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NCERT Class 10 Mathematics Exercise 8.1 Solutions: Introduction to Trigonometry

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise 8.1 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.

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How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.

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Ex 8.1 Q1
In △ABC right-angled at B, AB = 24 cm and BC = 7 cm. Find (i) sin A, cos A (ii) sin C, cos C.
1. In △ABC, the angle at B is 90°, so AC is the hypotenuse.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
2. Use the Pythagorean theorem to find AC: AC² = AB² + BC².
3. Substitute the given values: AC² = 24² + 7² = 576 + 49 = 625.
  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.
4. Take the square root to find AC: AC = √625 = 25 cm.
5. (i) For angle A, the opposite side is BC and the adjacent side is AB.
6. Calculate sin A = Opposite/Hypotenuse = BC/AC = 7/25.
  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. Calculate cos A = Adjacent/Hypotenuse = AB/AC = 24/25.
  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.
8. (ii) For angle C, the opposite side is AB and the adjacent side is BC.
9. Calculate sin C = Opposite/Hypotenuse = AB/AC = 24/25.
  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. Calculate cos C = Adjacent/Hypotenuse = BC/AC = 7/25.
  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) sin A = 7/25, cos A = 24/25; (ii) sin C = 24/25, cos C = 7/25
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Ex 8.1 Q2
In Fig. 8.13 (△PQR right-angled at Q, PQ = 12 cm, PR = 13 cm), find tan P − cot R.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. In △PQR, ∠Q = 90°, PQ = 12 cm, PR = 13 cm.
2. By Pythagoras theorem, QR² = PR² − PQ² = 13² − 12² = 169 − 144 = 25.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
3. So QR = 5 cm.
4. tan P = opposite/adjacent = QR/PQ = 5/12.
  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. cot R = adjacent/opposite = QR/PQ = 5/12.
  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. tan P − cot R = 5/12 − 5/12 = 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.
Answer: 0
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Ex 8.1 Q3
sin A = 3/4. Find cos A and tan A.
1. Use the identity sin² A + cos² A = 1.
  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. Substitute sin A = 3/4 to get (3/4)² + cos² A = 1.
  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.
3. This gives 9/16 + cos² A = 1, so cos² A = 1 − 9/16 = 7/16.
  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.
4. Taking the square root, cos A = √7/4 (taking the positive value for an acute angle).
  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. Use the definition tan A = sin A / cos 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.
6. Substitute the values to get tan A = (3/4) / (√7/4).
  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. Simplify the fraction to get tan A = 3/√7.
  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: cos A = √7/4, tan A = 3/√7
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Ex 8.1 Q4
15 cot A = 8. Find sin A and sec A.
1. Given 15 cot A = 8, so cot A = 8/15.
  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. In a right triangle, cot A = adjacent/opposite, so let adjacent = 8 and opposite = 15.
  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. By Pythagoras theorem, hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = 17.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
4. sin A = opposite/hypotenuse = 15/17.
  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. sec A = hypotenuse/adjacent = 17/8.
  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: sin A = 15/17; sec A = 17/8
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Ex 8.1 Q5
sec θ = 13/12. Find all the other trigonometric ratios.
1. Given sec θ = 13/12, we know that sec θ = hypotenuse / adjacent side.
  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. Let the hypotenuse be 13 and the adjacent side be 12.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
3. Using the Pythagorean theorem, opposite side = √(13² − 12²) = √(169 − 144) = √25 = 5.
4. Now, sin θ = opposite / hypotenuse = 5/13.
  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. cos θ = adjacent / hypotenuse = 12/13.
  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. tan θ = opposite / adjacent = 5/12.
  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. cot θ = adjacent / opposite = 12/5.
  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.
8. cosec θ = hypotenuse / opposite = 13/5.
  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: sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cot θ = 12/5, cosec θ = 13/5
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Ex 8.1 Q6
A and B are acute angles with cos A = cos B. Show that A = B.
1. Draw a right triangle with angles A and B as its two acute angles: △ABC right-angled at C, so AB is the hypotenuse.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
2. Then cos A = AC/AB and cos B = BC/AB.
  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. cos A = cos B gives AC/AB = BC/AB, so AC = BC.
  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. In △ABC, the angles opposite the equal sides AC and BC are equal: ∠B = ∠A.
Answer: A = B
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Ex 8.1 Q7
cot θ = 7/8. Evaluate (i) (1 + sin θ)(1 − sin θ)/[(1 + cos θ)(1 − cos θ)] (ii) cot²θ.
1. Given cot θ = 7/8, we have adjacent side = 7 and opposite side = 8.
