Exercise 8.1 — Introduction to Trigonometry

1. 12 marksNCERT Cl-10 Maths, Ex 8.1, Q1

   In △ABC, right-angled at B, AB = 24 cm and BC = 7 cm. Determine (i) sin A and cos A, (ii) sin C and cos C.

   Hint. Find the hypotenuse first. Then be careful: what is 'opposite' for angle A is 'adjacent' for angle C.

   Step 1 — Find the hypotenuse. The right angle is at B, so AC is the hypotenuse. AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625 AC = 25 cm

   Step 2 — Ratios for angle A. Standing at A: the side opposite is BC = 7, the side adjacent is AB = 24, hypotenuse AC = 25. sin A = 7/25 cos A = 24/25

   Step 3 — Ratios for angle C. Standing at C the two legs swap roles: opposite is now AB = 24, adjacent is BC = 7. sin C = 24/25 cos C = 7/25

   ✦ Answer: sin A = 7/25, cos A = 24/25, sin C = 24/25, cos C = 7/25

   Notice that sin A = cos C and cos A = sin C. That is no accident — A and C are the two acute angles of a right triangle, so they add to 90°.

   Where students slip. Using the same 'opposite' side for both angles. Opposite and adjacent are defined relative to the angle you are standing at, so they swap when you move from A to C.

   Another way. Spot the triple: 7, 24, 25 is a standard Pythagorean triple, so the hypotenuse can be written down without squaring anything.

2. 22 marksNCERT Cl-10 Maths, Ex 8.1, Q2 (Fig. 8.13)

   In the figure, △PQR is right-angled at Q with PQ = 12 cm and PR = 13 cm. Find tan P − cot R.

   Hint. Work out QR first. Then write both ratios in terms of the sides and look at them before subtracting.

   Step 1 — Find the third side. PR is the hypotenuse, since the right angle is at Q. QR² = PR² − PQ² = 169 − 144 = 25, so QR = 5 cm

   Step 2 — Write tan P. At P, the opposite side is QR = 5 and the adjacent side is PQ = 12. tan P = QR/PQ = 5/12

   Step 3 — Write cot R. At R, the opposite side is PQ = 12 and the adjacent side is QR = 5. Cotangent is adjacent over opposite, so cot R = QR/PQ = 5/12

   Step 4 — Subtract. tan P − cot R = 5/12 − 5/12 = 0

   ✦ Answer: 0

   The two ratios were bound to be equal, because P and R are complementary angles and cot is the co-ratio of tan.

   Where students slip. Computing cot R as tan R inverted but taking the sides from P's viewpoint, giving 12/5 and an answer of −7/12. Re-identify opposite and adjacent from scratch each time you change vertex.

   Another way. Recognise the (5, 12, 13) triple to skip the Pythagoras step, then note that in any right triangle tan of one acute angle always equals cot of the other — so the answer is 0 before any arithmetic.

3. 33 marksNCERT Cl-10 Maths, Ex 8.1, Q3

   If sin A = 3/4, calculate cos A and tan A.

   Hint. sin A = 3/4 means opposite = 3 and hypotenuse = 4 in a suitable right triangle. Get the adjacent side from Pythagoras.

   Step 1 — Build the triangle. sin A = opposite/hypotenuse = 3/4, so take a right triangle with the side opposite A equal to 3k and the hypotenuse 4k.

   Step 2 — Find the adjacent side. adjacent² = (4k)² − (3k)² = 16k² − 9k² = 7k² adjacent = k√7

   Step 3 — Read off the ratios. cos A = adjacent/hypotenuse = k√7 / 4k = √7/4 tan A = opposite/adjacent = 3k / k√7 = 3/√7

   The k cancels every time, which is why the answer does not depend on the triangle's size.

   Check. sin²A + cos²A = 9/16 + 7/16 = 1 ✓

   ✦ Answer: cos A = √7/4 and tan A = 3/√7 (equivalently 3√7/7)

   Where students slip. Assuming the adjacent side must be a whole number and forcing √7 to 2.65 too early. Leave surds exact — the answer is expected in surd form.

   Another way. Use the identity directly: cos A = √(1 − sin²A) = √(1 − 9/16) = √7/4, then tan A = sin A / cos A = (3/4) ÷ (√7/4) = 3/√7. No triangle needed.

