INTRODUCTION TO TRIGONOMETRY

EXERCISE 8.1

Q1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A

Solution:

In ∆ABC, right-angled at B,

Given:
AB = 24 cm,
BC = 7 cm

To find:
sin A and cos A

Step 1: Find AC using the Pythagoras theorem
AC² = AB² + BC²
  = 24² + 7²
  = 576 + 49
  = 625
⇒ AC = 625 = 25 cm

Step 2: Use definitions of trigonometric ratios

sin A = opposite side / hypotenuse = BC / AC = 7 / 25
cos A = adjacent side / hypotenuse = AB / AC = 24 / 25

Final Answer:
sin A = 7 / 25
cos A = 24 / 25

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(ii) sin C, cos C

Solution:

In ∆ABC, right-angled at B,

Given:
AB = 24 cm,
BC = 7 cm,
AC = 25 cm (from the previous calculation using Pythagoras theorem)

To find:
sin C and cos C

Step 1: Use definitions of trigonometric ratios

sin C = opposite side / hypotenuse = AB / AC = 24 / 25
cos C = adjacent side / hypotenuse = BC / AC = 7 / 25

Final Answer:
sin C = 24 / 25
cos C = 7 / 25

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Q2: In Fig. 8.13 (refer book), find tan P – cot R

Solution:

In ∆PQR, right-angled at Q

Given:
PQ = 12 cm,
PR = 13 cm

We need to find tan P – cot R

Step 1: Find QR using the Pythagoras theorem

PR² = PQ² + QR²
13² = 12² + QR²
169 = 144 + QR²
QR² = 25
QR = 5 cm

Step 2: Find tan P and cot R

tan P = opposite side / adjacent side = QR / PQ = 5 / 12

cot R = adjacent side / opposite side = QR / PQ = 5 / 12

Step 3: Calculate tan P – cot R

tan P – cot R = (5 / 12) – (5 / 12) = 0

Final Answer: tan P – cot R = 0

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Q3: If sin A = 3/4, calculate cos A and tan A.

Solution:

Given:
sin A = 3 / 4

We have to calculate cos A and tan A.

Step 1: Find cos A using the identity
sin² A + cos² A = 1

(3 / 4)² + cos² A = 1
9 / 16 + cos² A = 1
cos² A = 1 – 9 / 16
cos² A = 16 / 16 – 9 / 16 = 7 / 16

cos A = √(7 / 16) = √7 / 4

Step 2: Find tan A
tan A = sin A / cos A = (3 / 4) ÷ (√7 / 4)
tan A = 3 / √7

Final Answer:
cos A = √7 / 4
tan A = 3 / √7

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Q4: Given 15 cot A = 8, find sin A and sec A.

Solution:

Given:
15 cot A = 8

We have to calculate sin A and sec A.

Step 1: Find cot A

cot A = 8 / 15

cot A = adjacent side / opposite side

Let adjacent side = 8 and opposite side = 15

Step 2: Find the hypotenuse using Pythagoras theorem

hypotenuse² = 8² + 15²
= 64 + 225
= 289

hypotenuse = 17

Step 3: Find sin A and sec A

sin A = opposite side / hypotenuse = 15 / 17
sec A = hypotenuse / adjacent side = 17 / 8

Final Answer:
sin A = 15 / 17
sec A = 17 / 8

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Q5: Given sec θ = 13 / 12, calculate all other trigonometric ratios.

Solution:

Step 1: Write the definition of sec θ

sec θ = hypotenuse / adjacent side

Given sec θ = 13 / 12, we take:
hypotenuse = 13
adjacent side = 12

Step 2: Find the opposite side using Pythagoras theorem

opposite side² = hypotenuse² – adjacent side²
= 13² – 12²
= 169 – 144
= 25

opposite side = 5

Step 3: Find all the trigonometric ratios

1. sin θ = opposite side / hypotenuse = 5 / 13

2. cos θ = adjacent side / hypotenuse = 12 / 13

3. tan θ = opposite side / adjacent side = 5 / 12

4. cot θ = adjacent side / opposite side = 12 / 5

5. sec θ = 13 / 12 (already given)

6. cosec θ = hypotenuse / opposite side = 13 / 5

Final Answer:

sin θ = 5 / 13
cos θ = 12 / 13
tan θ = 5 / 12
cot θ = 12 / 5
sec θ = 13 / 12 (already given)
cosec θ = 13 / 5

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Q6: If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Solution:

We are given:
cos A = cos B

For acute angles, the cosine function is one-to-one.
That means if the cosine values of two acute angles are equal, then the angles themselves must be equal.