  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. Hypotenuse = √(7² + 8²) = √(49 + 64) = √113.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
3. sin θ = opposite/hypotenuse = 8/√113 and cos θ = adjacent/hypotenuse = 7/√113.
  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) The expression is (1 + sin θ)(1 − sin θ)/[(1 + cos θ)(1 − cos θ)].
  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. (i) This simplifies to (1 − sin²θ)/(1 − cos²θ).
  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. (i) Using identities, 1 − sin²θ = cos²θ and 1 − cos²θ = sin²θ.
  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.
7. (i) So the expression becomes cos²θ/sin²θ = cot²θ.
  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.
8. (i) cot²θ = (7/8)² = 49/64.
  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.
9. (ii) cot²θ = (7/8)² = 49/64.
  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.
Answer: (i) 49/64; (ii) 49/64
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Ex 8.1 Q8
3 cot A = 4. Check whether (1 − tan²A)/(1 + tan²A) = cos²A − sin²A.
1. Given 3 cot A = 4, so cot A = 4/3.
  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 cot A = adjacent/opposite, let adjacent side = 4 and opposite side = 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.
3. Hypotenuse = √(4² + 3²) = √(16 + 9) = √25 = 5.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
4. Therefore, tan A = opposite/adjacent = 3/4, sin A = 3/5, and cos A = 4/5.
  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. Calculate LHS: (1 − tan²A)/(1 + tan²A) = (1 − (3/4)²)/(1 + (3/4)²).
  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. Simplify numerator: 1 − 9/16 = 7/16.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. Simplify denominator: 1 + 9/16 = 25/16.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
8. LHS = (7/16) ÷ (25/16) = 7/25.
9. Calculate RHS: cos²A − sin²A = (4/5)² − (3/5)².
  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.
10. Simplify RHS: 16/25 − 9/25 = 7/25.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
11. Since LHS = RHS = 7/25, the identity is verified.
Answer: 7/25, yes
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Ex 8.1 Q9
In △ABC right-angled at B, tan A = 1/√3. Find (i) sin A cos C + cos A sin C (ii) cos A cos C − sin A sin C.
1. Given tan A = 1/√3, we identify the sides of the right triangle as opposite = 1, adjacent = √3, and hypotenuse = 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. From these sides, sin A = 1/2 and cos A = √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.
3. Since the triangle is right-angled at B, angle C is complementary to angle A, so sin C = cos A = √3/2 and cos C = sin A = 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.
4. (i) Substitute the values into sin A cos C + cos A sin C to get (1/2)(1/2) + (√3/2)(√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.
5. (i) Calculate the sum as 1/4 + 3/4, which equals 1.
6. (ii) Substitute the values into cos A cos C − sin A sin C to get (√3/2)(1/2) − (1/2)(√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.
7. (ii) Calculate the difference as √3/4 − √3/4, which equals 0.
Answer: (i) 1; (ii) 0
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Ex 8.1 Q10
In △PQR right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find sin P, cos P and tan P.
1. Let QR = x cm, so PR = (25 − x) cm.
2. Apply Pythagoras theorem in △PQR: PQ² + QR² = PR².
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
3. Substitute values: 5² + x² = (25 − x)².
  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.
4. Expand and simplify: 25 + x² = 625 − 50x + x².
  why?
  Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.
5. Cancel x² and solve: 50x = 600, so x = 12 cm.
6. Thus QR = 12 cm and PR = 25 − 12 = 13 cm.
7. sin P = opposite/hypotenuse = QR/PR = 12/13.
  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.
8. cos P = adjacent/hypotenuse = PQ/PR = 5/13.
  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. tan P = opposite/adjacent = QR/PQ = 12/5.
  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: sin P = 12/13, cos P = 5/13, tan P = 12/5
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Ex 8.1 Q11
True or false? Justify: (i) tan A is always less than 1 (ii) sec A = 12/5 for some angle A (iii) cos A is short for the cosecant of A (iv) cot A is cot times A (v) sin θ = 4/3 for some angle θ.
1. (i) False: tan 45° = 1 and tan 60° = √3, which is more than 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.
2. (ii) True: sec A = hypotenuse/adjacent side, and a hypotenuse of 12 with an adjacent side of 5 is possible (the third side is √119), so sec A = 12/5 for some 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.
3. (iii) False: cos A is short for the cosine of A; cosec A is the cosecant.
  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. (iv) False: cot A is one symbol, the cotangent of angle A, not cot multiplied by 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. (v) False: sin θ = opposite/hypotenuse, and the hypotenuse is the longest side, so sin θ cannot be more than 1; 4/3 is more than 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.
Answer: (i) False; (ii) True; (iii) False; (iv) False; (v) False
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Other exercises in this chapter

Exercise 8.2 — 4 questions
Exercise 8.3 — 13 questions

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NCERT Class 10 Mathematics Exercise 8.1 Solutions: Introduction to Trigonometry

Ex 8.1 Q1

Ex 8.1 Q2

Ex 8.1 Q3

Ex 8.1 Q4

Ex 8.1 Q5

Ex 8.1 Q6

Ex 8.1 Q7

Ex 8.1 Q8

Ex 8.1 Q9

Ex 8.1 Q10

Ex 8.1 Q11