CBSE Class 11 Applied Mathematics Trigonometry Notes Set 01

Trigonometric Ratios

Trigonometric Ratios of an Acute Angle

Let us take a right triangle ABC, right-angled at B. Here ∠CAB is an acute angle. BC is the side opposite to angle A. AC is the hypotenuse of the right triangle and the side AB is a side of ∠A, so we call it the side adjacent to angle A.
Note that the positions of the sides change when we consider angle C in place of A.
The trigonometric ratios of the angle A in right triangle ABC are defined as follows:
\( \sin A = \frac{BC}{AC} \), \( \cos A = \frac{AB}{AC} \), \( \tan A = \frac{BC}{AB} \)
\( \text{cosec } A = \frac{AC}{BC} \), \( \sec A = \frac{AC}{AB} \), \( \cot A = \frac{AB}{BC} \)
Note that the ratios cosec A, sec A and cot A are respectively the reciprocals of the ratios sin A, cos A and tan A.
Also, observe that \( \tan A = \frac{BC}{AB} = \frac{BC/AC}{AB/AC} = \frac{\sin A}{\cos A} \) and \( \cot A = \frac{\cos A}{\sin A} \).
So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the lengths of its sides. The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.

Trigonometric Identities

Fundamental Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angle involved.
In Δ ABC, right-angled at B, we have AB² + BC² = AC² … (i)
Dividing each term of (i) by AC², we get \( \frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = \frac{AC^2}{AC^2} \), i.e. \( \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = \left(\frac{AC}{AC}\right)^2 \)
i.e. (cos A)² + (sin A)² = 1, i.e. cos² A + sin² A = 1 … (ii)
This is true for all A such that 0° ≤ A ≤ 90°. So, this is a trigonometric identity.
Let us now divide (i) by AB². We get \( \frac{AB^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AC^2}{AB^2} \), or \( \left(\frac{AB}{AB}\right)^2 + \left(\frac{BC}{AB}\right)^2 = \left(\frac{AC}{AB}\right)^2 \)
i.e. 1 + tan² A = sec² A … (iii)
Since tan A and sec A are not defined for A = 90°, (iii) is true for all A such that 0° ≤ A < 90°.
Again, dividing (i) by BC², we get \( \frac{AB^2}{BC^2} + \frac{BC^2}{BC^2} = \frac{AC^2}{BC^2} \), i.e. \( \left(\frac{AB}{BC}\right)^2 + \left(\frac{BC}{BC}\right)^2 = \left(\frac{AC}{BC}\right)^2 \)
i.e. cot² A + 1 = cosec² A … (iv)
Note that cosec A and cot A are not defined for A = 0°. Therefore (iv) is true for all A such that 0° < A ≤ 90°.
Using these identities, we can express each trigonometric ratio in terms of the other trigonometric ratios, i.e. if any one of the ratios is known, we can also determine the values of the other trigonometric ratios.

Example 1. Express the ratios cos A, tan A and sec A in terms of sin A.
Answer:
Since cos² A + sin² A = 1, therefore cos² A = 1 − sin² A
i.e. \( \cos A = \pm\sqrt{1 - \sin^2 A} \). Since A is an acute angle, cos A is positive. This gives \( \cos A = \sqrt{1 - \sin^2 A} \)
Hence \( \tan A = \frac{\sin A}{\cos A} = \frac{\sin A}{\sqrt{1 - \sin^2 A}} \)
and \( \sec A = \frac{1}{\cos A} = \frac{1}{\sqrt{1 - \sin^2 A}} \)

Example 2. Prove that sec A (1 − sin A)(sec A + tan A) = 1.
Answer:
LHS = sec A (1 − sin A)(sec A + tan A) \( = \frac{1}{\cos A}(1 - \sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) \)
\( = \frac{(1 - \sin A)(1 + \sin A)}{\cos^2 A} = \frac{1 - \sin^2 A}{\cos^2 A} = \frac{\cos^2 A}{\cos^2 A} = 1 \) = RHS

Example 3. Prove that \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A \).
Answer:
LHS \( = \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} \)
\( = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} \)
= sec A + tan A = RHS

Heights and Distances

Line of Sight, Angle of Elevation and Angle of Depression

The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.
The angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e. the case when we raise our head to look at the object.

The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e. the case when we lower our head to look at the point being viewed.

Example 4. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
Answer:

First let us draw a simple diagram to represent the problem. Here AB represents the tower, CB is the distance of the point from the tower and ∠ACB is the angle of elevation. We need to determine the height of the tower, i.e. AB.
Also, ACB is a triangle, right-angled at B. To solve the problem, we choose the trigonometric ratio tan 60°, as the ratio involves AB and BC.
Now \( \tan 60^\circ = \frac{AB}{BC} \), i.e. \( \sqrt{3} = \frac{AB}{15} \), i.e. \( AB = 15\sqrt{3} \)
Hence, the height of the tower is \( 15\sqrt{3} \) m.