Therefore:
∠A = ∠B

Final Answer:
∠A = ∠B

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Q7:
If cot θ = 7 / 8, evaluate:
(1 + sin θ)(1 – sin θ) / (1 + cos θ)(1 – cos θ)

Solution:

Step 1: Let us assume a right-angled triangle
Since cot θ = adjacent / opposite = 7 / 8,
Let:

• adjacent side = 7

• opposite side = 8

Then, hypotenuse = √(7² + 8²) = √(49 + 64) = √113

Step 2: Find sin θ and cos θ

• sin θ = opposite / hypotenuse = 8 / √113

• cos θ = adjacent / hypotenuse = 7 / √113

Step 3: Evaluate the expression
Expression:
(1 + sin θ)(1 – sin θ) / (1 + cos θ)(1 – cos θ)

= (1 – sin² θ) / (1 – cos² θ)
= cos² θ / sin² θ
= (7 / √113)² / (8 / √113)²
= 49 / 64

Final Answer:
49 / 64

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Q7:

(ii) cot² θ

Solution:

Given:
cot θ = 7 / 8

Step 1: Use the identity
cot² θ = (cot θ)²

Step 2: Substitute the value of cot θ
cot² θ = (7 / 8)²
cot² θ = 49 / 64

Final Answer:
cot² θ = 49 / 64

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Q8:
If 3 cot A = 4, check whether
(1 – tan² A) / (1 + tan² A) = cos² A – sin² A or not.

Solution

Step 1: Find cot A
3 cot A = 4
⇒ cot A = 4 / 3

cot A = adjacent / opposite = 4 / 3

Let:

• adjacent side = 4

• opposite side = 3

• hypotenuse = √(4² + 3²) = √(16 + 9) = √25 = 5

Now we find:

• sin A = 3 / 5

• cos A = 4 / 5

• tan A = 3 / 4

Left-Hand Side (LHS):

(1 – tan² A) / (1 + tan² A)
= (1 – (3/4)²) / (1 + (3/4)²)
= (1 – 9/16) / (1 + 9/16)
= (16/16 – 9/16) / (16/16 + 9/16)
= (7/16) / (25/16)
= 7 / 25

Right-Hand Side (RHS):

cos² A – sin² A
= (4/5)² – (3/5)²
= 16/25 – 9/25
= 7 / 25

Conclusion:

LHS = RHS
Therefore, the given identity is true.

Final Answer: Yes, the identity holds true.

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Q9: In triangle ABC, right-angled at B, if tan A = 1 / √3, find the value of:

(i) sin A cos C + cos A sin C

Solution

Step 1: Use the given
tan A = 1 / √3
So, tan A = opposite / adjacent = 1 / √3

Let:

• side opposite to ∠A = 1

• side adjacent to ∠A = √3

• hypotenuse = √[(1)² + (√3)²] = √(1 + 3) = √4 = 2

Then:

• sin A = opposite / hypotenuse = 1 / 2

• cos A = adjacent / hypotenuse = √3 / 2

In right-angled triangle ABC, since B is the right angle,
∠C = 90° – ∠A
So we can use the cofunction identities:

• sin C = cos A = √3 / 2

• cos C = sin A = 1 / 2

Now calculate:

sin A cos C + cos A sin C
= (1 / 2)(1 / 2) + (√3 / 2)(√3 / 2)
= 1 / 4 + 3 / 4
= 4 / 4
= 1

Final Answer: 1

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Q9:

(ii) cos A cos C – sin A sin C

Solution:

In triangle ABC, right-angled at B, and
tan A = 1 / √3

From previous steps:

• sin A = 1 / 2

• cos A = √3 / 2

• sin C = √3 / 2

• cos C = 1 / 2

Now calculate:

cos A cos C – sin A sin C
= (√3 / 2)(1 / 2) – (1 / 2)(√3 / 2)
= (√3 / 4) – (√3 / 4)
= 0

Final Answer: 0

This is a standard identity:
cos(A + C) = cos A cos C – sin A sin C
Since A + C = 90°, cos 90° = 0

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Q10: In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm.
Determine the values of sin P, cos P and tan P.