4. 43 marksNCERT Cl-10 Maths, Ex 8.1, Q4

   Given 15 cot A = 8, find sin A and sec A.

   Hint. First isolate cot A, then turn it into a triangle. cot is adjacent over opposite.

   Step 1 — Isolate the ratio. 15 cot A = 8, so cot A = 8/15

   Step 2 — Read it as a triangle. cot A = adjacent/opposite = 8/15, so take adjacent = 8k and opposite = 15k.

   Step 3 — Find the hypotenuse. hyp² = (8k)² + (15k)² = 64k² + 225k² = 289k² hyp = 17k, since 289 = 17²

   Step 4 — Read off what was asked. sin A = opposite/hypotenuse = 15k/17k = 15/17 sec A = hypotenuse/adjacent = 17k/8k = 17/8

   Check. cos A = 8/17, and sin²A + cos²A = 225/289 + 64/289 = 1 ✓

   ✦ Answer: sin A = 15/17 and sec A = 17/8

   Where students slip. Reading cot A = 8/15 as opposite/adjacent and building the triangle the wrong way up, which gives sin A = 8/17. Cotangent is the reciprocal of tangent, so it is adjacent over opposite.

   Another way. (8, 15, 17) is a standard Pythagorean triple worth recognising — it appears again in Chapter 6's practice problems and saves the squaring step here.

5. 53 marksNCERT Cl-10 Maths, Ex 8.1, Q5

   Given sec θ = 13/12, calculate all the other trigonometric ratios.

   Hint. sec is hypotenuse over adjacent. Once you have all three sides, every ratio is a reading, not a calculation.

   Step 1 — Interpret sec θ. sec θ = hypotenuse/adjacent = 13/12, so take hypotenuse = 13k and adjacent = 12k.

   Step 2 — Find the opposite side. opposite² = (13k)² − (12k)² = 169k² − 144k² = 25k² opposite = 5k

   Step 3 — Now read off all six, since every side is known. sin θ = 5/13 cos θ = 12/13 (the reciprocal of the given sec θ) tan θ = 5/12 cosec θ = 13/5 cot θ = 12/5

   Check. sin²θ + cos²θ = 25/169 + 144/169 = 169/169 = 1 ✓

   ✦ Answer: sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5

   Where students slip. Taking cos θ = 13/12. A cosine can never exceed 1, so that value is impossible — sec and cos are reciprocals, so cos θ = 12/13.

   Another way. Get cos θ = 12/13 by reciprocal, then sin θ = √(1 − 144/169) = 5/13, and build the remaining four by reciprocals and quotients. The triangle just makes it visual.

6. 63 marksNCERT Cl-10 Maths, Ex 8.1, Q6

   If ∠A and ∠B are acute angles such that cos A = cos B, show that ∠A = ∠B.

   Hint. Put both angles into one right triangle. Equal cosines force two sides into the same ratio — then look for an isosceles triangle.

   Step 1 — Set up a single triangle containing both angles. Draw △ABC right-angled at C, so that ∠A and ∠B are its two acute angles.

   Step 2 — Write the two cosines. cos A = AC/AB and cos B = BC/AB

   Step 3 — Use the hypothesis. cos A = cos B gives AC/AB = BC/AB

   The denominators are the same side AB, so AC = BC

   Step 4 — Draw the geometric conclusion. In △ABC the sides AC and BC are equal, so the triangle is isosceles. Angles opposite equal sides are equal, and the angle opposite AC is ∠B while the angle opposite BC is ∠A. Therefore

   ∠A = ∠B

   ✦ Answer: Proved.

   The underlying reason is that cosine is strictly decreasing on 0° to 90°, so it never takes the same value twice — but the isosceles argument is the one expected here.

   Where students slip. Writing 'cos A = cos B therefore A = B' and stopping. That is the statement to be proved, not a step; the marks are for the isosceles-triangle reasoning that establishes it.

   Another way. The same argument works with any ratio. If sin A = sin B, or tan A = tan B, you reach an isosceles triangle in exactly the same way and the conclusion is identical.

