Class 10 Maths Chapter 8 - Introduction to Trigonometry

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📐 Trigonometric Ratios — Quick Reference Table

In a right-angled triangle with angle θ: Opposite side = P (Perpendicular), Adjacent side = B (Base), Hypotenuse = H

Three Identities: sin²θ + cos²θ = 1 | 1 + tan²θ = sec²θ | 1 + cot²θ = cosec²θ

Angle (θ)	0°	30°	45°	60°	90°
sin θ	0	1/2	1/√2	√3/2	1
cos θ	1	√3/2	1/√2	1/2	0
tan θ	0	1/√3	1	√3	Not defined
cosec θ	Not defined	2	√2	2/√3	1
sec θ	1	2/√3	√2	2	Not defined
cot θ	Not defined	√3	1	1/√3	0

📖 Exercise 8.1 — Trigonometric Ratios

Q1. In △ABC, right-angled at B, AB = 24 cm and BC = 7 cm. Find: (i) sin A and cos A (ii) sin C and cos C ▼

Find the Hypotenuse

The triangle is right-angled at B, so AB and BC are the two legs, and AC is the hypotenuse. We use the Pythagoras theorem to find AC.

AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625 So, AC = √625 = 25 cm

Part (i) — Ratios for Angle A

For angle A, the side opposite to A is BC = 7 cm, the side adjacent to A is AB = 24 cm, and the hypotenuse is AC = 25 cm.

sin A = Opposite/Hypotenuse = BC/AC = 7/25 cos A = Adjacent/Hypotenuse = AB/AC = 24/25

Part (ii) — Ratios for Angle C

For angle C, the side opposite to C is AB = 24 cm, and the side adjacent to C is BC = 7 cm. The hypotenuse remains AC = 25 cm.

sin C = Opposite/Hypotenuse = AB/AC = 24/25 cos C = Adjacent/Hypotenuse = BC/AC = 7/25

sin A = 7/25, cos A = 24/25 | sin C = 24/25, cos C = 7/25

Q2. In △ABC, right-angled at B, if sin A = 3/4, find all six trigonometric ratios of angle A. ▼

Identify the sides from sin A

We are given sin A = 3/4. Since sin A = Opposite/Hypotenuse, we can say the opposite side (BC) = 3k and the hypotenuse (AC) = 4k, where k is a common factor.

Find the third side using Pythagoras theorem

The adjacent side (AB) is unknown. Apply the Pythagoras theorem: AB² = AC² − BC²

AB² = (4k)² − (3k)² = 16k² − 9k² = 7k² So, AB = √7 · k

Write all six trigonometric ratios

Now, using BC = 3k (opposite), AB = √7k (adjacent), AC = 4k (hypotenuse):

sin A = 3/4 cosec A = 4/3 cos A = √7/4 sec A = 4/√7 tan A = 3/√7 cot A = √7/3

sin A = 3/4, cos A = √7/4, tan A = 3/√7, cosec A = 4/3, sec A = 4/√7, cot A = √7/3

Q3. In △PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P, and tan P. ▼

Set up a variable

Let QR = x cm. Then, since PR + QR = 25, we get PR = (25 − x) cm.

Apply Pythagoras theorem

The triangle is right-angled at Q, so PQ and QR are the legs and PR is the hypotenuse. Therefore PR² = PQ² + QR².

(25 − x)² = 5² + x² 625 − 50x + x² = 25 + x² 625 − 50x = 25 50x = 600 x = 12

Find all three sides

Now we can find all the sides: QR = 12 cm, PR = 25 − 12 = 13 cm, PQ = 5 cm.

Calculate the required ratios for angle P

For angle P, the opposite side is QR = 12, adjacent side is PQ = 5, and hypotenuse is PR = 13.

sin P = QR/PR = 12/13 cos P = PQ/PR = 5/13 tan P = QR/PQ = 12/5

sin P = 12/13, cos P = 5/13, tan P = 12/5

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

Find cot A and identify the sides

From 15 cot A = 8, we get cot A = 8/15. Since cot A = Adjacent/Opposite, the adjacent side (base) = 8k and the opposite side = 15k.