Solution:

Let us consider the right-angled triangle ∆PQR, with right angle at Q.

So, ∠Q = 90°.

Given:

• PQ = 5 cm

• PR + QR = 25 cm

Let us assume:

• QR = x,

• Then PR = 25 – x

By Pythagoras Theorem in ∆PQR:
PR² = PQ² + QR²
⇒ (25 – x)² = 5² + x²
⇒ 625 – 50x + x² = 25 + x²
Cancel x² from both sides:
⇒ 625 – 50x = 25
⇒ 600 = 50x
⇒ x = 12

So, QR = 12 cm,
Then PR = 25 – 12 = 13 cm

Now find the trigonometric ratios of angle P:

In triangle PQR:

• sin P = Opposite / Hypotenuse = QR / PR = 12 / 13

• cos P = Adjacent / Hypotenuse = PQ / PR = 5 / 13

• tan P = Opposite / Adjacent = QR / PQ = 12 / 5

Final Answers:

• sin P = 12 / 13

• cos P = 5 / 13

• tan P = 12 / 5

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Q11: State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

Answer: False.

Justification:

The value of tan A = Opposite side / Adjacent side in a right-angled triangle.

• The value of tan A depends on the angle A.

• If A = 45°, then tan A = 1.

• If A < 45°, then tan A < 1.

• If A > 45°, then tan A > 1.

So, tan A is not always less than 1.
It can be equal to or greater than 1, depending on the measure of angle A.

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Q11:

(ii) sec A = 12 / 5 for some value of angle A.

Answer: True

Justification:

We know:

sec A = hypotenuse / adjacent side

If sec A = 12 / 5,
then hypotenuse = 12 and adjacent side = 5

This is possible because in a right-angled triangle,
hypotenuse is always greater than the other sides

Therefore, the given value is valid

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Q11:

(iii) cos A is the abbreviation used for the cosecant of angle A.

Solution:

Answer: False

Justification:

• cos A stands for cosine of angle A, not cosecant.

• The correct abbreviation for cosecant of angle A is cosec A (or csc A).

• So, cos A and cosec A are two different trigonometric ratios.

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Q11:

(iv) cot A is the product of cot and A.

Answer: False

Justification:

• cot A is not the product of cot × A.

• It means cotangent of angle A, which is a trigonometric ratio.

• Just like sin A, cos A, or tan A, it refers to a function applied to the angle A — not multiplication.

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Q11:

(v) sin θ = 4/3 for some angle θ.

Answer: False

Justification:

• The value of sin θ must always lie between –1 and 1, inclusive.
  That is: −1 ≤ sin θ ≤ 1

• But here, sin θ = 4/3, which is greater than 1.

• This is not possible for any real angle θ.

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EXERCISE 8.2

Q1. Evaluate the following :

(i) sin 60° × cos 30° + sin 30° × cos 60°

Solution:
We use the known trigonometric values:
sin 60° = √3/2
cos 30° = √3/2
sin 30° = 1/2
cos 60° = 1/2

Now substitute:
= (√3/2 × √3/2) + (1/2 × 1/2)
= 3/4 + 1/4
= 4/4
= 1

Final Answer: 1

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Q1:

(ii) 2 tan² 45° + cos² 30° – sin² 60°

Solution:

Step 1: Use standard trigonometric values

• tan 45° = 1 → tan² 45° = 1² = 1

• cos 30° = √3 / 2 → cos² 30° = (√3 / 2)² = 3 / 4

• sin 60° = √3 / 2 → sin² 60° = (√3 / 2)² = 3 / 4

Step 2: Substitute the values into the expression

= 2 × 1 + 3/4 – 3/4
= 2 + 3/4 – 3/4
= 2

Final Answer: 2

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Q1:

(iii)

cos 45°
———————
sec 30° + cosec 30°

Solution:

Step 1: Use standard trigonometric values
cos 45° = 1 / √2
sec 30° = 2 / √3
cosec 30° = 2

Step 2: Substitute the values
= (1 / √2) ÷ (2 / √3 + 2)

Step 3: Multiply numerator and denominator appropriately
= 1 / √2 × 1 / (2 / √3 + 2)
= 1 / [√2 × (2 / √3 + 2)]

