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NCERT Class 10 Mathematics Exercise 8.2 Solutions: Introduction to Trigonometry

Step-by-step working for every question of Exercise 8.2 in NCERT's Class 10 Mathematics book, chapter “Introduction to Trigonometry”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 8.2 Q1

   Evaluate: (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan²45° + cos²30° − sin²60° (iii) cos 45°/(sec 30° + cosec 30°) (iv) (sin 30° + tan 45° − cosec 60°)/(sec 30° + cos 60° + cot 45°) (v) (5 cos²60° + 4 sec²30° − tan²45°)/(sin²30° + cos²30°).

   1. (i) Substitute values: sin 60° = √3/2, cos 30° = √3/2, sin 30° = 1/2, cos 60° = 1/2.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   2. (i) Calculate: (√3/2 × √3/2) + (1/2 × 1/2) = 3/4 + 1/4 = 1.
   3. (ii) Substitute values: tan 45° = 1, cos 30° = √3/2, sin 60° = √3/2.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   4. (ii) Calculate: 2(1)² + (√3/2)² − (√3/2)² = 2 + 3/4 − 3/4 = 2.
   5. (iii) Substitute values: cos 45° = 1/√2, sec 30° = 2/√3, cosec 30° = 2.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   6. (iii) Simplify denominator: 2/√3 + 2 = (2 + 2√3)/√3.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   7. (iii) Divide: (1/√2) ÷ [(2 + 2√3)/√3] = √3 / [√2(2 + 2√3)].
   8. (iii) Rationalize and simplify to get (3√2 − √6)/8.
      why?

      Multiplying the top and bottom of a fraction by the same non-zero number keeps its value. Using the conjugate (a − √b for a + √b) clears the root from the denominator, because (a + √b)(a − √b) = a² − b.

   9. (iv) Substitute values: sin 30° = 1/2, tan 45° = 1, cosec 60° = 2/√3, sec 30° = 2/√3, cos 60° = 1/2, cot 45° = 1.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   10. (iv) Numerator: 1/2 + 1 − 2/√3 = 3/2 − 2/√3 = (3√3 − 4)/(2√3).
   11. (iv) Denominator: 2/√3 + 1/2 + 1 = 2/√3 + 3/2 = (4 + 3√3)/(2√3).
   12. (iv) Divide and rationalize to get (43 − 24√3)/11.
       why?

       Multiplying the top and bottom of a fraction by the same non-zero number keeps its value. Using the conjugate (a − √b for a + √b) clears the root from the denominator, because (a + √b)(a − √b) = a² − b.

   13. (v) Substitute values: cos 60° = 1/2, sec 30° = 2/√3, tan 45° = 1, sin 30° = 1/2.
       why?

       In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   14. (v) Numerator: 5(1/4) + 4(4/3) − 1 = 5/4 + 16/3 − 1 = 67/12.
   15. (v) Denominator: (1/2)² + (√3/2)² = 1/4 + 3/4 = 1.
   16. (v) Result: (67/12) / 1 = 67/12.

   Answer: (i) 1; (ii) 2; (iii) (3√2 - √6)/8; (iv) (43 - 24√3)/11; (v) 67/12

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2. Ex 8.2 Q2

   Choose and justify: (i) 2 tan 30°/(1 + tan²30°) = sin 60°, cos 60°, tan 60° or sin 30°? (ii) (1 − tan²45°)/(1 + tan²45°) = tan 90°, 1, sin 45° or 0? (iii) sin 2A = 2 sin A when A = 0°, 30°, 45° or 60°? (iv) 2 tan 30°/(1 − tan²30°) = cos 60°, sin 60°, tan 60° or sin 30°?

   1. (i) Substitute tan 30° = 1/√3 into the expression 2 tan 30°/(1 + tan²30°).
      why?

      sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.