Find the hypotenuse

We use the Pythagoras theorem to find the hypotenuse.

Hypotenuse² = (8k)² + (15k)² = 64k² + 225k² = 289k² Hypotenuse = √289k² = 17k

Find sin A and sec A

Using Opposite = 15k, Adjacent = 8k, Hypotenuse = 17k:

sin A = Opposite/Hypotenuse = 15k/17k = 15/17 sec A = Hypotenuse/Adjacent = 17k/8k = 17/8

sin A = 15/17 and sec A = 17/8

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

Identify the sides from sec θ

Since sec θ = Hypotenuse/Adjacent = 13/12, we take the hypotenuse = 13k and the adjacent side (base) = 12k.

Find the opposite side

Using the Pythagoras theorem, the opposite side (perpendicular) is:

Opposite² = Hypotenuse² − Adjacent² = (13k)² − (12k)² = 169k² − 144k² = 25k² Opposite = 5k

Write all five remaining ratios

Now with Opposite = 5k, Adjacent = 12k, Hypotenuse = 13k:

sin θ = 5/13 cos θ = 12/13 tan θ = 5/12 cosec θ = 13/5 cot θ = 12/5

sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5

Q6. If ∠A and ∠B are acute angles such that sin A = cos B, then prove that A + B = 90°. ▼

Write the given condition

We are given that sin A = cos B. Our goal is to show that A + B = 90°.

Use the complementary angle identity

We know that cos B = sin(90° − B). So we can write the given condition as:

sin A = sin(90° − B)

Compare both sides

Since sine is a one-to-one function for acute angles, if sin A = sin(90° − B), then the angles must be equal.

A = 90° − B ∴ A + B = 90° (Proved ✓)

Since sin A = sin(90° − B), we get A = 90° − B, i.e., A + B = 90°. Proved.

Q7. If cot θ = 7/8, evaluate: (i) (1 + sin θ)(1 − sin θ) / [(1 + cos θ)(1 − cos θ)] (ii) cot²θ ▼

Find all sides from cot θ = 7/8

Since cot θ = Adjacent/Opposite = 7/8, we take adjacent = 7k and opposite = 8k. The hypotenuse is found by Pythagoras theorem.

Hypotenuse = √(7k)² + (8k)² = √(49 + 64)k² = √113 · k

Simplify Part (i) using an algebraic identity

Notice that (1 + sin θ)(1 − sin θ) = 1 − sin²θ, and (1 + cos θ)(1 − cos θ) = 1 − cos²θ. Using sin²θ + cos²θ = 1, we know 1 − sin²θ = cos²θ and 1 − cos²θ = sin²θ.

Expression = cos²θ / sin²θ = cot²θ = (7/8)² = 49/64

Part (ii) — Find cot²θ directly

cot θ = 7/8, so cot²θ is simply the square of 7/8.

cot²θ = (7/8)² = 49/64

Part (i) = 49/64 Part (ii) = 49/64

Q8. If 3 cot A = 4, check whether (1 − tan²A)/(1 + tan²A) = cos²A − sin²A or not. ▼

Find the sides from cot A

From 3 cot A = 4, we get cot A = 4/3. So adjacent = 4k, opposite = 3k, and hypotenuse = √(16 + 9)k = 5k. Therefore:

sin A = 3/5, cos A = 4/5, tan A = 3/4

Calculate the Left Hand Side (LHS)

We substitute tan A = 3/4 into the LHS expression.

LHS = (1 − (3/4)²) / (1 + (3/4)²) = (1 − 9/16) / (1 + 9/16) = (7/16) / (25/16) = 7/25

Calculate the Right Hand Side (RHS)

We substitute the values of sin A and cos A into the RHS.

RHS = cos²A − sin²A = (4/5)² − (3/5)² = 16/25 − 9/25 = 7/25

Compare LHS and RHS

Since LHS = 7/25 and RHS = 7/25, both sides are equal.

Yes, LHS = RHS = 7/25. The equation holds true.

