Trigonometric Identities and Complementary Angles
Trigonometric Identities and Complementary Angles
Three identities and one complementary-angle rule solve nearly every NTPC trigonometry simplification. The exam asks for complementary-angle values, long products that collapse to 1, pairs such as sec θ + tan θ, and power expressions such as sin⁴ θ − cos⁴ θ.
1. Identity Box
sin² θ + cos² θ = 1
1 + tan² θ = sec² θ, so (sec θ + tan θ)(sec θ − tan θ) = 1
1 + cot² θ = cosec² θ, so (cosec θ + cot θ)(cosec θ − cot θ) = 1
| Derived result | Form |
|---|---|
| sin⁴ θ − cos⁴ θ | sin² θ − cos² θ = 2 sin² θ − 1 |
| sin⁴ θ + cos⁴ θ | 1 − 2 sin² θ cos² θ |
| sin⁶ θ + cos⁶ θ | 1 − 3 sin² θ cos² θ |
| (sin θ + cos θ)² | 1 + 2 sin θ cos θ |
| sec θ + tan θ = p | sec θ = (p² + 1)/(2p), tan θ = (p² − 1)/(2p), sin θ = (p² − 1)/(p² + 1) |
| √((1 − cos θ)/(1 + cos θ)) | cosec θ − cot θ (acute θ) |
2. Complementary Angles
| Rule | Example |
|---|---|
| sin(90° − θ) = cos θ, cos(90° − θ) = sin θ | sin 37° = cos 53° |
| tan(90° − θ) = cot θ, cot(90° − θ) = tan θ | tan 20° × tan 70° = 1 |
| sec(90° − θ) = cosec θ, cosec(90° − θ) = sec θ | cosec 50° = sec 40° |
| If sin A = cos B (acute) | A + B = 90° |
tan 1° × tan 2° × ... × tan 89° = 1 (pairs give 1, and tan 45° = 1)
cos 1° × cos 2° × ... × cos 90° = 0 (cos 90° = 0)
✗ sin⁴ θ − cos⁴ θ = (sin θ − cos θ)⁴ | ✓ Factor as a difference of squares: (sin² θ − cos² θ)(sin² θ + cos² θ) = sin² θ − cos² θ
✗ sec θ + tan θ = 2, so sec θ − tan θ = −2 | ✓ Their product is 1: sec θ − tan θ = 1/2
हिंदी नोट: sin² θ + cos² θ = 1, sec² θ − tan² θ = 1 और cosec² θ − cot² θ = 1। पूरक कोणों में sin(90° − θ) = cos θ और tan(90° − θ) = cot θ। sec θ + tan θ दिया हो तो sec θ − tan θ उसका व्युत्क्रम होता है।
Exam Pointer: Verified NTPC patterns: complementary angles (April 2016 CBT-1), long products simplified with identities (August 2025 UG CBT-1, June 2025 Graduate CBT-1), sec θ + tan θ given (August and September 2025 UG CBT-1) and sin⁴ θ − cos⁴ θ given (June 2025 Graduate CBT-1).
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Complementary angles
[PYQ: NTPC CBT-1 6-Apr-2016 Shift-1]
EXAM LEVEL
Q. Find sin 37°/cos 53° + tan 10° tan 20° tan 70° tan 80°.
sin 37° = cos 53°, so the first term is 1. tan 10° tan 80° = 1 and tan 20° tan 70° = 1. Total = 2.
Answer: 2
EXAMATLAS LEVEL
Q. If sin 3θ = cos(θ − 6°), with both angles acute, find θ. Also evaluate (cos² 20° + cos² 70°)/(sin² 59° + sin² 31°).
3θ + θ − 6° = 90° gives θ = 24°. cos 70° = sin 20° and sin 31° = cos 59°, so the expression is 1/1 = 1.
