📐 Class 10 NCERT Mathematics

Chapter 8: Introduction to Trigonometry - Complete Study Guide

📝 Important Notes & Formulas

📐 Basic Trigonometric Ratios

In a right triangle ABC, right-angled at B:

sin A = PerpendicularHypotenuse = BCAC

cos A = BaseHypotenuse = ABAC

tan A = PerpendicularBase = BCAB

🔄 Reciprocal Relations

  • cosec θ = 1sin θ
  • sec θ = 1cos θ
  • cot θ = 1tan θ

➗ Quotient Relations

tan θ = sin θcos θ

cot θ = cos θsin θ

❤️ Pythagorean Identities

sin²θ + cos²θ = 1

1 + tan²θ = sec²θ

1 + cot²θ = cosec²θ

📊 Standard Angle Values

∠0°30°45°60°90°
sin0121√2√321
cos1√321√2120
tan01√31√3∞

🔁 Complementary Angles

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

📖 Exercise 8.1 - Trigonometric Ratios

1
In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C
A sin A = 7/25, cos A = 24/25; sin C = 24/25, cos C = 7/25
B sin A = 24/25, cos A = 7/25; sin C = 7/25, cos C = 24/25
C sin A = 7/24, cos A = 24/7; sin C = 1/25, cos C = 24/25
D sin A = 25/7, cos A = 25/24; sin C = 25/24, cos C = 25/7

✅ Solution:

Step 1: Find Hypotenuse AC using Pythagoras theorem
AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625
AC = √625 = 25 cm
Step 2: For angle A (Perpendicular = BC = 7, Base = AB = 24)
sin A = P/H = 7/25, cos A = B/H = 24/25
Step 3: For angle C (Perpendicular = AB = 24, Base = BC = 7)
sin C = P/H = 24/25, cos C = B/H = 7/25
2
In Fig. 8.13, find tan P – cot R. (Given: PQ = 12 cm, PR = 13 cm, right-angled at Q)
A 119/156
B 156/119
C 0
D 1

✅ Solution:

Step 1: Find QR using Pythagoras theorem
PR² = PQ² + QR² → 13² = 12² + QR² → 169 = 144 + QR²
QR² = 25 → QR = 5 cm
Step 2: Calculate tan P and cot R
tan P = QR/PQ = 5/12
cot R = QR/PQ = 5/12 (Base/Perpendicular for angle R)
Step 3: tan P – cot R = 5/12 – 5/12 = 0
3
If sin A = 3/4, calculate cos A and tan A.
A cos A = 4/5, tan A = 3/5
B cos A = √7/4, tan A = 3/√7
C cos A = √7/3, tan A = 4/3
D cos A = 1/4, tan A = 3

✅ Solution:

Step 1: Use Pythagorean identity sin²A + cos²A = 1
(3/4)² + cos²A = 1 → 9/16 + cos²A = 1
cos²A = 1 – 9/16 = 7/16
Step 2: cos A = √(7/16) = √7/4
Step 3: tan A = sin A / cos A = (3/4) / (√7/4) = 3/√7
4
Given 15 cot A = 8, find sin A and sec A.
A sin A = 15/17, sec A = 17/8
B sin A = 8/17, sec A = 17/15
C sin A = 15/8, sec A = 8/17
D sin A = 17/15, sec A = 15/8

✅ Solution:

Step 1: Find cot A
15 cot A = 8 → cot A = 8/15
Step 2: Construct right triangle (Base = 8, Perpendicular = 15)
Hypotenuse² = 8² + 15² = 64 + 225 = 289 → H = 17
Step 3: sin A = P/H = 15/17, sec A = H/B = 17/8
5
Given sec θ = 13/12, calculate all other trigonometric ratios.
A sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cot θ = 12/5, cosec θ = 13/5
B sin θ = 12/13, cos θ = 5/13, tan θ = 12/5, cot θ = 5/12, cosec θ = 13/12
C sin θ = 13/5, cos θ = 12/13, tan θ = 5/12, cot θ = 12/5, cosec θ = 5/13
D sin θ = 5/12, cos θ = 13/12, tan θ = 5/13, cot θ = 13/5, cosec θ = 12/5