Q9. In △ABC right-angled at B, tan A = 1/√3. Find the value of: (i) sin A · cos C + cos A · sin C (ii) cos A · cos C − sin A · sin C ▼

Find the angles A and C

We know tan 30° = 1/√3. So angle A = 30°. Since the triangle is right-angled at B, angle B = 90°, and the angles must add up to 180°. Therefore, angle C = 180° − 90° − 30° = 60°.

Evaluate Part (i)

Substituting A = 30° and C = 60°:

sin 30° · cos 60° + cos 30° · sin 60° = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 4/4 = 1

Notice this equals sin(30° + 60°) = sin 90° = 1. This is the sine addition formula!

Evaluate Part (ii)

Substituting A = 30° and C = 60°:

cos 30° · cos 60° − sin 30° · sin 60° = (√3/2)(1/2) − (1/2)(√3/2) = √3/4 − √3/4 = 0

Part (i) = 1 Part (ii) = 0

Q10. In △PQR right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine the angles ∠QPR and ∠PRQ. ▼

Identify the sides relative to angle P

For angle P (∠QPR): the side PQ = 3 cm is the adjacent side, PR = 6 cm is the hypotenuse, and QR is the opposite side.

Use cos P to find angle P

Since we know the adjacent side and hypotenuse, we use cos P.

cos P = PQ/PR = 3/6 = 1/2 We know that cos 60° = 1/2 ∴ ∠QPR (angle P) = 60°

Find angle R using the angle sum property

Since the sum of all angles in a triangle is 180°, and angle Q = 90°, angle P = 60°:

∠PRQ = 180° − 90° − 60° = 30°

∠QPR = 60° and ∠PRQ = 30°

Q11. State True or False and justify: (i) tan A is always less than 1 (ii) sec A = 12/5 is possible (iii) cos A = cosecant of A (iv) cot A = product of cot and A (v) sin θ = 4/3 for some angle θ ▼

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

✗

FALSE. tan A = Opposite/Adjacent. When the opposite side is longer than the adjacent side, tan A becomes greater than 1. For example, tan 60° = √3 ≈ 1.73, which is clearly greater than 1.

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

✓

TRUE. sec A = Hypotenuse/Adjacent. The hypotenuse is always greater than the adjacent side, so sec A is always ≥ 1. Since 12/5 = 2.4 > 1, this is perfectly valid. We can form a right triangle with adjacent = 5, hypotenuse = 12, and opposite = √(144−25) = √119.

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

✗

FALSE. "cos A" is the abbreviation for the cosine of angle A, not cosecant. The cosecant of angle A is written as "cosec A" (or "csc A"). These are two completely different trigonometric ratios.

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

✗

FALSE. "cot A" is a single trigonometric function (cotangent) applied to the angle A. It is not a product of two separate quantities "cot" and "A". The word "cot" has no meaning by itself — it only makes sense when applied to an angle.

(v) "sin θ = 4/3 for some angle θ"

✗

FALSE. sin θ = Opposite/Hypotenuse. Since the hypotenuse is always the longest side in a right triangle, the opposite side can never be longer than the hypotenuse. Therefore, sin θ can never be greater than 1. Since 4/3 ≈ 1.33 > 1, this is not possible.

📖 Exercise 8.2 — Trigonometric Ratios of Specific Angles

Standard values to remember: sin 30° = 1/2, cos 30° = √3/2, sin 45° = cos 45° = 1/√2, sin 60° = √3/2, cos 60° = 1/2, tan 30° = 1/√3, tan 45° = 1, tan 60° = √3

Q1. Evaluate: (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan²45° + cos²30° − sin²60° ▼

Part (i): sin 60° cos 30° + sin 30° cos 60°

Substitute the standard values

We substitute the known values of sin 60°, cos 30°, sin 30°, and cos 60°.

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

Observation: This result equals sin(60° + 30°) = sin 90° = 1, which demonstrates the sine addition formula.

Part (ii): 2 tan²45° + cos²30° − sin²60°

Substitute the standard values

We know tan 45° = 1, cos 30° = √3/2, and sin 60° = √3/2. We substitute these values.

