Chapter 8: Introduction to Trigonometry Class 10 Maths NCERT Solutions

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Class 10 Maths NCERT Solutions - Chapter wise

1

Chapter 1: Real Numbers

2

Chapter 2: Polynomials

3

Chapter 3 : Pair of Linear Equations in Two Variables

4

Chapter 4 : Quadratic Equations

5

Chapter 5 : Arithmetic Progressions

6

Chapter 6 : Triangles

7

Chapter 7 : Coordinate Geometry

8

Chapter 8: Introduction to Trigonometry

9

Chapter 9: Some Applications of Trigonometry

10

Chapter 10 : Circles

11

Chapter 11 : Areas Related to Circles

12

Chapter 12: Surface Areas and Volumes

13

Chapter 13: Statistics

14

Chapter 14 : Probability

Chapter Notes

Chapter 8: Introduction to Trigonometry — Exercise 8.1

Trigonometry is the secret language of triangles! Mastering this exercise helps you understand how engineers and architects measure heights and distances without ever using a physical measuring tape.

Q1. In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A (ii) sin C, cos C

  • The Logic: In trigonometry, everything starts with the Pythagorean Theorem. First, we find the missing side (Hypotenuse). Then, we use the “Side Ratios.” Human Tip: Remember that “Opposite” and “Adjacent” sides swap places depending on whether you are looking from Angle A or Angle C!

  • The Steps:

    1. Find Hypotenuse (AC): AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625.

    2. AC = root(625) = 25 cm.

    3. For Angle A: Opposite = BC (7), Adjacent = AB (24), Hypotenuse = 25.

      • sin A = Opposite / Hypotenuse = 7 / 25

      • cos A = Adjacent / Hypotenuse = 24 / 25

    4. For Angle C: Opposite = AB (24), Adjacent = BC (7), Hypotenuse = 25.

      • sin C = Opposite / Hypotenuse = 24 / 25

      • cos C = Adjacent / Hypotenuse = 7 / 25

  • The Result: (i) sin A = 7/25, cos A = 24/25; (ii) sin C = 24/25, cos C = 7/25

Q2. In the given figure, find tan P – cot R.

  • The Logic: Just like Q1, find the missing side first. Pro-Tip: tan is (Opposite/Adjacent) and cot is (Adjacent/Opposite). They are basically mirrors of each other!

  • The Steps:

    1. In triangle PQR (right-angled at Q), PR = 13 cm, PQ = 12 cm.

    2. QR^2 = PR^2 – PQ^2 = 13^2 – 12^2 = 169 – 144 = 25.

    3. QR = 5 cm.

    4. tan P (looking from P) = Opposite/Adjacent = QR/PQ = 5/12.

    5. cot R (looking from R) = Adjacent/Opposite = QR/PQ = 5/12.

    6. tan P – cot R = 5/12 – 5/12 = 0.

  • The Result: tan P – cot R = 0

Q3. If sin A = 3/4, calculate cos A and tan A.

  • The Logic: Think of “3/4” as a ratio. Let Opposite = 3k and Hypotenuse = 4k. Use Pythagoras to find the Adjacent side.

  • The Steps:

    1. sin A = 3 / 4 (Opposite / Hypotenuse).

    2. Adjacent^2 = Hypotenuse^2 – Opposite^2 = 4^2 – 3^2 = 16 – 9 = 7.

    3. Adjacent = root 7.

    4. cos A = Adjacent / Hypotenuse = (root 7) / 4.

    5. tan A = Opposite / Adjacent = 3 / (root 7).

  • The Result: cos A = (root 7)/4 and tan A = 3/(root 7)

Q4. Given 15 cot A = 8, find sin A and sec A.

  • The Logic: Don’t let the 15 scare you. Just move it to the other side: cot A = 8 / 15. This gives you the ratio for (Adjacent / Opposite).

  • The Steps:

    1. cot A = 8 / 15. Let Adjacent = 8 and Opposite = 15.

    2. Hypotenuse^2 = 15^2 + 8^2 = 225 + 64 = 289.

    3. Hypotenuse = root(289) = 17.

    4. sin A = Opposite / Hypotenuse = 15 / 17.

    5. sec A = Hypotenuse / Adjacent = 17 / 8.