7. 74 marksNCERT Cl-10 Maths, Ex 8.1, Q7

   If cot θ = 7/8, evaluate (i) (1 + sin θ)(1 − sin θ) / [(1 + cos θ)(1 − cos θ)], and (ii) cot²θ.

   Hint. Do not compute sin θ and cos θ yet. Each bracket pair is a difference of squares — simplify first and the answer falls out.

   (i) The fraction.

   Step 1 — Recognise the difference of squares in both the top and the bottom. (1 + sin θ)(1 − sin θ) = 1 − sin²θ (1 + cos θ)(1 − cos θ) = 1 − cos²θ

   Step 2 — Apply the first identity. Since sin²θ + cos²θ = 1, we have 1 − sin²θ = cos²θ and 1 − cos²θ = sin²θ.

   So the fraction is cos²θ / sin²θ = cot²θ.

   Step 3 — Substitute the given value. cot θ = 7/8, so cot²θ = 49/64

   ✦ (i) = 49/64

   (ii) cot²θ.

   cot²θ = (7/8)² = 49/64

   ✦ (ii) = 49/64

   The two parts are equal, which is the point of the question — part (i) is cot²θ wearing a disguise.

   Where students slip. Building the triangle (adjacent 7, opposite 8, hypotenuse √113) and grinding the fraction out with surds. It works, but it is several times the labour and invites arithmetic slips.

   Another way. If you do want the check: hypotenuse = √113, so sin θ = 8/√113 and cos θ = 7/√113. Then (1 − 64/113)/(1 − 49/113) = (49/113)/(64/113) = 49/64 ✓

8. 84 marksNCERT Cl-10 Maths, Ex 8.1, Q8

   If 3 cot A = 4, check whether (1 − tan²A)/(1 + tan²A) = cos²A − sin²A.

   Hint. Evaluate the two sides separately from the triangle and compare. The question asks you to check, so both sides must be computed.

   Step 1 — Get the ratios. 3 cot A = 4 gives cot A = 4/3, so tan A = 3/4. Taking adjacent = 4k and opposite = 3k, the hypotenuse is 5k, since 3-4-5 is a Pythagorean triple.

   So sin A = 3/5 and cos A = 4/5.

   Step 2 — Evaluate the left-hand side. tan²A = 9/16 LHS = (1 − 9/16)/(1 + 9/16) = (7/16)/(25/16)

   The sixteenths cancel, so LHS = 7/25

   Step 3 — Evaluate the right-hand side. RHS = cos²A − sin²A = 16/25 − 9/25 = 7/25

   Step 4 — Compare. 7/25 = 7/25

   ✦ Answer: Yes, the equality holds — both sides equal 7/25.

   This is in fact an identity, true for every acute A, not a coincidence of these numbers. Dividing the top and bottom of the left side by sec²A converts it to cos²A − sin²A directly.

   Where students slip. Evaluating one side and asserting the other must match. The instruction is 'check whether', so both sides have to be worked out independently — that is where the marks are.

   Another way. Prove it in general instead of numerically: (1 − tan²A)/(1 + tan²A) = (1 − sin²A/cos²A)/sec²A = ((cos²A − sin²A)/cos²A) × cos²A = cos²A − sin²A.

9. 94 marksNCERT Cl-10 Maths, Ex 8.1, Q9

   In △ABC, right-angled at B, tan A = 1/√3. Find the value of (i) sin A cos C + cos A sin C, and (ii) cos A cos C − sin A sin C.

   Hint. tan A = 1/√3 identifies angle A exactly. Since the triangle is right-angled at B, that fixes angle C too.

   Step 1 — Identify the angles. tan A = 1/√3 is the value of tan 30°, so ∠A = 30°. The angles of △ABC sum to 180° and ∠B = 90°, so ∠C = 180° − 90° − 30° = 60°.

   Step 2 — Write down the four ratios needed. sin A = sin 30° = 1/2 cos A = cos 30° = √3/2 sin C = sin 60° = √3/2 cos C = cos 60° = 1/2

   Step 3 — Evaluate part (i). sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1

   ✦ (i) = 1

   Step 4 — Evaluate part (ii). cos A cos C − sin A sin C = (√3/2)(1/2) − (1/2)(√3/2) = √3/4 − √3/4 = 0

   ✦ (ii) = 0

   The two expressions are the addition formulas for sin(A + C) and cos(A + C), which you meet in Class 11. Here A + C = 90°, so they come to sin 90° = 1 and cos 90° = 0.

