CBSE Class 10 Mathematics Trigonometry Notes Set 11

Trigonometric Ratios of Some Specific Angles

Certain angles are more useful in problems and applications than others. In this section, we find the values of the trigonometric ratios for the five angles \( 0^\circ, 30^\circ, 45^\circ, 60^\circ \) and \( 90^\circ \).

Trigonometric Ratios of 45°

Start with a right angled isosceles triangle, because in this triangle each acute angle measures \( 45^\circ \). Let \( AB = BC = a \) units. By Pythagoras theorem in \( \triangle ABC \), \( AC^2 = AB^2 + BC^2 = a^2 + a^2 = 2a^2 \), so \( AC = \sqrt{2}\,a \).

With respect to \( \angle C = 45^\circ \) (Perpendicular \( = AB \), Base \( = BC \)):
\( \sin 45^\circ = \frac{AB}{AC} = \frac{a}{\sqrt{2}a} = \frac{1}{\sqrt{2}} \), \( \cos 45^\circ = \frac{BC}{AC} = \frac{a}{\sqrt{2}a} = \frac{1}{\sqrt{2}} \), \( \tan 45^\circ = \frac{AB}{BC} = \frac{a}{a} = 1 \).
Similarly, \( \cot 45^\circ = 1 \), \( \sec 45^\circ = \sqrt{2} \) and \( \operatorname{cosec} 45^\circ = \sqrt{2} \).

Trigonometric Ratios of 60° and 30°

In an equilateral triangle each angle measures \( 60^\circ \). So we take an equilateral \( \triangle ABC \) with \( AB = BC = AC = 2a \) units and draw \( AD \perp BC \). In \( \triangle ABD \) and \( \triangle ADC \): \( AB = AC \) (sides of an equilateral triangle), \( AD = AD \) (common) and \( \angle ADB = \angle ADC = 90^\circ \). Hence \( \triangle ABD \cong \triangle ACD \) (RHS congruence criterion).

By C.P.C.T., \( BD = DC = \frac{1}{2}BC = a \) unit and \( \angle BAD = \angle CAD = \frac{1}{2}\angle BAC = 30^\circ \). By Pythagoras theorem in \( \triangle ABD \): \( AD^2 = AB^2 - BD^2 = (2a)^2 - a^2 = 3a^2 \), so \( AD = \sqrt{3}\,a \).

In \( \triangle ABD \): \( \sin 60^\circ = \frac{AD}{AB} = \frac{\sqrt{3}a}{2a} = \frac{\sqrt{3}}{2} \), \( \cos 60^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2} \), \( \tan 60^\circ = \frac{AD}{BD} = \frac{\sqrt{3}a}{a} = \sqrt{3} \). Similarly, \( \cot 60^\circ = \frac{1}{\sqrt{3}} \), \( \sec 60^\circ = 2 \) and \( \operatorname{cosec} 60^\circ = \frac{2}{\sqrt{3}} \).

Also, \( \sin 30^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2} \), \( \cos 30^\circ = \frac{AD}{AB} = \frac{\sqrt{3}a}{2a} = \frac{\sqrt{3}}{2} \), \( \tan 30^\circ = \frac{BD}{AD} = \frac{a}{\sqrt{3}a} = \frac{1}{\sqrt{3}} \). Similarly, \( \cot 30^\circ = \sqrt{3} \), \( \sec 30^\circ = \frac{2}{\sqrt{3}} \) and \( \operatorname{cosec} 30^\circ = 2 \).

Trigonometric Ratios of 0° and 90°

Take a right triangle ABC and keep decreasing the angle B. As \( \angle B \) decreases, the side AC keeps decreasing and the hypotenuse AB comes closer and closer to BC. Eventually, when \( \angle B \) becomes zero, the length of AC becomes 0 and AB coincides with BC. Then:

\( \sin 0^\circ = \frac{AC}{AB} = \frac{0}{AB} = 0 \), \( \cos 0^\circ = \frac{BC}{AB} = 1 \) (since AB and BC coincide), \( \tan 0^\circ = \frac{AC}{BC} = \frac{0}{BC} = 0 \), \( \sec 0^\circ = \frac{1}{\cos 0^\circ} = 1 \).
\( \cot 0^\circ \) is undefined since it is the reciprocal of 0, and \( \operatorname{cosec} 0^\circ \) is undefined for the same reason. Thus only 4 ratios are well defined at \( 0^\circ \).

