Trigonometry

10 practice questions on Trigonometry for NDA, with answers

Shishya's practice questions, written with AI. Each answer was checked by an automated second pass, not by a person.

1. 1.If sin θ = 3/5 and θ lies in the second quadrant, what is the value of tan θ?

   • (A)3/4
   • (B)-3/4
   • (C)4/3
   • (D)-4/3
   Show the answer and solution

   Answer: (B) -3/4

   Solution: Given: sin θ = 3/5 and θ is in the second quadrant. Step 1: Find cos θ using the identity sin² θ + cos² θ = 1. (3/5)² + cos² θ = 1 9/25 + cos² θ = 1 cos² θ = 1 - 9/25 = 16/25 cos θ = ±4/5 Step 2: Determine the sign of cos θ. In the second quadrant, sine is positive and cosine is negative. Therefore, cos θ = -4/5 Step 3: Calculate tan θ. tan θ = sin θ / cos θ tan θ = (3/5) / (-4/5) tan θ = (3/5) × (-5/4) tan θ = -3/4 Answer: B

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2. 2.In a triangle ABC, if a = 7, b = 24 and c = 25, what is the value of sin C?

   • (A)7/25
   • (B)24/25
   • (C)1
   • (D)7/24
   Show the answer and solution

   Answer: (C) 1

   Solution: Given: a = 7, b = 24, c = 25 where a, b, c are sides opposite to angles A, B, C respectively. Step 1: Check if the triangle is right-angled. a² + b² = 7² + 24² = 49 + 576 = 625 c² = 25² = 625 Since a² + b² = c², the triangle is right-angled at C. Step 2: Find sin C. In a right-angled triangle, the angle opposite to the hypotenuse is 90°. Therefore, C = 90° sin C = sin 90° = 1 Alternatively, using sine rule: sin C / c = sin A / a But since C = 90°, sin C = 1 Answer: C

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3. 3.If sin θ = 3/5 and θ is in the first quadrant, what is the value of tan θ?

   • (A)3/4
   • (B)4/3
   • (C)5/4
   • (D)4/5
   Show the answer and solution

   Answer: (A) 3/4

   Solution: Given sin θ = 3/5 and θ is in the first quadrant. Step 1: Use the Pythagorean identity sin² θ + cos² θ = 1. (3/5)² + cos² θ = 1 9/25 + cos² θ = 1 cos² θ = 1 - 9/25 = 16/25 Step 2: Since θ is in the first quadrant, cos θ is positive. cos θ = 4/5 Step 3: Calculate tan θ = sin θ / cos θ. tan θ = (3/5) / (4/5) = 3/5 × 5/4 = 3/4 Therefore, tan θ = 3/4.

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4. 4.If tan A + cot A = 2, then what is the value of tan² A + cot² A?

   • (A)2
   • (B)4
   • (C)6
   • (D)8
   Show the answer and solution

   Answer: (A) 2

   Solution: Given: tan A + cot A = 2 Step 1: Square both sides. (tan A + cot A)² = 2² tan² A + 2 tan A cot A + cot² A = 4 Step 2: Simplify tan A cot A. Since tan A = 1/cot A, we have tan A × cot A = 1 Step 3: Substitute into the equation. tan² A + 2(1) + cot² A = 4 tan² A + cot² A + 2 = 4 tan² A + cot² A = 4 - 2 tan² A + cot² A = 2 Answer: A

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5. 5.If tan⁻¹(1/2) + tan⁻¹(1/3) = tan⁻¹(x), then the value of x is:

   • (A)5/6
   • (B)1
   • (C)6/5
   • (D)5/7
   Show the answer and solution

   Answer: (B) 1

   Solution: Using the formula: tan⁻¹A + tan⁻¹B = tan⁻¹[(A + B)/(1 - AB)], valid when AB < 1 Here A = 1/2 and B = 1/3 Step 1: Check if AB < 1 AB = (1/2)(1/3) = 1/6 < 1 ✓ Step 2: Apply the formula tan⁻¹(1/2) + tan⁻¹(1/3) = tan⁻¹[(1/2 + 1/3)/(1 - 1/6)] Step 3: Simplify numerator 1/2 + 1/3 = 3/6 + 2/6 = 5/6 Step 4: Simplify denominator 1 - 1/6 = 5/6 Step 5: Compute the ratio (5/6)/(5/6) = 1 Therefore, x = 1.

