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Unit-8(Examples)

Introduction to Trigonometry

**Example-1 :-** Given tan A = 4/3, find the other trigonometric ratios of the angle A.

Give that: tan A = 4/3 cot A = 1/tan A = 3/4 sec² A = 1 + tan² A = 1 + (4/3)² = 1 + 16/9 = 25/9 sec A = 5/3 cos A = 1/sec A = 3/5 sin² A = 1 - cos² A = 1 - (3/5)² = 1 - 9/25 = 16/25 sin A = 4/5 cosec A = 1/sin A = 5/4

**Example-2 :-** If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.

Let us consider two right triangles ABC and PQR where sin B = sin Q see in figure We have sin B = AC/AB sin Q = PR/PQ Then, AC/AB = PR/PQ Therefore, AC/PR = AB/PQ = k .......(1) Now using Pythagoras theorem, From (1) and (2), we have AC/PR = AB/PQ = BC/QR Δ ACB ~ Δ PRQ and therefore, ∠ B = ∠ Q.

**Example-3 :-** Consider Δ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ (see Figure). Determine the values of
(i) cos² θ + sin² θ, (ii) cos² θ – sin² θ

In Δ ACB, we have AB = 29 units and BC = 21 units AC² = AB² - BC² = (29)² - (21)² = (29 - 21)(29 + 21) = 8 x 50 = 400 AC = 20 units so, sin θ = AC/AB = 20/29, cos θ = 21/29

Now, (i) cos² θ + sin² θ = (21/29)² + (20/29)² = 441/841 + 400/841 = 841/841 = 1

(ii)cos² θ – sin² θ = (21/29)² - (20/29)² = 441/841 - 400/841 = 41/841

**Example-4 :-** In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.

In Δ ABC, tan A = BC/AB = 1 (see in figure) i.e., BC = AB Let AB = BC = k, where k is a positive number AC² = AB² + BC² = k² + k² = 2k² AC = √2k Therefore, sin A = BC/AC = k/√2k = 1/√2 cos A = AB/AC = 1/√2 so, 2 sin A . cos A = 2 (1/√2) x (1/√2) = 1 which is the required value.

**Example-5 :-** In Δ OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Figure).
Determine the values of sin Q and cos Q.

In Δ OPQ, we have OQ² = OP² + PQ² (1 + PQ)² = OP² + PQ² i.e., 1 + PQ² + 2PQ = OP² + PQ² i.e., 1 + 2PQ = 7² i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm so, sin Q = 7/25 and cos Q = 24/25

**Example-6 :-** In Δ ABC, right-angled at B, AB = 5 cm and ∠ ACB = 30° (see Figure). Determine the lengths of the sides BC and AC.

To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB. Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore AB/BC = tan C 5/BC = tan 30° 5/BC = 1/√3 BC = 5√3 cm To find the length of the side AC, we consider sin 30° = AB/AC 1/2 = 5/AC AC = 10cm

**Example-7 :-** In Δ PQR, right-angled at Q (see Figure), PQ = 3 cm and PR = 6 cm. Determine ∠ QPR and ∠ PRQ.

Given PQ = 3 cm and PR = 6 cm Therefore, PQ/PR = sin R sin R = 3/6 = 1/2 so, ∠ PRQ = 30° ∠ QPR = 60°

**Example-8 :-** If sin (A – B) = 1/2, cos (A + B) = 1/2, 0° < A + B ≤ 90°, A > B, find A and B.

Since, sin (A – B) = 1/2, sin (A - B) = sin 30° Therefore, A – B = 30° .....(1) Also, since cos (A + B) = 1/2, cos (A + B) = 60° Therefore, A + B = 60° ......(2) Solving (1) and (2), A - B + A + B = 30° + 60° 2A = 90° A = 90°/2 A = 45° Put in (1) equation A - B = 30° 45° - B = 30° -B = 30° - 45° -B = -15° B = 15° we get : A = 45° and B = 15°.

**Example-9 :-** Evaluate : tan 65°/cot 25°

We know : cot A = tan (90° – A) So, cot 25° = tan (90° – 25°) = tan 65° i.e., tan 65°/cot 25° = tan 65°/tan 65° = 1

**Example-10 :-** If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

We are given that sin 3A = cos (A – 26°)..... (1) Since sin 3A = cos (90° – 3A), we can write (1) as cos (90° – 3A) = cos (A – 26°) Since 90° – 3A and A – 26° are both acute angles, Therefore, 90° – 3A = A – 26° which gives A = 29°

**Example-11 :-** Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

cot 85° + cos 75° = cot (90° – 5°) + cos (90° – 15°) = tan 5° + sin 15°

**Example-12 :-** Express the ratios cos A, tan A and sec A in terms of sin A.

Since cos² A + sin² A = 1, Therefore, cos² A = 1 – sin² A, i.e., cos A = √1 - sin² A tan A = sin A/cos A = sin A/√1 - sin² A sec A = 1/cos A = 1/√1 - sin² A

**Example-13 :-** Prove that : sec A (1 – sin A)(sec A + tan A) = 1.

**Example-14 :-** Prove that :

**Example-15 :-** Prove that :
using the identity sec² θ = 1 + tan² θ.

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