   2. (i) The numerator becomes 2/√3 and the denominator becomes 1 + 1/3 = 4/3.
   3. (i) Simplifying (2/√3) ÷ (4/3) gives 1/√3 × 3/4 = 3/(4√3) = √3/2.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   4. (i) Since sin 60° = √3/2, the correct option is (A).
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   5. (ii) Substitute tan 45° = 1 into the expression (1 − tan²45°)/(1 + tan²45°).
      why?

      sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.

   6. (ii) The numerator becomes 1 − 1 = 0 and the denominator becomes 1 + 1 = 2.
   7. (ii) The value is 0/2 = 0, which corresponds to option (D).
   8. (iii) For A = 0°, sin 2A = sin 0° = 0 and 2 sin A = 2 × 0 = 0, so the equation holds.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   9. (iii) For A = 30°, sin 60° = √3/2 but 2 sin 30° = 1, so it does not hold.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   10. (iii) For A = 45°, sin 90° = 1 but 2 sin 45° = √2, so it does not hold.
       why?

       In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   11. (iii) For A = 60°, sin 120° = √3/2 but 2 sin 60° = √3, so it does not hold.
       why?

       In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   12. (iii) The equation is true only for A = 0°, which is option (A).
   13. (iv) Substitute tan 30° = 1/√3 into the expression 2 tan 30°/(1 − tan²30°).
       why?

       sin²θ + cos²θ = 1 (Pythagoras on a right triangle with hypotenuse 1). Dividing it by cos²θ gives 1 + tan²θ = sec²θ, and by sin²θ gives 1 + cot²θ = cosec²θ.

   14. (iv) The numerator is 2/√3 and the denominator is 1 − 1/3 = 2/3.
   15. (iv) Simplifying (2/√3) ÷ (2/3) gives 2/√3 × 3/2 = √3.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   16. (iv) Since tan 60° = √3, the correct option is (C).
       why?

       In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   Answer: (i) (A); (ii) (D); (iii) (A); (iv) (C)

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3. Ex 8.2 Q3

   tan(A + B) = √3 and tan(A − B) = 1/√3 with 0° < A + B ≤ 90° and A > B. Find A and B.

   1. Given tan(A + B) = √3, we know that tan 60° = √3.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   2. Since 0° < A + B ≤ 90°, the only possible value is A + B = 60°.
   3. Given tan(A − B) = 1/√3, we know that tan 30° = 1/√3.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   4. Since A > B, the angle A − B is positive, so A − B = 30°.
   5. Adding the two equations (A + B) + (A − B) = 60° + 30° gives 2A = 90°.
   6. Solving for A, we get A = 45°.
   7. Substituting A = 45° into A + B = 60° gives 45° + B = 60°.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   8. Solving for B, we get B = 15°.

   Answer: A = 45°, B = 15°

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4. Ex 8.2 Q4

   True or false? Justify: (i) sin(A + B) = sin A + sin B (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 not defined for A = 0°.

   1. (i) Take A = 30° and B = 60°.
   2. (i) sin(30° + 60°) = sin 90° = 1.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   3. (i) sin 30° + sin 60° = 1/2 + √3/2 ≈ 1.366.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   4. (i) Since 1 ≠ 1.366, the statement is False.
   5. (ii) Consider θ from 0° to 90°.
   6. (ii) sin 0° = 0, sin 30° = 0.5, sin 60° ≈ 0.866, sin 90° = 1.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   7. (ii) The value increases as θ increases, so the statement is True.
   8. (iii) Consider θ from 0° to 90°.
   9. (iii) cos 0° = 1, cos 30° ≈ 0.866, cos 60° = 0.5, cos 90° = 0.
      why?

      In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   10. (iii) The value decreases as θ increases, so the statement is False.
   11. (iv) Take θ = 0°.
   12. (iv) sin 0° = 0 and cos 0° = 1.
       why?

       In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   13. (iv) Since 0 ≠ 1, the statement is False.
   14. (v) cot A = cos A / sin A.
       why?

       In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   15. (v) For A = 0°, sin 0° = 0.
       why?

       In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.

   16. (v) Division by zero is undefined, so the statement is True.

   Answer: (i) False; (ii) True; (iii) False; (iv) False; (v) True

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