Answer: 24°; 1
Pattern 2: Simplifying products with identities
[PYQ: NTPC UG CBT-1 18-Aug-2025 Shift-2 | NTPC Graduate CBT-1 12-Jun-2025 Shift-3]
EXAM LEVEL
Q. Simplify (1 − sin² θ) sec² θ.
1 − sin² θ = cos² θ and cos² θ × sec² θ = 1.
Answer: 1
EXAMATLAS LEVEL
Q. Simplify (sec θ − cos θ)(cosec θ − sin θ)(tan θ + cot θ).
sec θ − cos θ = (1 − cos² θ)/cos θ = sin² θ/cos θ, and cosec θ − sin θ = cos² θ/sin θ, so their product is sin θ cos θ. Also tan θ + cot θ = (sin² θ + cos² θ)/(sin θ cos θ) = 1/(sin θ cos θ). The whole product is 1. Check at 45°: (√2 − 1/√2)² × 2 = (1/2) × 2 = 1.
Answer: 1
Pattern 3: sec θ + tan θ and cosec θ − cot θ pairs
[PYQ: NTPC UG CBT-1 14-Aug-2025 Shift-2 | NTPC UG CBT-1 4-Sep-2025 Shift-2]
EXAM LEVEL
Q. If sec θ + tan θ = 2, find sec θ − tan θ, sec θ and tan θ.
The product is 1, so sec θ − tan θ = 1/2. Adding, 2 sec θ = 5/2, so sec θ = 5/4; subtracting, tan θ = 3/4.
Answer: 1/2; 5/4 and 3/4
EXAMATLAS LEVEL
Q. If cosec θ − cot θ = 1/3, find cos θ and sin θ.
cosec θ + cot θ = 3. Adding, cosec θ = 5/3, so sin θ = 3/5; subtracting, cot θ = 4/3, so cos θ = cot θ × sin θ = 4/5.
Answer: cos θ = 4/5, sin θ = 3/5
Pattern 4: Power expressions: sin⁴ θ − cos⁴ θ and sin θ ± cos θ
[PYQ: NTPC Graduate CBT-1 12-Jun-2025 Shift-2]
EXAM LEVEL
Q. If sin θ = 3/5, find sin⁴ θ − cos⁴ θ.
sin⁴ θ − cos⁴ θ = sin² θ − cos² θ = 9/25 − 16/25 = −7/25.
Answer: −7/25
EXAMATLAS LEVEL
Q. If sin θ + cos θ = 7/5 and sin θ > cos θ, find sin θ cos θ, sin θ − cos θ and sin⁴ θ − cos⁴ θ.
Squaring, 1 + 2 sin θ cos θ = 49/25, so sin θ cos θ = 12/25. (sin θ − cos θ)² = 1 − 24/25 = 1/25, so sin θ − cos θ = 1/5. Then sin⁴ θ − cos⁴ θ = (sin θ − cos θ)(sin θ + cos θ) = 7/25.
Answer: 12/25, 1/5, 7/25
CBT-2 LEVEL
Pattern 5: Sixth powers and fourth powers from sin θ cos θ
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ.
sin⁶ θ + cos⁶ θ = 1 − 3 sin² θ cos² θ, so the sum is 1.
Answer: 1
EXAMATLAS LEVEL
Q. If sin θ cos θ = 1/4, find sin⁴ θ + cos⁴ θ and sin⁶ θ + cos⁶ θ.
sin² θ cos² θ = 1/16. sin⁴ θ + cos⁴ θ = 1 − 2/16 = 7/8. sin⁶ θ + cos⁶ θ = 1 − 3/16 = 13/16.
Answer: 7/8 and 13/16
60-Second Revision
- Three Pythagorean identities; each gives a product-equals-1 pair.
- Complementary angles: a ratio of θ equals the co-ratio of 90° − θ.
- sin⁴ θ − cos⁴ θ = sin² θ − cos² θ.
- From sin θ + cos θ, square to get sin θ cos θ.