Similarly, keep increasing \( \angle B \) till its measure reaches \( 90^\circ \). As B increases, BC decreases; eventually, when \( \angle B = 90^\circ \), BC becomes 0 and AC coincides with AB. Then:

\( \sin 90^\circ = \frac{AC}{AB} = 1 \) (since AB and AC coincide), \( \cos 90^\circ = \frac{BC}{AB} = \frac{0}{AB} = 0 \), \( \cot 90^\circ = 0 \), \( \operatorname{cosec} 90^\circ = 1 \).
\( \tan 90^\circ \) is not defined since it becomes \( \frac{1}{0} \), and \( \sec 90^\circ \) is not defined.

Trigonometric Table

Knowledge Hub

1. \( \tan \theta \) and \( \cot \theta \) are not defined for \( \theta = 90^\circ \) and \( 0^\circ \) respectively.
2. \( \operatorname{cosec} \theta \) and \( \sec \theta \) are not defined when \( \theta = 0^\circ \) and \( 90^\circ \) respectively.
3. \( \sin \theta = \cos \theta \) for \( \theta = 45^\circ \).
4. As \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \), \( \sin \theta \) increases from 0 to 1.
5. As \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \), \( \cos \theta \) decreases from 1 to 0.
6. As \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \), \( \tan \theta \) increases from 0 to \( \infty \).
7. \( \sin \theta \) and \( \cos \theta \) cannot be greater than one numerically.
8. \( \sec \theta \) and \( \operatorname{cosec} \theta \) cannot be less than one numerically.
9. \( \tan \theta \) and \( \cot \theta \) can have any value.

Solved Examples

Example 1. If \( (x - 2)\sin^2 30^\circ + (x - 3)\tan^2 60^\circ - x\cos^2 45^\circ = \frac{17}{4} \), find the value of x.
Answer:
\( (x - 2)\left( \frac{1}{2} \right)^2 + (x - 3)(\sqrt{3})^2 - x\left( \frac{1}{\sqrt{2}} \right)^2 = \frac{17}{4} \)
\( \Rightarrow \frac{x - 2}{4} + 3(x - 3) - \frac{x}{2} = \frac{17}{4} \)
\( \Rightarrow \frac{x - 2}{4} + 3x - 9 - \frac{x}{2} = \frac{17}{4} \Rightarrow \frac{x - 2 + 12x - 36 - 2x}{4} = \frac{17}{4} \)
\( \Rightarrow x + 12x - 2x = 17 + 36 + 2 \Rightarrow 11x = 55 \Rightarrow x = 5 \).

Example 2. Find the values of A and B if \( \cos(3A + B) = \frac{1}{2} \) and \( \sin(3A - B) = \frac{1}{2} \).
Answer:
\( \cos(3A + B) = \frac{1}{2} = \cos 60^\circ \), so \( 3A + B = 60^\circ \) ....(i)
\( \sin(3A - B) = \frac{1}{2} = \sin 30^\circ \), so \( 3A - B = 30^\circ \) ....(ii)
Solving (i) and (ii), we get \( A = 15^\circ \) and \( B = 15^\circ \).

Trigonometric Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all the values of the angle involved.

(a) \( \sin^2 \theta + \cos^2 \theta = 1 \)
Proof. Let ABC be a right angled triangle, right angled at C and \( \angle B = \theta \). Then \( \sin \theta = \frac{AC}{AB} \) and \( \cos \theta = \frac{BC}{AB} \).

\( \therefore \sin^2 \theta + \cos^2 \theta = \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AC^2 + BC^2}{AB^2} \). By Pythagoras theorem \( AC^2 + BC^2 = AB^2 \), so \( \sin^2 \theta + \cos^2 \theta = \frac{AB^2}{AB^2} = 1 \).

(b) \( 1 + \tan^2 \theta = \sec^2 \theta \)
Proof. In the same triangle ABC, \( \tan \theta = \frac{AC}{BC} \) and \( \sec \theta = \frac{AB}{BC} \). Now, \( 1 + \tan^2 \theta = 1 + \frac{AC^2}{BC^2} = \frac{BC^2 + AC^2}{BC^2} = \frac{AB^2}{BC^2} \) (by Pythagoras theorem) \( = \left( \frac{AB}{BC} \right)^2 = \sec^2 \theta \).

(c) \( 1 + \cot^2 \theta = \operatorname{cosec}^2 \theta \)
Proof. We have \( \cot \theta = \frac{BC}{AC} \) and \( \operatorname{cosec} \theta = \frac{AB}{AC} \). Now, \( 1 + \cot^2 \theta = 1 + \frac{BC^2}{AC^2} = \frac{AC^2 + BC^2}{AC^2} = \frac{AB^2}{AC^2} = \left( \frac{AB}{AC} \right)^2 = \operatorname{cosec}^2 \theta \).

Conversion Table

Each trigonometric ratio can be written in terms of every other ratio of the same acute angle \( \theta \).