = 1 \)
• \( 1 + \tan^2 \theta = \sec^2 \theta \)
• \( 1 + \cot^2 \theta = \csc^2 \theta \)

Relations Between All Six Trigonometric Ratios

i) Given: In triangle ABC, \( \angle B = 90^\circ \), \( AB = 12 \text{ cm} \), and \( BC = 5 \) cm . Find all trigonometric ratios.

Since \( \angle B = 90^\circ \), sides \( AB \) and \( BC \) are perpendicular. To find the trigonometric ratios of the acute angles, first calculate the hypotenuse \( AC \) using the Pythagorean theorem:

\[ AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

Let’s find the trigonometric ratios for \( \angle A \):

• \( \sin A = \frac{opposite\, side}{hypotenuse} = \frac{BC}{AC} = \frac{5}{13} \)
• \( \cos A = \frac{adjacent\, side}{hypotenuse} = \frac{AB}{AC} = \frac{12}{13} \)
• \( \tan A = \frac{opposite\, side}{adjacent\, side} = \frac{BC}{AB} = \frac{5}{12} \)
• \( \cot A = \frac{1}{\tan A} = \frac{12}{5} \)
• \( \sec A = \frac{1}{\cos A} = \frac{13}{12} \)
• \( \csc A = \frac{1}{\sin A} = \frac{13}{5} \)

Similarly, for \( \angle C \):

• \( \sin C = \frac{AB}{AC} = \frac{12}{13} \)
• \( \cos C = \frac{BC}{AC} = \frac{5}{13} \)
• \( \tan C = \frac{AB}{BC} = \frac{12}{5} \)
• \( \cot C = \frac{1}{\tan C} = \frac{5}{12} \)
• \( \sec C = \frac{1}{\cos C} = \frac{13}{5} \)
• \( \csc C = \frac{1}{\sin C} = \frac{13}{12} \)

ii) Given: \( \cos \theta = \frac{3}{5} \), find all other trigonometric ratios.

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we find \( \sin \theta \):

\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

Then:

• \( \sin \theta = \frac{4}{5} \)
• \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{4/5}{3/5} = \frac{4}{3} \)
• \( \cot \theta = \frac{1}{\tan \theta} = \frac{3}{4} \)
• \( \sec \theta = \frac{1}{\cos \theta} = \frac{5}{3} \)
• \( \csc \theta = \frac{1}{\sin \theta} = \frac{5}{4} \)