= 2(1)² + (√3/2)² − (√3/2)² = 2(1) + 3/4 − 3/4 = 2 + 0 = 2

Part (i) = 1 Part (ii) = 2

Q2. Choose the correct option and justify: (i) 2tan30°/(1+tan²30°) (ii) (1−tan²45°)/(1+tan²45°) (iii) sin 2A = 2 sin A is true when A = ? (iv) 2tan30°/(1−tan²30°) ▼

Part (i): 2 tan 30° / (1 + tan²30°)

Substitute tan 30° = 1/√3

= 2(1/√3) / (1 + (1/√3)²) = (2/√3) / (1 + 1/3) = (2/√3) / (4/3) = (2/√3) × (3/4) = 6 / (4√3) = 3/(2√3) = (3√3)/(2×3) = √3/2 = sin 60°

Answer: sin 60°

Part (ii): (1 − tan²45°) / (1 + tan²45°)

Substitute tan 45° = 1

= (1 − 1²) / (1 + 1²) = 0 / 2 = 0 = cos 90°

Answer: cos 90°

Part (iii): sin 2A = 2 sin A is true when A = ?

Expand sin 2A using the double angle formula

We know that sin 2A = 2 sin A cos A. So the equation becomes:

2 sin A cos A = 2 sin A Dividing both sides by 2 sin A (assuming sin A ≠ 0): cos A = 1 → A = 0°

Answer: A = 0°

Part (iv): 2 tan 30° / (1 − tan²30°)

Substitute tan 30° = 1/√3

= 2(1/√3) / (1 − (1/√3)²) = (2/√3) / (1 − 1/3) = (2/√3) / (2/3) = (2/√3) × (3/2) = 3/√3 = √3 = tan 60°

Answer: tan 60°

Q3. If tan(A + B) = √3 and tan(A − B) = 1/√3, where 0° < A + B ≤ 90° and A > B, find A and B. ▼

Find the value of (A + B)

We know that tan 60° = √3. Since tan(A + B) = √3, the angle (A + B) must equal 60°.

A + B = 60° ... (equation 1)

Find the value of (A − B)

We know that tan 30° = 1/√3. Since tan(A − B) = 1/√3, the angle (A − B) must equal 30°.

A − B = 30° ... (equation 2)

Solve the two equations

Add equation 1 and equation 2 to find A. Then subtract to find B.

Adding: 2A = 90° → A = 45° Subtracting: 2B = 30° → B = 15°

A = 45° and B = 15°

Q4. State True or False and justify: (i) sin(A+B) = sinA + sinB (ii) sin θ increases as θ increases (iii) cos θ increases as θ increases (iv) sin θ = cos θ for all θ (v) cot A is undefined for A = 0° ▼

(i) sin(A + B) = sin A + sin B

✗

FALSE. Let us check with A = 30°, B = 60°. Then sin(30° + 60°) = sin 90° = 1. But sin 30° + sin 60° = 1/2 + √3/2 = (1 + √3)/2 ≈ 1.37. Since 1 ≠ 1.37, the statement is false. Trigonometric functions do not distribute over addition like that.

(ii) The value of sin θ increases as θ increases (0° to 90°)

✓

TRUE. As the angle increases from 0° to 90°, the sine value increases steadily: sin 0° = 0, sin 30° = 0.5, sin 45° ≈ 0.707, sin 60° ≈ 0.866, sin 90° = 1. So yes, sin θ increases as θ increases in this range.

(iii) The value of cos θ increases as θ increases (0° to 90°)

✗

FALSE. It is the opposite — cos θ decreases as θ increases. We can verify: cos 0° = 1, cos 30° = √3/2 ≈ 0.866, cos 45° ≈ 0.707, cos 60° = 0.5, cos 90° = 0. The values go down, not up.

(iv) sin θ = cos θ for all values of θ

✗

FALSE. sin θ = cos θ is only true for one specific angle, which is θ = 45°, where both equal 1/√2. For all other angles they are different. For example, sin 30° = 1/2 but cos 30° = √3/2.

