Class 10 trigonometry exercise 8.1

Given: A right-angled triangle ABC, with the right angle at B. AB = 24 cm and BC = 7 cm.

Step 1: Find the hypotenuse (AC) using the Pythagorean theorem.
AC² = AB² + BC²
AC² = 24² + 7²
AC² = 576 + 49 = 625
AC = √625 = 25 cm.

(i) For angle A:
Opposite side = BC = 7 cm
Adjacent side = AB = 24 cm
Hypotenuse = AC = 25 cm
sin A = ^Opposite⁄_Hypotenuse = ^BC⁄_AC = 725
cos A = ^Adjacent⁄_Hypotenuse = ^AB⁄_AC = 2425

(ii) For angle C:
Opposite side = AB = 24 cm
Adjacent side = BC = 7 cm
Hypotenuse = AC = 25 cm
sin C = ^Opposite⁄_Hypotenuse = ^AB⁄_AC = 2425
cos C = ^Adjacent⁄_Hypotenuse = ^BC⁄_AC = 725

Given: A right-angled triangle PQR, right-angled at Q. PQ = 12 cm and PR = 13 cm.

Step 1: Find the side QR using the Pythagorean theorem.
PR² = PQ² + QR²
13² = 12² + QR²
169 = 144 + QR²
QR² = 169 – 144 = 25
QR = √25 = 5 cm.

Step 2: Calculate tan P.
For angle P, Opposite side = QR = 5 cm, Adjacent side = PQ = 12 cm.
tan P = ^Opposite⁄_Adjacent = ^QR⁄_PQ = 512.

Step 3: Calculate cot R.
For angle R, Adjacent side = QR = 5 cm, Opposite side = PQ = 12 cm.
cot R = ^Adjacent⁄_Opposite = ^QR⁄_PQ = 512.

Step 4: Find tan P − cot R.
tan P – cot R = 512 – 512 = 0.

Answer: The value of tan P − cot R is 0.

Given: sin A = 34.

We know sin A = ^Opposite⁄_Hypotenuse. So, let the opposite side be 3k and the hypotenuse be 4k for some positive constant k.

Step 1: Find the adjacent side using the Pythagorean theorem.
Hypotenuse² = Opposite² + Adjacent²
(4k)² = (3k)² + Adjacent²
16k² = 9k² + Adjacent²
Adjacent² = 16k² – 9k² = 7k²
Adjacent = √(7k²) = k√7.

Step 2: Calculate cos A and tan A.
cos A = ^Adjacent⁄_Hypotenuse = ^k√7⁄_4k = √74.
tan A = ^Opposite⁄_Adjacent = ^3k⁄_k√7 = 3√7.

Given: 15 cot A = 8, which means cot A = 815.

We know cot A = ^Adjacent⁄_Opposite. So, let the adjacent side be 8k and the opposite side be 15k.

Step 1: Find the hypotenuse using the Pythagorean theorem.
Hypotenuse² = Opposite² + Adjacent²
Hypotenuse² = (15k)² + (8k)²
Hypotenuse² = 225k² + 64k² = 289k²
Hypotenuse = √(289k²) = 17k.

Step 2: Calculate sin A and sec A.
sin A = ^Opposite⁄_Hypotenuse = ^15k⁄_17k = 1517.
sec A = ^Hypotenuse⁄_Adjacent = ^17k⁄_8k = 178.

Given: sec θ = 1312.

We know sec θ = ^Hypotenuse⁄_Adjacent. Let Hypotenuse = 13k and Adjacent = 12k.

Step 1: Find the opposite side using the Pythagorean theorem.
Hypotenuse² = Opposite² + Adjacent²
(13k)² = Opposite² + (12k)²
169k² = Opposite² + 144k²
Opposite² = 169k² – 144k² = 25k²
Opposite = √(25k²) = 5k.

Step 2: Calculate all other ratios.
sin θ = ^Opposite⁄_Hypotenuse = ^5k⁄_13k = 513.
cos θ = ^Adjacent⁄_Hypotenuse = ^12k⁄_13k = 1213.
tan θ = ^Opposite⁄_Adjacent = ^5k⁄_12k = 512.
cosec θ = ^Hypotenuse⁄_Opposite = ^13k⁄_5k = 135.
cot θ = ^Adjacent⁄_Opposite = ^12k⁄_5k = 125.

Step 1: Construct a triangle.
Consider a right-angled triangle ABC, where the right angle is at C (∠C = 90°). This makes ∠A and ∠B acute angles.

Step 2: Write the expressions for cos A and cos B.
In ΔABC:
cos A = ^{Adjacent side to A}⁄_Hypotenuse = ^AC⁄_AB.
cos B = ^{Adjacent side to B}⁄_Hypotenuse = ^BC⁄_AB.

Step 3: Use the given condition.
We are given that cos A = cos B.
Therefore, ACAB = BCAB.
Multiplying both sides by AB, we get: AC = BC.