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6. 6.If sin θ + cos θ = √2, then the value of sin θ · cos θ is:

   • (A)1/2
   • (B)1
   • (C)√2
   • (D)1/4
   Show the answer and solution

   Answer: (A) 1/2

   Solution: Squaring both sides: (sin θ + cos θ)² = 2, thus sin²θ + cos²θ + 2 sin θ cos θ = 2. Since sin²θ + cos²θ = 1, we get 1 + 2 sin θ cos θ = 2, giving sin θ cos θ = 1/2.

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7. 7.If sin θ = 3/5 and θ is an acute angle, then the value of tan θ is:

   • (A)3/4
   • (B)4/3
   • (C)5/4
   • (D)4/5
   Show the answer and solution

   Answer: (A) 3/4

   Solution: Given sin θ = 3/5 and θ is acute. Step 1: Use the identity sin²θ + cos²θ = 1 (3/5)² + cos²θ = 1 9/25 + cos²θ = 1 cos²θ = 1 - 9/25 = 16/25 cos θ = 4/5 (positive since θ is acute) Step 2: Find tan θ tan θ = sin θ / cos θ = (3/5) / (4/5) = 3/5 × 5/4 = 3/4 Therefore, tan θ = 3/4.

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8. 8.In a triangle ABC, if a = 7 cm, b = 5 cm and ∠C = 60°, then the length of side c is:

   • (A)√39 cm
   • (B)√69 cm
   • (C)√59 cm
   • (D)√49 cm
   Show the answer and solution

   Answer: (A) √39 cm

   Solution: Using the cosine rule: c² = a² + b² - 2ab cos C Given: a = 7 cm, b = 5 cm, ∠C = 60° Step 1: Substitute values c² = 7² + 5² - 2(7)(5)cos 60° c² = 49 + 25 - 70 × (1/2) c² = 74 - 35 c² = 39 Step 2: Take square root c = √39 cm Therefore, the length of side c is √39 cm.

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9. 9.The value of sin 15° × cos 15° is:

   • (A)1/2
   • (B)√3/4
   • (C)1/4
   • (D)1/√2
   Show the answer and solution

   Answer: (C) 1/4

   Solution: We need to find sin 15° × cos 15°. Step 1: Use the double angle formula: sin 2A = 2 sin A cos A. Therefore, sin A cos A = (1/2) sin 2A. Step 2: Apply this formula with A = 15°. sin 15° × cos 15° = (1/2) sin (2 × 15°) = (1/2) sin 30° Step 3: We know sin 30° = 1/2. sin 15° × cos 15° = (1/2) × (1/2) = 1/4 Therefore, the value is 1/4.

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10. 10.If A + B + C = 180°, then the value of sin 2A + sin 2B + sin 2C is equal to:

    • (A)2 sin A sin B sin C
    • (B)4 sin A sin B sin C
    • (C)sin A + sin B + sin C
    • (D)cos A + cos B + cos C
    Show the answer and solution

    Answer: (B) 4 sin A sin B sin C

    Solution: Given A + B + C = 180° Step 1: Write sin 2A + sin 2B + sin 2C Using sum-to-product formula: sin P + sin Q = 2 sin((P+Q)/2) cos((P-Q)/2) sin 2A + sin 2B = 2 sin(A + B) cos(A - B) Step 2: Since A + B + C = 180°, we have A + B = 180° - C sin(A + B) = sin(180° - C) = sin C So: sin 2A + sin 2B = 2 sin C cos(A - B) Step 3: Add sin 2C: sin 2A + sin 2B + sin 2C = 2 sin C cos(A - B) + 2 sin C cos C = 2 sin C [cos(A - B) + cos C] Step 4: Since C = 180° - A - B: cos C = cos(180° - A - B) = -cos(A + B) Using sum-to-product: cos(A - B) + cos(A + B) = 2 cos A cos B Therefore: cos(A - B) + cos C = cos(A - B) - cos(A + B) = 2 sin A sin B Step 5: Substitute back: sin 2A + sin 2B + sin 2C = 2 sin C × 2 sin A sin B = 4 sin A sin B sin C The answer is 4 sin A sin B sin C.

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