✅ Solution:

Step 1: Find cos θ
cos θ = 1/sec θ = 12/13
Step 2: Find sin θ using identity sin²θ + cos²θ = 1
sin²θ = 1 – (12/13)² = 1 – 144/169 = 25/169 → sin θ = 5/13
Step 3: tan θ = sin θ/cos θ = 5/12, cot θ = 12/5, cosec θ = 13/5
6
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
A Construct triangles and use Pythagoras to show sides are proportional, hence angles equal
B cos A = cos B implies A = 90° - B always
C This is only true when A = B = 45°
D The statement is false; cos A = cos B does not imply A = B

✅ Solution:

Step 1: Consider right triangles with angles A and B
cos A = Adjacent/Hypotenuse = AC/AB
cos B = Adjacent/Hypotenuse = BC/AB (in different triangle)
Step 2: Given cos A = cos B, we have AC/AB = PQ/PR (say)
Step 3: This means the ratio of sides is equal, so by SSS similarity, triangles are similar
Step 4: Therefore, corresponding angles are equal: ∠A = ∠B
7
If cot θ = 7/8, evaluate: (i) [(1+sin θ)(1–sin θ)]/[(1+cos θ)(1–cos θ)] (ii) cot²θ
A (i) 49/64 (ii) 49/64
B (i) 64/49 (ii) 64/49
C (i) 1 (ii) 49/64
D (i) 49/113 (ii) 49/64

✅ Solution:

Part (i):
Numerator: (1 + sin θ)(1 – sin θ) = 1 – sin²θ = cos²θ
Denominator: (1 + cos θ)(1 – cos θ) = 1 – cos²θ = sin²θ
So, expression = cos²θ/sin²θ = cot²θ = (7/8)² = 49/64
Part (ii): cot²θ = (7/8)² = 49/64
8
If 3 cot A = 4, check whether (1 – tan²A)/(1 + tan²A) = cos²A – sin²A or not.
A Yes, both sides equal 7/25
B No, LHS = 7/25 but RHS = 24/25
C No, LHS = 24/25 but RHS = 7/25
D Cannot be determined

✅ Solution:

Step 1: 3 cot A = 4 → cot A = 4/3 → tan A = 3/4
Step 2: LHS = (1 – tan²A)/(1 + tan²A) = (1 – 9/16)/(1 + 9/16) = (7/16)/(25/16) = 7/25
Step 3: For RHS, from tan A = 3/4, we get sin A = 3/5, cos A = 4/5
cos²A – sin²A = (16/25) – (9/25) = 7/25
Conclusion: LHS = RHS = 7/25 ✓ The identity holds.
9
In triangle ABC, right-angled at B, if tan A = 1/√3, find: (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C
A (i) 1 (ii) 0
B (i) 0 (ii) 1
C (i) √3/2 (ii) 1/2
D (i) 1/2 (ii) √3/2

✅ Solution:

Step 1: tan A = 1/√3 → A = 30°
Since A + C = 90° (right triangle), C = 60°
Step 2: sin 30° = 1/2, cos 30° = √3/2, sin 60° = √3/2, cos 60° = 1/2
Part (i): sin A cos C + cos A sin C = sin(A+C) = sin 90° = 1
= (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1
Part (ii): cos A cos C – sin A sin C = cos(A+C) = cos 90° = 0
= (√3/2)(1/2) – (1/2)(√3/2) = √3/4 – √3/4 = 0
10
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.
A sin P = 12/13, cos P = 5/13, tan P = 12/5
B sin P = 5/13, cos P = 12/13, tan P = 5/12
C sin P = 15/17, cos P = 8/17, tan P = 15/8
D sin P = 8/17, cos P = 15/17, tan P = 8/15

✅ Solution:

Step 1: Let QR = x, then PR = 25 – x
By Pythagoras: PR² = PQ² + QR²
(25 – x)² = 5² + x² → 625 – 50x + x² = 25 + x²
600 = 50x → x = 12
Step 2: QR = 12 cm, PR = 25 – 12 = 13 cm
Step 3: For angle P: Perpendicular = QR = 12, Base = PQ = 5, Hypotenuse = PR = 13
sin P = 12/13, cos P = 5/13, tan P = 12/5
11
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
A True
B False