Step 4: Conclude using properties of a triangle.
In a triangle, angles opposite to equal sides are equal. Since side AC = side BC, the angles opposite to these sides must be equal.
Angle opposite to AC is ∠B.
Angle opposite to BC is ∠A.
Therefore, ∠A = ∠B. (Proved)

(i) Evaluate (1 + sin θ)(1 − sin θ)(1 + cos θ)(1 − cos θ)

Method 1: Using trigonometric identities (Simpler Method)
First, simplify the expression. Using the identity (a+b)(a-b) = a² – b²:
Numerator: (1 + sin θ)(1 – sin θ) = 1 – sin²θ.
Denominator: (1 + cos θ)(1 – cos θ) = 1 – cos²θ.
The expression becomes 1 – sin²θ1 – cos²θ.
Using the identity sin²θ + cos²θ = 1, we have:
1 – sin²θ = cos²θ
1 – cos²θ = sin²θ
So, the expression is cos²θsin²θ = (^{cos θ}⁄_{sin θ})² = cot²θ.
Given cot θ = 78, so cot²θ = (78)² = 4964.

(ii) Evaluate cot²θ

This is simply the square of cot θ.
cot²θ = (cot θ)² = (78)² = 4964.

Given: 3 cot A = 4, so cot A = 43.

From this, tan A = ¹⁄_{cot A} = 34.

Step 1: Calculate the Left Hand Side (LHS).
LHS = 1 − tan²A1 + tan²A = 1 − (³⁄₄)²1 + (³⁄₄)² = 1 − ⁹⁄₁₆1 + ⁹⁄₁₆
LHS = (^16-9⁄₁₆)(¹⁶⁺⁹⁄₁₆) = ⁷⁄₁₆²⁵⁄₁₆ = 725.

Step 2: Calculate the Right Hand Side (RHS).
First, find sin A and cos A. From cot A = ⁴⁄₃ = ^Adjacent⁄_Opposite.
Let Adjacent = 4k, Opposite = 3k.
Hypotenuse² = (4k)² + (3k)² = 16k² + 9k² = 25k².
Hypotenuse = √(25k²) = 5k.
So, sin A = ^Opposite⁄_Hypotenuse = ^3k⁄_5k = 35.
And cos A = ^Adjacent⁄_Hypotenuse = ^4k⁄_5k = 45.
RHS = cos²A − sin²A = (45)² – (35)²
RHS = 1625 – 925 = 725.

Step 3: Compare LHS and RHS.
Since LHS = 725 and RHS = 725, the statement is true.

Given: tan A = 1√3 in a triangle ABC right-angled at B.

We know tan A = ^Opposite⁄_Adjacent = ^BC⁄_AB. Let BC = k and AB = k√3.

Step 1: Find the hypotenuse AC.
AC² = AB² + BC² = (k√3)² + (k)² = 3k² + k² = 4k².
AC = √(4k²) = 2k.

Step 2: Find the required trigonometric ratios.
For angle A: sin A = ^BC⁄_AC = ^k⁄_2k = 12 and cos A = ^AB⁄_AC = ^k√3⁄_2k = √32.
For angle C: sin C = ^AB⁄_AC = ^k√3⁄_2k = √32 and cos C = ^BC⁄_AC = ^k⁄_2k = 12.

(i) Calculate sin A cos C + cos A sin C.
= (12) × (12) + (√32) × (√32)
= 14 + 34 = 44 = 1.

(ii) Calculate cos A cos C − sin A sin C.
= (√32) × (12) – (12) × (√32)
= √34 – √34 = 0.

Given: ΔPQR is right-angled at Q. PQ = 5 cm and PR + QR = 25 cm.

Step 1: Express one side in terms of the other.
Let QR = x cm. Then PR = (25 – x) cm.

Step 2: Use the Pythagorean theorem to find x.
PR² = PQ² + QR²
(25 – x)² = 5² + x²
625 – 50x + x² = 25 + x²
625 – 25 = 50x
600 = 50x
x = ⁶⁰⁰⁄₅₀ = 12.

Step 3: Determine the lengths of the sides.
QR = x = 12 cm.
PR = 25 – x = 25 – 12 = 13 cm.

Step 4: Calculate the required ratios for angle P.
For angle P:
Opposite side = QR = 12 cm
Adjacent side = PQ = 5 cm
Hypotenuse = PR = 13 cm
sin P = ^Opposite⁄_Hypotenuse = 1213.
cos P = ^Adjacent⁄_Hypotenuse = 513.
tan P = ^Opposite⁄_Adjacent = 125.

(i) The value of tan A is always less than 1.
False.
Justification: tan A = ^Opposite⁄_Adjacent. The opposite side can be longer than the adjacent side. For example, if Opposite = 4 and Adjacent = 3, then tan A = 4/3, which is greater than 1.

(ii) sec A = 125 for some value of angle A.
True.
Justification: sec A = ^Hypotenuse⁄_Adjacent. Since the hypotenuse is always the longest side in a right-angled triangle, the value of sec A must be greater than or equal to 1. As 12/5 = 2.4 > 1, this is possible.

(iii) cos A is the abbreviation used for the cosecant of angle A.
False.
Justification: cos A is the abbreviation for cosine of angle A. The abbreviation for cosecant of angle A is cosec A.

(iv) cot A is the product of cot and A.
False.
Justification: cot A is a single term representing the cotangent function of the angle A. ‘cot’ by itself has no meaning; it is not a variable to be multiplied by A.

(v) sin θ = 43 for some angle θ.
False.
Justification: sin θ = ^Opposite⁄_Hypotenuse. In a right-angled triangle, the hypotenuse is always the longest side, so the opposite side can never be greater than the hypotenuse. Therefore, sin θ must always be less than or equal to 1. Since 4/3 > 1, this is not possible.