This expression simplifies to:

(3√2 – √6) / 8

Final Answer: (3√2 – √6) / 8

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Q1:

(iv)

(sin 30° + tan 45° – cosec 60°)
—————————————
(sec 30° + cos 60° + cot 45°)

Solution:

Step 1: Use standard trigonometric values

• sin 30° = 1 / 2

• tan 45° = 1

• cosec 60° = 2 / √3

• sec 30° = 2 / √3

• cos 60° = 1 / 2

• cot 45° = 1

Step 2: Substitute values

Numerator:
= (1/2) + 1 – (2 / √3)

Denominator:
= (2 / √3) + (1/2) + 1

Step 3: Simplify both parts

Numerator: 1/2 + 1 = 3/2 So, 3/2 – (2 / √3)

Denominator: 1/2 + 1 = 3/2 So, 3/2 + (2 / √3)

Now we write:

(3/2 – 2 / √3)
——————
(3/2 + 2 / √3)

Step 4: Rationalize

Multiply numerator and denominator by the conjugate of the denominator:
(3/2 – 2 / √3) ÷ (3/2 + 2 / √3) × (3/2 – 2 / √3) / (3/2 – 2 / √3)

This becomes:

[(3/2 – 2 / √3)]²
———————
[(3/2)² – (2 / √3)²]

Numerator:

(3/2)² = 9/4
(2 / √3)² = 4 / 3
So:

Numerator = 9/4 – 2 × 3/2 × 2/√3 + 4/3
(You will eventually get: 43 – 24√3)

Denominator:
(9/4) – (4 / 3)

LCM of 4 and 3 is 12:
= (27 – 16) / 12 = 11 / 12

So the whole expression becomes:

(43 – 24√3) / 11

Final Answer: (43 – 24√3) / 11

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Q1:

(v)

(5 cos² 60° + 4 sec² 30° – tan² 45°)
———————————————
(sin² 30° + cos² 30°)

Solution:

Step 1: Use standard trigonometric values

• cos 60° = 1 / 2 → cos² 60° = 1 / 4

• sec 30° = 2 / √3 → sec² 30° = 4 / 3

• tan 45° = 1 → tan² 45° = 1

• sin 30° = 1 / 2 → sin² 30° = 1 / 4

• cos 30° = √3 / 2 → cos² 30° = 3 / 4

Step 2: Substitute values

Numerator:

5 × (1 / 4) + 4 × (4 / 3) – 1
= 5 / 4 + 16 / 3 – 1

Find LCM of 4 and 3 = 12
= (15 / 12) + (64 / 12) – (12 / 12)
= (15 + 64 – 12) / 12
= 67 / 12

Denominator:

sin² 30° + cos² 30°
= 1 / 4 + 3 / 4 = 4 / 4 = 1

Step 3: Final Result

Numerator = 67 / 12
Denominator = 1

So, full expression = 67 / 12

Final Answer: 67 / 12

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Q2: Choose the correct option and justify your choice :

(i)

(2 × tan 30°) ÷ (1 + tan² 30°)

Solution:

Step 1: Use standard values

• tan 30° = 1 / √3

• tan² 30° = 1 / 3

Step 2: Substitute

= (2 × 1 / √3) ÷ (1 + 1 / 3)
= (2 / √3) ÷ (4 / 3)

Step 3: Simplify

= (2 / √3) × (3 / 4)
= 6 / (4√3)
= 3 / (2√3)

Now rationalize:

= (3 / 2√3) × (√3 / √3)
= (3√3) / (2 × 3)
= √3 / 2

Final Answer: √3 / 2
And we know:
sin 60° = √3 / 2

Answer: (A) sin 60°

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Q2:

(ii) (1 − tan² 45°) / (1 + tan² 45°)

Solution:

tan 45° = 1
⇒ tan² 45° = 1

Now substitute:

(1 − 1) / (1 + 1) = 0 / 2 = 0

Answer: (D) 0

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Q2:

(iii) sin 2A = 2 sin A is true when A

Solution:

We know that:
sin 2A = 2 sin A cos A

So the equation becomes:
2 sin A cos A = 2 sin A

Divide both sides by 2 sin A (assuming sin A ≠ 0):
cos A = 1

cos A = 1 when A = 0°

Answer: (A) 0°

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Q2:

(iv) (2 tan 30°) / (1 − tan² 30°)

Solution:

We know:
tan 30° = 1/√3
⇒ tan² 30° = 1/3

Now substitute:

= (2 × 1/√3) / (1 − 1/3)
= (2/√3) / (2/3)
= (2/√3) × (3/2)
= 3/√3
= √3

But tan 60° = √3

Answer: (C) tan 60°

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Q3:

If tan(A + B) = √3 and tan(A – B) = 1/√3; 0° < A + B ≤ 90° and A > B
Find: A and B

Solution:

We are given:
• tan(A + B) = √3
• tan(A – B) = 1/√3

Step 1: Use known values of tangent
We know:
• tan(60°) = √3
• tan(30°) = 1/√3

So, by comparing:
A + B = 60°
A – B = 30°

Step 2: Solve the two equations

Equation 1: A + B = 60°
Equation 2: A – B = 30°

Add the two equations:
(A + B) + (A – B) = 60° + 30°
2A = 90°
A = 45°

Substitute A = 45° into equation 1:
45° + B = 60°
B = 15°

Final Answer: A = 45°, B = 15°

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Q4: (i) State whether the following are true or false. Justify your answer.

Statement: sin(A + B) = sin A + sin B

Answer: False

Justification:
There is a standard trigonometric identity for sin(A + B):

sin(A + B) = sin A × cos B + cos A × sin B

This is not equal to sin A + sin B.

So, the statement is false because it ignores the multiplication with cosine terms that are part of the correct identity.

Example to verify:

Let A = 30°, B = 60°
Then,

Left side:
sin(A + B) = sin(90°) = 1

Right side:
sin A + sin B = sin(30°) + sin(60°) = 0.5 + 0.866 = 1.366

Since 1 ≠ 1.366, the two sides are not equal.

Conclusion: The statement is false.

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Q4: (ii) The value of sin q increases as q increases.

Answer:
True (only in the interval 0° ≤ θ ≤ 90°)

Justification:
In the interval from 0° to 90°, the sine of an angle increases as the angle increases.

Examples:
• sin 0° = 0
• sin 30° = 0.5
• sin 60° ≈ 0.866
• sin 90° = 1

As you can see, sin θ increases as θ increases between 0° and 90°.

Conclusion:
The statement is true within the interval 0° ≤ θ ≤ 90°.

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Q4:

(iii) The value of cos θ increases as θ increases.

Answer:
False (in the interval 0° ≤ θ ≤ 90°)

Justification:
In the interval from 0° to 90°, the cosine of an angle decreases as the angle increases.

Examples:
• cos 0° = 1
• cos 30° ≈ 0.866
• cos 60° = 0.5
• cos 90° = 0

As you can see, cos θ decreases as θ increases between 0° and 90°.

Conclusion: The statement is false.

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Q4: (iv)
sin θ = cos θ for all values of θ.

Answer:
False

Justification:
sin θ and cos θ are not equal for all values of θ.
They are only equal at a specific angle.

Let’s test with some examples:

• sin 0° = 0, cos 0° = 1 → not equal
• sin 45° = 0.707, cos 45° = 0.707 → equal
• sin 60° = 0.866, cos 60° = 0.5 → not equal

So, sin θ = cos θ only when θ = 45°, not for all values of θ.

Conclusion: The statement is false.

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Q4: (v) cot A is not defined for A = 0°.

Answer: True

Justification:
cot A = cos A ÷ sin A

At A = 0°:
• sin 0° = 0
• cos 0° = 1

So,
cot 0° = 1 ÷ 0 → Not defined (division by zero is not allowed)

Conclusion:
The statement is true because cot A is not defined when A = 0°.

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EXERCISE 8.3

Q1:

Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A.

Solution:

We use trigonometric identities to express each ratio in terms of cot A.

1. sin A = 1 ÷ √(1 + cot²A)

2. tan A = 1 ÷ cot A

3. sec A = √(1 + cot²A) ÷ cot A

Final Answer:

• sin A = 1 / √(1 + cot²A)
• tan A = 1 / cot A
• sec A = √(1 + cot²A) / cot A

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Q2: Write all the other trigonometric ratios of ∠A in terms of sec A.

Solution:

We express all the 6 trigonometric ratios in terms of sec A using standard identities.