(v) cot A is not defined for A = 0°

✓

TRUE. cot A = cos A / sin A. When A = 0°, sin 0° = 0. So cot 0° = cos 0° / sin 0° = 1/0, which is undefined (division by zero is not allowed in mathematics).

📖 Exercise 8.3 — Trigonometric Ratios of Complementary Angles

Key Rule: Two angles are complementary if they add up to 90°. For complementary angles:

sin θ = cos(90° − θ) | cos θ = sin(90° − θ) | tan θ = cot(90° − θ) | sec θ = cosec(90° − θ)

Q1. Evaluate: (i) sin 18°/cos 72° (ii) tan 26°/cot 64° (iii) cos 48° − sin 42° (iv) cosec 31° − sec 59° ▼

Part (i): sin 18° / cos 72°

Notice that 18° + 72° = 90°, so these are complementary angles. Using the identity cos 72° = cos(90° − 18°) = sin 18°:

sin 18° / cos 72° = sin 18° / sin 18° = 1

Part (i) = 1

Part (ii): tan 26° / cot 64°

Since 26° + 64° = 90°, they are complementary. Using cot 64° = cot(90° − 26°) = tan 26°:

tan 26° / cot 64° = tan 26° / tan 26° = 1

Part (ii) = 1

Part (iii): cos 48° − sin 42°

Since 48° + 42° = 90°, using sin 42° = sin(90° − 48°) = cos 48°:

cos 48° − sin 42° = cos 48° − cos 48° = 0

Part (iii) = 0

Part (iv): cosec 31° − sec 59°

Since 31° + 59° = 90°, using sec 59° = sec(90° − 31°) = cosec 31°:

cosec 31° − sec 59° = cosec 31° − cosec 31° = 0

Part (iv) = 0

Q2. Show that: (i) tan 48° tan 23° tan 42° tan 67° = 1 (ii) cos 38° cos 52° − sin 38° sin 52° = 0 ▼

Part (i): tan 48° · tan 23° · tan 42° · tan 67° = 1

Identify complementary pairs

Notice that 48° + 42° = 90° and 23° + 67° = 90°. So 48° and 42° are complementary, and 23° and 67° are complementary.

Convert using complementary angle identity

Using tan θ = cot(90° − θ): tan 48° = cot 42° and tan 23° = cot 67°. Now substitute:

= cot 42° · cot 67° · tan 42° · tan 67° = (tan 42° · cot 42°) × (tan 67° · cot 67°) = 1 × 1 = 1 ✓

Key idea: tan θ × cot θ = tan θ × (1/tan θ) = 1 always.

Part (ii): cos 38° cos 52° − sin 38° sin 52° = 0

Use complementary angle relations

Since 38° + 52° = 90°, we have cos 38° = sin 52° and sin 38° = cos 52°. Substitute these:

= sin 52° · cos 52° − cos 52° · sin 52° = sin 52° cos 52° − sin 52° cos 52° = 0 ✓

Q3. If tan 2A = cot(A − 18°), where 2A is an acute angle, find the value of A. ▼

Convert tan to cot using complementary angle identity

We know that tan θ = cot(90° − θ). So we rewrite tan 2A as:

tan 2A = cot(90° − 2A)

Set the two cot expressions equal

Since the left side now equals cot(90° − 2A) and the right side is cot(A − 18°), and both cot expressions are equal:

90° − 2A = A − 18°

Solve for A

90° + 18° = A + 2A 108° = 3A A = 108°/3 = 36°

A = 36°

Q4. If tan A = cot B, prove that A + B = 90°. ▼

Use the complementary angle identity for cot

We know that cot B = tan(90° − B). So we substitute this in the given equation:

tan A = tan(90° − B)

Compare both angles

Since both sides have the tangent function and the angles must be equal (for acute angles where tan is one-to-one):

A = 90° − B ∴ A + B = 90° (Proved ✓)

Since tan A = tan(90° − B), we get A = 90° − B, so A + B = 90°. Proved.