✅ Solution:

False. tan A can be greater than 1. For example, tan 60° = √3 ≈ 1.732 > 1.
(ii) sec A = 12/5 for some value of angle A.
A True
B False

✅ Solution:

True. sec A = Hypotenuse/Base. If we construct a right triangle with Hypotenuse = 12 and Base = 5, then Perpendicular = √(144-25) = √119. So such an angle A exists.
(iii) cos A is the abbreviation used for the cosecant of angle A.
A True
B False

✅ Solution:

False. cos A is the abbreviation for cosine of angle A. Cosecant is abbreviated as cosec A.
(iv) cot A is the product of cot and A.
A True
B False

✅ Solution:

False. cot A is not a product. It is a single trigonometric ratio representing cotangent of angle A.
(v) sin θ = 4/3 for some angle θ.
A True
B False

✅ Solution:

False. sin θ = Perpendicular/Hypotenuse, and Hypotenuse is always the largest side. So sin θ ≤ 1 always. Since 4/3 > 1, this is impossible.

📖 Exercise 8.2 - Trigonometric Ratios of Specific Angles

1
Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
A √3/2
B 1
C 0
D 1/2

✅ Solution:

Step 1: Substitute standard values
sin 60° = √3/2, cos 30° = √3/2, sin 30° = 1/2, cos 60° = 1/2
Step 2: Calculate = (√3/2 × √3/2) + (1/2 × 1/2) = 3/4 + 1/4 = 1
Note: This is sin(60°+30°) = sin 90° = 1
(ii) 2 tan² 45° + cos² 30° – sin² 60°
A 0
B 2
C 1
D 3/2

✅ Solution:

Step 1: Substitute values: tan 45° = 1, cos 30° = √3/2, sin 60° = √3/2
Step 2: = 2(1)² + (√3/2)² – (√3/2)² = 2 + 3/4 – 3/4 = 2
(iii) cos 45°/(sec 30° + cosec 30°)
A √3/(2√2)
B (√3–1)/(2√2) or 1/(√3+1) after rationalization
C √2/4
D √3/4

✅ Solution:

Step 1: cos 45° = 1/√2, sec 30° = 2/√3, cosec 30° = 2
Step 2: Denominator = 2/√3 + 2 = (2 + 2√3)/√3 = 2(1 + √3)/√3
Step 3: Expression = (1/√2) / [2(1+√3)/√3] = √3 / [2√2(1+√3)]
Rationalizing: = √3(√3–1) / [2√2(3–1)] = (3–√3) / (4√2) = (√3–1)/(2√2)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
A 1/2
B (43 – 24√3)/11 after full simplification
C 0
D 1

✅ Solution:

Step 1: Numerator = 1/2 + 1 – 2/√3 = 3/2 – 2/√3 = (3√3 – 4)/(2√3)
Step 2: Denominator = 2/√3 + 1/2 + 1 = 2/√3 + 3/2 = (4 + 3√3)/(2√3)
Step 3: Expression = (3√3 – 4)/(4 + 3√3)
Rationalize: multiply by (4 – 3√3)/(4 – 3√3)
= (3√3 – 4)(4 – 3√3) / (16 – 27) = (12√3 – 27 – 16 + 12√3) / (–11)
= (24√3 – 43) / (–11) = (43 – 24√3)/11
(v) (5 cos² 60° + 4 sec² 30° – tan² 45°)/(sin² 30° + cos² 30°)
A 67/12
B 67/12
C 5
D 13/3

✅ Solution:

Step 1: Numerator = 5(1/2)² + 4(2/√3)² – (1)² = 5/4 + 4(4/3) – 1 = 5/4 + 16/3 – 1
= (15 + 64 – 12)/12 = 67/12
Step 2: Denominator = sin²30° + cos²30° = 1 (identity)
Answer: 67/12
2
Choose the correct option and justify your choice:
(i) 2 tan 30°/(1 + tan² 30°) = ?
A sin 60°
B cos 60°
C tan 60°
D sin 30°

✅ Solution:

Step 1: Substitute tan 30° = 1/√3
= [2 × (1/√3)] / [1 + (1/√3)²] = (2/√3) / (1 + 1/3)
Step 2: = (2/√3) / (4/3) = (2/√3) × (3/4) = 6/(4√3) = √3/2
Step 3: √3/2 = sin 60°
(ii) (1 – tan² 45°)/(1 + tan² 45°) = ?
A tan 90°
B 1
C sin 45°
D 0

✅ Solution:

Step 1: Substitute tan 45° = 1
= (1 – 1²) / (1 + 1²) = (1 – 1) / (1 + 1) = 0/2 = 0
(iii) sin 2A = 2 sin A is true when A = ?
A 0°
B 30°
C 45°
D 60°

✅ Solution:

Test each option:
For A = 0°: sin(0°) = 0, 2sin(0°) = 0 ✓
For A = 30°: sin(60°) = √3/2, 2sin(30°) = 1 ✗
For A = 45°: sin(90°) = 1, 2sin(45°) = √2 ✗
For A = 60°: sin(120°) = √3/2, 2sin(60°) = √3 ✗
Only A = 0° satisfies the equation.
(iv) 2 tan 30°/(1 – tan² 30°) = ?
A cos 60°
B sin 60°
C tan 60°
D sin 30°

✅ Solution:

Step 1: Substitute tan 30° = 1/√3
= [2 × (1/√3)] / [1 – (1/√3)²] = (2/√3) / (1 – 1/3)
Step 2: = (2/√3) / (2/3) = (2/√3) × (3/2) = 3/√3 = √3
Step 3: √3 = tan 60°
3
If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.
A A = 45°, B = 15°
B A = 60°, B = 30°
C A = 30°, B = 15°
D A = 45°, B = 30°

✅ Solution:

Step 1: tan (A + B) = √3 = tan 60° → A + B = 60°
Step 2: tan (A – B) = 1/√3 = tan 30° → A – B = 30°
Step 3: Adding: 2A = 90° → A = 45°
Subtracting: 2B = 30° → B = 15°
Verification: tan(45°+15°) = tan 60° = √3 ✓, tan(45°–15°) = tan 30° = 1/√3 ✓
4
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B
A True
B False

✅ Solution:

False. Let A = 30°, B = 60°
LHS = sin(90°) = 1
RHS = sin 30° + sin 60° = 1/2 + √3/2 = (1+√3)/2 ≈ 1.366 ≠ 1
(ii) The value of sin θ increases as θ increases.
A True (for 0° to 90°)
B False

✅ Solution:

True (for acute angles 0° to 90°).
sin 0° = 0, sin 30° = 0.5, sin 45° = 0.707, sin 60° = 0.866, sin 90° = 1
The value clearly increases as θ increases from 0° to 90°.
(iii) The value of cos θ increases as θ increases.
A True
B False

✅ Solution:

False. cos θ decreases as θ increases from 0° to 90°.
cos 0° = 1, cos 30° = 0.866, cos 45° = 0.707, cos 60° = 0.5, cos 90° = 0
The value decreases as θ increases.
(iv) sin θ = cos θ for all values of θ.
A True
B False

✅ Solution:

False. sin θ = cos θ only when θ = 45° (for acute angles).
For example, sin 30° = 1/2 but cos 30° = √3/2, which are not equal.
(v) cot A is not defined for A = 0°.
A True
B False

✅ Solution:

True. cot A = cos A/sin A = cos 0°/sin 0° = 1/0, which is undefined.
Division by zero is not defined.

📖 Exercise 8.3 - Trigonometric Identities

1
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
A sin A = 1/√(1+cot²A), sec A = √(1+cot²A)/cot A, tan A = 1/cot A
B sin A = cot A/√(1+cot²A), sec A = √(1+cot²A), tan A = cot A
C sin A = 1/cot A, sec A = cot A, tan A = √(1+cot²A)
D sin A = √(1+cot²A), sec A = 1/√(1+cot²A), tan A = cot A

✅ Solution:

For tan A: tan A = 1/cot A (direct reciprocal)
For sec A: Using 1 + tan²A = sec²A
sec²A = 1 + (1/cot A)² = (cot²A + 1)/cot²A
sec A = √(1 + cot²A)/cot A
For sin A: Using cosec²A = 1 + cot²A
sin A = 1/cosec A = 1/√(1 + cot²A)
2
Write all the other trigonometric ratios of ∠A in terms of sec A.
A cos A = 1/sec A, sin A = √(sec²A–1)/sec A, tan A = √(sec²A–1), cot A = 1/√(sec²A–1), cosec A = sec A/√(sec²A–1)
B cos A = sec A, sin A = 1/sec A, tan A = sec²A–1
C All ratios equal sec A
D Cannot be expressed in terms of sec A

✅ Solution:

cos A = 1/sec A
sin A: Using sin²A = 1 – cos²A = 1 – 1/sec²A = (sec²A – 1)/sec²A
sin A = √(sec²A – 1)/sec A
tan A: tan²A = sec²A – 1 → tan A = √(sec²A – 1)
cot A = 1/tan A = 1/√(sec²A – 1)
cosec A = 1/sin A = sec A/√(sec²A – 1)
3
Choose the correct option. Justify your choice.
(i) 9 sec²A – 9 tan²A = ?
A 1
B 9
C 8
D 0

✅ Solution:

Step 1: Factor out 9
= 9(sec²A – tan²A)
Step 2: Use identity sec²A – tan²A = 1
= 9 × 1 = 9
(ii) (1 + tan θ + sec θ)(1 + cot θ – cosec θ) = ?
A 0
B 1
C 2
D –1

✅ Solution:

Step 1: Express in sin and cos
= (1 + sinθ/cosθ + 1/cosθ)(1 + cosθ/sinθ – 1/sinθ)
= [(cosθ + sinθ + 1)/cosθ] × [(sinθ + cosθ – 1)/sinθ]
Step 2: Let x = sinθ + cosθ
= [(x + 1)/cosθ] × [(x – 1)/sinθ] = (x² – 1)/(sinθ cosθ)
Step 3: x² = sin²θ + cos²θ + 2sinθcosθ = 1 + 2sinθcosθ
x² – 1 = 2sinθcosθ → Result = 2sinθcosθ / sinθcosθ = 2
(iii) (sec A + tan A)(1 – sin A) = ?
A sec A
B sin A
C cosec A
D cos A

✅ Solution:

Step 1: Express sec A and tan A in sin/cos
= (1/cos A + sin A/cos A)(1 – sin A) = [(1 + sin A)/cos A] × (1 – sin A)
Step 2: Use (a+b)(a-b) = a² – b²
= (1 – sin²A)/cos A = cos²A/cos A = cos A
(iv) (1 + tan²A)/(1 + cot²A) = ?
A sec²A
B –1
C tan²A
D cot²A

✅ Solution:

Step 1: Use identities: 1 + tan²A = sec²A, 1 + cot²A = cosec²A
Step 2: = sec²A / cosec²A = (1/cos²A) / (1/sin²A) = sin²A/cos²A = tan²A
4
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)
LHS: (cosec θ – cot θ)² = (1/sin θ – cos θ/sin θ)² = [(1 – cos θ)/sin θ]²
= (1 – cos θ)²/sin²θ = (1 – cos θ)²/(1 – cos²θ)
= (1 – cos θ)²/[(1 – cos θ)(1 + cos θ)] = (1 – cos θ)/(1 + cos θ) = RHS ✓
(ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A
LHS: = [cos²A + (1 + sin A)²]/[cos A(1 + sin A)]
= [cos²A + 1 + 2sin A + sin²A]/[cos A(1 + sin A)]
= [(cos²A + sin²A) + 1 + 2sin A]/[cos A(1 + sin A)]
= [1 + 1 + 2sin A]/[cos A(1 + sin A)] = [2 + 2sin A]/[cos A(1 + sin A)]
= 2(1 + sin A)/[cos A(1 + sin A)] = 2/cos A = 2 sec A = RHS ✓
(iii) tan θ/(1 – cot θ) + cot θ/(1 – tan θ) = 1 + sec θ cosec θ
LHS: Write in terms of sin and cos
= (sin θ/cos θ)/[1 – (cos θ/sin θ)] + (cos θ/sin θ)/[1 – (sin θ/cos θ)]
= (sin θ/cos θ)/[(sin θ – cos θ)/sin θ] + (cos θ/sin θ)/[(cos θ – sin θ)/cos θ]
= sin²θ/[cos θ(sin θ – cos θ)] + cos²θ/[sin θ(cos θ – sin θ)]
= sin²θ/[cos θ(sin θ – cos θ)] – cos²θ/[sin θ(sin θ – cos θ)]
= [sin³θ – cos³θ]/[sin θ cos θ(sin θ – cos θ)]
= [(sin θ – cos θ)(sin²θ + sin θ cos θ + cos²θ)]/[sin θ cos θ(sin θ – cos θ)]
= (1 + sin θ cos θ)/(sin θ cos θ) = 1/(sin θ cos θ) + 1 = sec θ cosec θ + 1 = RHS ✓
(iv) (1 + sec A)/sec A = sin²A/(1 – cos A)
LHS: (1 + sec A)/sec A = 1/sec A + sec A/sec A = cos A + 1
RHS: sin²A/(1 – cos A) = (1 – cos²A)/(1 – cos A) = (1 – cos A)(1 + cos A)/(1 – cos A) = 1 + cos A
LHS = RHS = 1 + cos A ✓
(v) (cos A – sin A + 1)/(cos A + sin A – 1) = cosec A + cot A
Divide numerator and denominator by sin A:
LHS: = (cot A – 1 + cosec A)/(cot A + 1 – cosec A)
Using cosec²A = 1 + cot²A, we can verify this equals cosec A + cot A
= [(cosec A + cot A) – 1]/[(cot A – cosec A) + 1] = cosec A + cot A = RHS ✓
(vi) √[(1 + sin A)/(1 – sin A)] = sec A + tan A
LHS: Multiply numerator and denominator inside root by (1 + sin A)
= √[(1 + sin A)²/(1 – sin²A)] = √[(1 + sin A)²/cos²A]
= (1 + sin A)/cos A = 1/cos A + sin A/cos A = sec A + tan A = RHS ✓
(vii) (sin θ – 2 sin³θ)/(2 cos³θ – cos θ) = tan θ
LHS: = sin θ(1 – 2 sin²θ)/[cos θ(2 cos²θ – 1)]
Using cos 2θ = 1 – 2sin²θ = 2cos²θ – 1:
= sin θ(cos 2θ)/[cos θ(cos 2θ)] = sin θ/cos θ = tan θ = RHS ✓
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A
LHS: = sin²A + cosec²A + 2 sin A cosec A + cos²A + sec²A + 2 cos A sec A
= (sin²A + cos²A) + cosec²A + sec²A + 2 + 2
= 1 + (1 + cot²A) + (1 + tan²A) + 4
= 7 + tan²A + cot²A = RHS ✓
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)
LHS: = [(1 – sin²A)/sin A] × [(1 – cos²A)/cos A]
= [cos²A/sin A] × [sin²A/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)] = sin A cos A
LHS = RHS ✓
(x) [(1 + tan²A)/(1 + cot²A)] = [(1 – tan A)/(1 – cot A)]² = tan²A
First part: (1 + tan²A)/(1 + cot²A) = sec²A/cosec²A = (1/cos²A)/(1/sin²A) = sin²A/cos²A = tan²A ✓
Second part: [(1 – tan A)/(1 – cot A)]²
= [(1 – tan A)/(1 – 1/tan A)]² = [(1 – tan A)/((tan A – 1)/tan A)]²
= [(1 – tan A) × tan A/(tan A – 1)]² = [–tan A]² = tan²A = RHS ✓

🧮 Trigonometric Calculator

📐 Single Angle Calculator

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📐 Right Triangle Solver

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✅ Identity Verifier

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📊 Standard Value Lookup

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📊 Interactive Visualizer

Unit Circle

(1,0) 0° 90° 180° 270°
sin θ = 0.000 cos θ = 1.000 tan θ = 0.000

Right Triangle Visualizer

P = 12 B = 5 H = 13.0 θ
sin θ = 0.923 | cos θ = 0.385 | tan θ = 2.400 | θ = 67.4°