1. cos A

By definition:

sec A = 1 / cos A
So,
cos A = 1 / sec A

2. sin A

Use identity:

sin²A + cos²A = 1
Substitute cos A = 1 / sec A:
sin²A + (1 / sec A)² = 1
sin²A + 1 / sec²A = 1
sin²A = 1 – 1 / sec²A
Take LCM:
sin²A = (sec²A – 1) / sec²A
Take square root:
sin A = √(sec²A – 1) / sec A

3. tan A

Use identity:

tan A = sin A / cos A
We already found:

• sin A = √(sec²A – 1) / sec A

• cos A = 1 / sec A

So:

tan A = [√(sec²A – 1) / sec A] ÷ [1 / sec A]
tan A = √(sec²A – 1)

4. cot A

cot A = 1 / tan A
We already have:
tan A = √(sec²A – 1)
So:
cot A = 1 / √(sec²A – 1)

5. cosec A

cosec A = 1 / sin A
We already have:
sin A = √(sec²A – 1) / sec A
So:
cosec A = sec A / √(sec²A – 1)

Final Answer (All Ratios in Terms of sec A):

• sin A = √(sec²A – 1) / sec A

• cos A = 1 / sec A

• tan A = √(sec²A – 1)

• cot A = 1 / √(sec²A – 1)

• cosec A = sec A / √(sec²A – 1)

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Q3: Choose the correct option. Justify your choice.

(i) 9 sec² A – 9 tan² A

Solution:

Step 1:
Use the trigonometric identity:
sec² A – tan² A = 1

Step 2:
Factor out 9:
9(sec² A – tan² A)

Step 3:
Substitute the identity:
9 × 1 = 9

Final Answer: (B) 9

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Q3: (ii) (1 + tan θ + sec θ)(1 + cot θ − cosec θ)

Solution:

First bracket:
1 + tan θ + sec θ = (cos θ + sin θ + 1) / cos θ

Second bracket:
1 + cot θ − cosec θ = (sin θ + cos θ − 1) / sin θ

Multiplying:
[(cos θ + sin θ + 1)(cos θ + sin θ − 1)] / (sin θ cos θ)

Using (a + b)(a − b) = a² − b²:
(cos θ + sin θ)² − 1 = (cos² θ + sin² θ + 2 sin θ cos θ) − 1
= 1 + 2 sin θ cos θ − 1
= 2 sin θ cos θ

So the expression becomes:
(2 sin θ cos θ) / (sin θ cos θ) = 2

Final Answer: (C) 2

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Q3: (iii) (sec A + tan A)(1 – sin A)

Solution:

= (1 / cos A + sin A / cos A) × (1 – sin A)

= (1 + sin A) / cos A × (1 – sin A)

= (1 + sin A)(1 – sin A) / cos A

= (1 – sin² A) / cos A

= cos² A / cos A

= cos A

Final Answer: (D) cos A

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Q3: (iv) (1 + tan² A) / (1 + cot² A)

Solution:

= sec² A / csc² A

= (1 / cos² A) ÷ (1 / sin² A)

= sin² A / cos² A

= tan² A

Answer: (D) tan² A

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Q4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosec θ − cot θ)² = (1 − cos θ) / (1 + cos θ)

Solution:

LHS:
(cosec θ − cot θ)²
= (1 / sin θ − cos θ / sin θ)²
= [(1 − cos θ) / sin θ]²

Since sin² θ = 1 − cos² θ = (1 − cos θ)(1 + cos θ):

LHS = (1 − cos θ)² / [(1 − cos θ)(1 + cos θ)]

Cancel (1 − cos θ) from numerator and denominator:

= (1 − cos θ) / (1 + cos θ)

RHS: (1 − cos θ) / (1 + cos θ)

Hence proved.

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Q4: (ii) cos A / (1 + sin A) + (1 + sin A) / cos A = 2 sec A

Solution:

cos A / (1 + sin A)
= multiply numerator and denominator by (1 − sin A):
= cos A(1 − sin A) / (1 − sin² A)
= cos A(1 − sin A) / cos² A
= (1 − sin A) / cos A

So the left hand side becomes:
(1 − sin A) / cos A + (1 + sin A) / cos A
= [(1 − sin A) + (1 + sin A)] / cos A
= [1 − sin A + 1 + sin A] / cos A
= 2 / cos A
= 2 sec A

Hence proved.