Q5. If sec 4A = cosec(A − 20°), where 4A is an acute angle, find the value of A. ▼

Convert sec to cosec using complementary angle identity

We know that sec θ = cosec(90° − θ). So we rewrite sec 4A as:

sec 4A = cosec(90° − 4A)

Set both cosec expressions equal and solve

Now comparing cosec(90° − 4A) = cosec(A − 20°), the angles inside must be equal:

90° − 4A = A − 20° 90° + 20° = A + 4A 110° = 5A A = 22°

A = 22°

Q6. If A, B, C are the interior angles of a triangle ABC, prove that sin((B+C)/2) = cos(A/2). ▼

Use the angle sum property of a triangle

We know that in any triangle, the sum of all interior angles is 180°. So A + B + C = 180°, which means B + C = 180° − A.

Divide both sides by 2

(B + C)/2 = (180° − A)/2 = 90° − A/2

Apply complementary angle identity

Now we take sin of both sides and use the identity sin(90° − θ) = cos θ:

sin((B + C)/2) = sin(90° − A/2) = cos(A/2) (Proved ✓)

Using angle sum property and complementary identity: sin((B+C)/2) = cos(A/2). Proved.

Q7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°. ▼

Convert sin 67° to an angle less than 45°

Since 67° > 45°, we write 67° = 90° − 23°, and use the identity sin(90° − θ) = cos θ:

sin 67° = sin(90° − 23°) = cos 23°

Convert cos 75° to an angle less than 45°

Since 75° > 45°, we write 75° = 90° − 15°, and use the identity cos(90° − θ) = sin θ:

cos 75° = cos(90° − 15°) = sin 15°

Write the final expression

sin 67° + cos 75° = cos 23° + sin 15°

📖 Exercise 8.4 — Trigonometric Identities

The three fundamental identities to use in proofs:

① sin²θ + cos²θ = 1 ② 1 + tan²θ = sec²θ ③ 1 + cot²θ = cosec²θ

Tip: To prove LHS = RHS, always simplify one side to match the other. Start from the more complex side.

Q1. Prove: (i) (1 − cos²θ) cosec²θ = 1 (ii) (1 + tan²θ)(1 + sin θ)(1 − sin θ) = 1 ▼

Part (i): (1 − cos²θ) cosec²θ = 1

Simplify (1 − cos²θ)

From the identity sin²θ + cos²θ = 1, we get 1 − cos²θ = sin²θ. We substitute this:

LHS = sin²θ × cosec²θ

Use the definition cosec θ = 1/sin θ

= sin²θ × (1/sin θ)² = sin²θ × (1/sin²θ) = 1 = RHS ✓

LHS = RHS = 1. Proved.

Part (ii): (1 + tan²θ)(1 + sin θ)(1 − sin θ) = 1

Simplify (1 + tan²θ)

Using the identity 1 + tan²θ = sec²θ:

LHS = sec²θ × (1 + sin θ)(1 − sin θ)

Simplify (1 + sin θ)(1 − sin θ)

Using the algebraic identity (a + b)(a − b) = a² − b²:

= sec²θ × (1 − sin²θ) = sec²θ × cos²θ

Substitute sec θ = 1/cos θ

= (1/cos²θ) × cos²θ = 1 = RHS ✓

LHS = RHS = 1. Proved.

Q2. Prove: cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A ▼

Take LCM of the two fractions

The LCM of (1 + sin A) and cos A is cos A(1 + sin A). We combine the two fractions:

LHS = [cos²A + (1 + sin A)²] / [cos A(1 + sin A)]

Expand the numerator

Expand (1 + sin A)² = 1 + 2 sin A + sin²A. Then add cos²A:

Numerator = cos²A + 1 + 2 sin A + sin²A = (sin²A + cos²A) + 1 + 2 sin A = 1 + 1 + 2 sin A [using sin²A + cos²A = 1] = 2 + 2 sin A = 2(1 + sin A)

Simplify the full expression

LHS = 2(1 + sin A) / [cos A(1 + sin A)] = 2 / cos A = 2 sec A = RHS ✓

LHS = RHS = 2 sec A. Proved.