   Where students slip. Taking ∠C = 60° but then using sin C = 1/2 by copying sin A's value. Look each ratio up separately — sin 60° and sin 30° are not the same number.

   Another way. Use the sides instead of the table. With opposite 1, adjacent √3 and hypotenuse 2 for angle A, the same two legs swap roles for angle C, and the arithmetic is identical.

10. 104 marksNCERT Cl-10 Maths, Ex 8.1, Q10

    In △PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.

    Hint. You cannot find PR and QR one at a time. Use Pythagoras to get a second relation between them, then solve the pair together.

    Step 1 — Write Pythagoras with the right angle at Q. PR² = PQ² + QR² PR² − QR² = PQ² = 25

    Step 2 — Factor the left side as a difference of squares. (PR − QR)(PR + QR) = 25

    Step 3 — Substitute the given sum. PR + QR = 25, so (PR − QR) × 25 = 25, giving PR − QR = 1

    This is the key move: factoring turns the quadratic relation into a linear one.

    Step 4 — Solve the pair. PR + QR = 25 PR − QR = 1 Adding: 2·PR = 26, so PR = 13 and QR = 12.

    Check. 5² + 12² = 25 + 144 = 169 = 13² ✓

    Step 5 — Read off the ratios at P. Opposite P is QR = 12, adjacent is PQ = 5, hypotenuse PR = 13. sin P = 12/13 cos P = 5/13 tan P = 12/5

    ✦ Answer: sin P = 12/13, cos P = 5/13, tan P = 12/5

    Where students slip. Guessing PR = 13 and QR = 12 from the familiar triple without justification. The numbers are right, but the marks are for deriving them — and a question with different data would defeat the guess.

    Another way. Substitute instead: let QR = x, so PR = 25 − x. Then (25 − x)² = 25 + x², which expands to 625 − 50x = 25, giving x = 12.

11. 115 marksNCERT Cl-10 Maths, Ex 8.1, Q11

    State whether each is true or false, and justify. (i) tan A is always less than 1. (ii) sec A = 12/5 for some value of angle A. (iii) cos A is the abbreviation used for the cosecant of angle A. (iv) cot A is the product of cot and A. (v) sin θ = 4/3 for some angle θ.

    Hint. For the false ones, a single counter-example is a complete justification. For the true ones, say why the value is attainable.

    (i) False. tan 60° = √3 ≈ 1.73, which is greater than 1. In fact tan A grows without bound as A approaches 90°, because the opposite side outgrows the adjacent one. A single counter-example settles it.

    (ii) True. sec A = 12/5 = 2.4. Secant is the reciprocal of cosine, and cos A can be any value from 0 to 1 for an acute angle, so sec A can be anything from 1 upwards. Since 2.4 ≥ 1, some acute angle has this secant — namely the angle whose cosine is 5/12.

    (iii) False. cos A is the abbreviation for the cosine of A. The cosecant is written cosec A. The two are different ratios: cos A = adjacent/hypotenuse, while cosec A = hypotenuse/opposite.

    (iv) False. cot A is a single indivisible symbol meaning the cotangent of the angle A. It is not cot multiplied by A — 'cot' on its own is not a number, so there is nothing to multiply. This also means you can never cancel the 'cot' from an expression.

    (v) False. sin θ = opposite/hypotenuse, and in a right triangle the hypotenuse is always the longest side. So the fraction can never exceed 1, whereas 4/3 ≈ 1.33 does.

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

    Where students slip. Giving the verdict without the reason. In a 'justify your answer' question the true/false word is worth almost nothing on its own — the counter-example or the range argument is what earns the mark.

    Another way. Parts (i), (ii) and (v) are all range questions. Committing one line to memory settles all three: sine and cosine live in [0, 1], secant and cosecant in [1, ∞), and tangent and cotangent range over all of [0, ∞) for acute angles.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (chapter 8 now runs to three exercises — 8.1, 8.2 and 8.3; trigonometric ratios of complementary angles were removed from this chapter, and the summary's six points make no mention of them). Questions are referenced from the NCERT textbook for identification.