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Q4: (iii) (tan θ) / (1 − cot θ) + (cot θ) / (1 − tan θ) = 1 + sec θ cosec θ

Solution:

LHS:

First term:
(tan θ) / (1 − cot θ)
= (sin θ / cos θ) ÷ (1 − cos θ / sin θ)
= (sin θ / cos θ) ÷ ((sin θ − cos θ) / sin θ)
= (sin θ / cos θ) × (sin θ / (sin θ − cos θ))
= sin² θ / [cos θ (sin θ − cos θ)]

Second term:
(cot θ) / (1 − tan θ)
= (cos θ / sin θ) ÷ (1 − sin θ / cos θ)
= (cos θ / sin θ) ÷ ((cos θ − sin θ) / cos θ)
= (cos θ / sin θ) × (cos θ / (cos θ − sin θ))
= − cos² θ / [sin θ (sin θ − cos θ)]

Adding the two terms:

= sin² θ / [cos θ (sin θ − cos θ)] − cos² θ / [sin θ (sin θ − cos θ)]
= [sin³ θ − cos³ θ] / [sin θ cos θ (sin θ − cos θ)]

Factor the numerator (difference of cubes):
sin³ θ − cos³ θ = (sin θ − cos θ)(sin² θ + sin θ cos θ + cos² θ)

Cancel (sin θ − cos θ) from numerator and denominator:
= [sin² θ + sin θ cos θ + cos² θ] / [sin θ cos θ]

Since sin² θ + cos² θ = 1:
= [1 + sin θ cos θ] / [sin θ cos θ]

Split the terms:
= 1 / (sin θ cos θ) + 1
= cosec θ sec θ + 1

RHS = 1 + sec θ cosec θ

Hence proved.

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Q4: (iv) (1 + sec A) / sec A = sin² A / (1 − cos A)

Solution:

LHS = (1 + sec A) / sec A
= 1 / sec A + sec A / sec A
= cos A + 1

RHS = sin² A / (1 − cos A)
Replace sin² A by (1 − cos² A):
= (1 − cos² A) / (1 − cos A)
= (1 − cos A)(1 + cos A) / (1 − cos A)
Cancel (1 − cos A):
= 1 + cos A

Therefore LHS = RHS = 1 + cos A.

Hence proved.

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Q4: (v) (cos A − sin A + 1) / (cos A + sin A − 1) = cosec A + cot A

Solution:

Start with the left hand side.

LHS = (cos A − sin A + 1) / (cos A + sin A − 1)

Multiply numerator and denominator by (cos A + sin A + 1):

LHS = [ (cos A − sin A + 1)(cos A + sin A + 1) ]
/ [ (cos A + sin A − 1)(cos A + sin A + 1) ]

Denominator simplifies as a difference of squares:
(cos A + sin A − 1)(cos A + sin A + 1) = (cos A + sin A)² − 1²
= (cos² A + sin² A + 2 sin A cos A) − 1
= 1 + 2 sin A cos A − 1
= 2 sin A cos A

Now expand the numerator:
(cos A − sin A + 1)(cos A + sin A + 1)

= cos² A + cos A sin A + cos A
− cos A sin A − sin² A − sin A

• cos A + sin A + 1

Cancel + cos A sin A and − cos A sin A, cancel − sin A and + sin A:

= cos² A − sin² A + 2 cos A + 1

Group and simplify the quadratic terms:
cos² A − sin² A + 1
= cos² A − sin² A + (sin² A + cos² A)
= 2 cos² A

So the numerator becomes:
2 cos² A + 2 cos A = 2 cos A (cos A + 1)

Therefore
LHS = [2 cos A (cos A + 1)] / [2 sin A cos A]
= (cos A + 1) / sin A

Split the fraction:
(cos A + 1) / sin A = cos A / sin A + 1 / sin A
= cot A + cosec A

Hence LHS = cot A + cosec A, which equals the right hand side.