Q3. Prove: tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ ▼

Replace tan and cot with sin and cos

Write tan θ = sin θ/cos θ and cot θ = cos θ/sin θ to work in a common language.

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

Simplify each denominator

First term denominator: 1 − cos θ/sin θ = (sin θ − cos θ)/sin θ Second term denominator: 1 − sin θ/cos θ = (cos θ − sin θ)/cos θ

Simplify each fraction

First term = (sin θ/cos θ) × (sin θ/(sin θ − cos θ)) = sin²θ / [cos θ(sin θ − cos θ)] Second term = (cos θ/sin θ) × (cos θ/(cos θ − sin θ)) = cos²θ / [sin θ(cos θ − sin θ)] = −cos²θ / [sin θ(sin θ − cos θ)]

Combine with common denominator sin θ cos θ(sin θ − cos θ)

LHS = [sin³θ − cos³θ] / [sin θ cos θ(sin θ − cos θ)]

Factor the numerator using a³ − b³ = (a−b)(a²+ab+b²)

sin³θ − cos³θ = (sin θ − cos θ)(sin²θ + sin θ cos θ + cos²θ) = (sin θ − cos θ)(1 + sin θ cos θ) [since sin²θ+cos²θ=1]

Cancel (sin θ − cos θ) and simplify

LHS = (1 + sin θ cos θ) / (sin θ cos θ) = 1/(sin θ cos θ) + 1 = cosec θ sec θ + 1 = 1 + sec θ cosec θ = RHS ✓

LHS = RHS = 1 + sec θ cosec θ. Proved.

Q4. Prove: (1 + sec A)/sec A = sin²A/(1 − cos A) ▼

Simplify the LHS

We write sec A = 1/cos A and simplify the LHS step by step.

LHS = (1 + 1/cos A) / (1/cos A) = [(cos A + 1)/cos A] / (1/cos A) = (cos A + 1)/cos A × cos A = cos A + 1 = 1 + cos A

Simplify the RHS

In the RHS, we use the identity sin²A = 1 − cos²A = (1 − cos A)(1 + cos A).

RHS = sin²A / (1 − cos A) = (1 − cos²A) / (1 − cos A) = (1 − cos A)(1 + cos A) / (1 − cos A) = 1 + cos A

Compare LHS and RHS

Both the LHS and RHS simplify to (1 + cos A).

LHS = RHS = 1 + cos A. Proved.

Q5. Prove: (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A (using identity cosec²A = 1 + cot²A) ▼

Divide numerator and denominator by sin A

Dividing every term by sin A converts the expression into cosec A and cot A terms.

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

Replace "1" using the identity cosec²A − cot²A = 1

We know that cosec²A − cot²A = 1, so 1 = cosec²A − cot²A = (cosec A − cot A)(cosec A + cot A). We substitute this "1" in the denominator only:

Denominator = cot A − cosec A + (cosec A − cot A)(cosec A + cot A) = (cosec A − cot A)[−1 + (cosec A + cot A)] = (cosec A − cot A)(cosec A + cot A − 1)

Cancel common factors

LHS = (cosec A + cot A − 1) / [(cosec A − cot A)(cosec A + cot A − 1)] = 1 / (cosec A − cot A)

Rationalise by multiplying by (cosec A + cot A)

= (cosec A + cot A) / [(cosec A − cot A)(cosec A + cot A)] = (cosec A + cot A) / (cosec²A − cot²A) = (cosec A + cot A) / 1 = cosec A + cot A = RHS ✓

LHS = RHS = cosec A + cot A. Proved.

Q6. Prove: √[(1 + sin A)/(1 − sin A)] = sec A + tan A ▼

Rationalise by multiplying inside the square root by (1 + sin A)

We multiply both numerator and denominator inside the square root by (1 + sin A) to eliminate the square root from the denominator.