Therefore
(cos A − sin A + 1) / (cos A + sin A − 1) = cosec A + cot A

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Q4: (vi) √[(1 + sin A) / (1 − sin A)] = sec A + tan A

Solution:

Start with the Left Hand Side:

LHS = √[(1 + sin A) / (1 − sin A)]

Multiply numerator and denominator inside the root by (1 + sin A):

LHS = √[ (1 + sin A)(1 + sin A) / (1 − sin A)(1 + sin A) ]

= √[ (1 + sin A)² / (1 − sin² A) ]

But 1 − sin² A = cos² A:

LHS = √[ (1 + sin A)² / cos² A ]

Now take the square root:

= (1 + sin A) / cos A

Split into two fractions:

= 1 / cos A + sin A / cos A

= sec A + tan A

Hence LHS = sec A + tan A = RHS.

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Q4: (vii) (sin θ − 2 sin³ θ) / (2 cos³ θ − cos θ) = tan θ

Solution:

Proof:

Step 1: Factor both numerator and denominator

Numerator: sin θ − 2 sin³ θ
= sin θ (1 − 2 sin² θ)

Denominator: 2 cos³ θ − cos θ
= cos θ (2 cos² θ − 1)

Step 2: Recall trigonometric identity

1 − 2 sin² θ = cos 2θ
2 cos² θ − 1 = cos 2θ

So we have:

Numerator: sin θ (cos 2θ)
Denominator: cos θ (cos 2θ)

Step 3: Cancel common factor

Now cancel out cos 2θ from both numerator and denominator:

[sin θ (cos 2θ)] / [cos θ (cos 2θ)]
= sin θ / cos θ

Step 4: Simplify

sin θ / cos θ = tan θ

Hence proved.

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Q4: (viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

Solution:

Proof:

Step 1: Expand both squares on the left-hand side

(sin A + cosec A)² = sin² A + 2 sin A cosec A + cosec² A

(cos A + sec A)² = cos² A + 2 cos A sec A + sec² A

Step 2: Add the two expanded expressions

Left-hand side becomes:

sin² A + cosec² A + 2 sin A cosec A

• cos² A + sec² A + 2 cos A sec A

Group like terms:

(sin² A + cos² A)

• (cosec² A + sec² A)

• (2 sin A cosec A + 2 cos A sec A)

Step 3: Use standard identities

sin² A + cos² A = 1
sin A × cosec A = 1
cos A × sec A = 1

So:

= 1

• cosec² A + sec² A

• 2(1 + 1)
  = 1 + cosec² A + sec² A + 4

Step 4: Use identities again

cosec² A = 1 + cot² A
sec² A = 1 + tan² A

So:

= 1 + (1 + cot² A) + (1 + tan² A) + 4
= 1 + 1 + cot² A + 1 + tan² A + 4
= 7 + tan² A + cot² A

Hence proved.

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Q4: (ix)

(cosec A − sin A)(sec A − cos A) = 1 / (tan A + cot A)
[Hint: Simplify LHS and RHS separately]

Solution:

LHS:
(cosec A − sin A)(sec A − cos A)
= (1/sin A − sin A)(1/cos A − cos A)
= (1 − sin²A)/sin A × (1 − cos²A)/cos A
= (cos²A / sin A) × (sin²A / cos A)
= (cos²A × sin²A) / (sin A × cos A)
= sin A × cos A

RHS:
1 / (tan A + cot A)
= 1 / (sin A / cos A + cos A / sin A)
= 1 / ((sin²A + cos²A) / (sin A × cos A))
= 1 / (1 / (sin A × cos A))
= sin A × cos A

Since LHS = RHS = sin A × cos A, the identity is proved.

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Q4: (x)

(1 + tan²A) / (1 + cot²A) = (1 − tan A)² / (1 − cot A)² = tan²A

Solution:

LHS1: (1 + tan²A) / (1 + cot²A)
= sec²A / csc²A
= (sec A / csc A)²
= ( (1/cos A) / (1/sin A) )²
= (sin A / cos A)²
= tan²A

LHS2: (1 − tan A)² / (1 − cot A)²
= [ (1 − tan A) / (1 − cot A) ]²

Replace cot A by 1/tan A: (1 − tan A) / (1 − 1/tan A)
= (1 − tan A) / ( (tan A − 1) / tan A )
= (1 − tan A) · (tan A / (tan A − 1))
Note: tan A − 1 = −(1 − tan A), so this equals (1 − tan A) · (tan A / (−(1 – tan A))) = −tan A
Squaring gives (−tan A)² = tan²A

Thus both expressions simplify to tan²A, and the identity is proved.

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