LHS = √[(1 + sin A)(1 + sin A) / ((1 − sin A)(1 + sin A))] = √[(1 + sin A)² / (1 − sin²A)]

Use the identity 1 − sin²A = cos²A

= √[(1 + sin A)² / cos²A] = (1 + sin A) / cos A [since both are positive for acute A]

Split into two separate fractions

= 1/cos A + sin A/cos A = sec A + tan A = RHS ✓

LHS = RHS = sec A + tan A. Proved.

Q7. Prove: (sin θ − 2sin³θ) / (2cos³θ − cos θ) = tan θ ▼

Factor out common terms from numerator and denominator

Take sin θ common from the numerator, and cos θ common from the denominator.

LHS = sin θ(1 − 2sin²θ) / cos θ(2cos²θ − 1)

Replace cos²θ in the denominator using sin²θ + cos²θ = 1

We know cos²θ = 1 − sin²θ. Substitute this in the denominator bracket:

2cos²θ − 1 = 2(1 − sin²θ) − 1 = 2 − 2sin²θ − 1 = 1 − 2sin²θ

Notice that both brackets are identical — cancel them

LHS = sin θ(1 − 2sin²θ) / cos θ(1 − 2sin²θ) = sin θ / cos θ = tan θ = RHS ✓

LHS = RHS = tan θ. Proved.

Q8. Prove: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A ▼

Expand both squared brackets using (a + b)² = a² + 2ab + b²

LHS = sin²A + 2·sinA·cosecA + cosec²A + cos²A + 2·cosA·secA + sec²A

Simplify the middle terms using sin A × cosec A = 1 and cos A × sec A = 1

Since cosec A = 1/sin A, we have sin A × cosec A = 1. Similarly cos A × sec A = 1.

= sin²A + 2(1) + cosec²A + cos²A + 2(1) + sec²A = (sin²A + cos²A) + cosec²A + sec²A + 4

Replace using known identities

Use: sin²A + cos²A = 1, sec²A = 1 + tan²A, and cosec²A = 1 + cot²A.

= 1 + (1 + cot²A) + (1 + tan²A) + 4 = 1 + 1 + cot²A + 1 + tan²A + 4 = 7 + tan²A + cot²A = RHS ✓

LHS = RHS = 7 + tan²A + cot²A. Proved.

Q9. Prove: (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A) ▼

Simplify the LHS — write cosec A and sec A as fractions

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

Simplify each bracket by taking LCM

First bracket: (1 − sin²A)/sin A = cos²A/sin A [using 1−sin²A = cos²A] Second bracket: (1 − cos²A)/cos A = sin²A/cos A [using 1−cos²A = sin²A]

Multiply the two simplified brackets

LHS = (cos²A/sin A) × (sin²A/cos A) = cos²A × sin²A / (sin A × cos A) = sin A cos A

Simplify the RHS

Write tan A = sin A/cos A and cot A = cos A/sin 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)) [since sin²A+cos²A = 1] = sin A cos A

Compare LHS and RHS

Both LHS and RHS equal sin A cos A.

LHS = RHS = sin A cos A. Proved.

Q10. Prove: (1 + tan²A)/(1 + cot²A) = ((1 − tan A)/(1 − cot A))² = tan²A ▼

Part 1: Prove (1 + tan²A)/(1 + cot²A) = tan²A

Replace (1 + tan²A) and (1 + cot²A) using identities

Use 1 + tan²A = sec²A and 1 + cot²A = cosec²A:

= sec²A / cosec²A = (1/cos²A) / (1/sin²A) = sin²A / cos²A = tan²A ✓

Part 2: Prove ((1 − tan A)/(1 − cot A))² = tan²A

Replace cot A with 1/tan A

= ((1 − tan A) / (1 − 1/tan A))²

Simplify the denominator by taking LCM

= ((1 − tan A) / ((tan A − 1)/tan A))² = ((1 − tan A) × tan A / (tan A − 1))² = (−tan A × (tan A − 1) / (tan A − 1))² [since (1−tanA) = −(tanA−1)] = (−tan A)² = tan²A ✓

Both expressions equal tan²A. Proved.

✓ Chapter 7 - Coordinate Geometry Completed!

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