  • The Result: sin A = 15/17 and sec A = 17/8

Q5. Given sec theta = 13/12, calculate all other trigonometric ratios.

  • The Logic: You have the Hypotenuse (13) and the Adjacent (12). This is a classic “5-12-13” triplet! Find the Opposite side (5) and fill in the blanks.

  • The Steps:

    1. Opposite = root(13^2 – 12^2) = 5.

    2. sin theta = 5/13

    3. cos theta = 12/13 (reciprocal of sec)

    4. tan theta = 5/12

    5. cot theta = 12/5

    6. cosec theta = 13/5

  • The Result: sin = 5/13, cos = 12/13, tan = 5/12, cot = 12/5, cosec = 13/5

Student Summary Table (Trigonometry Cheat Sheet)

Ratio Side Relation Human Tip (SOH CAH TOA)
sin Opposite / Hypotenuse Some People Have
cos Adjacent / Hypotenuse Curly Brown Hair
tan Opposite / Adjacent Through Proper Brushing
cosec 1 / sin Flip the sin ratio
sec 1 / cos Flip the cos ratio
cot 1 / tan Flip the tan ratio

Q6. If Angle A and Angle B are acute angles such that cos A = cos B, then show that Angle A = Angle B.

  • The Logic: We use two right-angled triangles (or one big one) and the definition of cos (Adjacent/Hypotenuse). If the ratios are equal, it implies the sides are proportional, leading us to equal angles.

  • The Steps:

    1. Consider a right triangle ABC, right-angled at C.

    2. cos A = AC / AB (Adjacent to A is AC)

    3. cos B = BC / AB (Adjacent to B is BC)

    4. Given: cos A = cos B

    5. AC / AB = BC / AB

    6. This gives: AC = BC

    7. In a triangle, angles opposite to equal sides are always equal.

  • The Result: Therefore, Angle A = Angle B. (Proved)

Q7. If cot theta = 7 / 8, evaluate: (i) (1 + sin theta)(1 – sin theta) / (1 + cos theta)(1 – cos theta), (ii) cot^2 theta

  • The Logic: Part (i) looks scary, but it’s just an identity! Remember (a+b)(a-b) = a^2 – b^2.

  • The Steps:

    1. The expression simplifies to: (1 – sin^2 theta) / (1 – cos^2 theta).

    2. Using identities (which you’ll learn soon): $1 – sin^2 = cos^2$ and $1 – cos^2 = sin^2$.

    3. The whole thing becomes cos^2 / sin^2, which is just cot^2 theta!

    4. Given cot theta = 7/8, so cot^2 theta = (7/8)^2 = 49 / 64.

  • The Result: (i) 49/64, (ii) 49/64

Q8. If 3 ob A = 4, check whether (1 – tan^2 A) / (1 + tan^2 A) = cos^2 A – sin^2 A or not.

  • The Logic: First, write 3 cot A = 4 as cot A = 4 / 3. This means Adjacent = 4 and Opposite = 3. Use Pythagoras to find the Hypotenuse (it’s a 3-4-5 triangle!).

  • The Steps:

    1. Adjacent = 4, Opposite = 3, Hypotenuse = 5.

    2. tan A = 3/4, sin A = 3/5, cos A = 4/5.

    3. LHS: (1 – (3/4)^2) / (1 + (3/4)^2) = (1 – 9/16) / (1 + 9/16) = (7/16) / (25/16) = 7 / 25.

    4. RHS: (4/5)^2 – (3/5)^2 = 16/25 – 9/25 = 7 / 25.

  • The Result: Yes, LHS = RHS.

Q9. In triangle ABC, right-angled at B, if tan A = 1 / root 3, find the value of:

(i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C

  • The Logic: tan A = 1 / root 3 means Opposite = 1 and Adjacent = root 3.

  • The Steps:

    1. Hypotenuse = root[(root 3)^2 + 1^2] = root(3 + 1) = 2.

    2. For Angle A: sin A = 1/2, cos A = (root 3)/2.

    3. For Angle C: sin C = (root 3)/2, cos C = 1/2.

    4. (i) (1/2 * 1/2) + (root 3/2 * root 3/2) = 1/4 + 3/4 = 1.

    5. (ii) (root 3/2 * 1/2) – (1/2 * root 3/2) = root 3/4 – root 3/4 = 0.

  • The Result: (i) 1, (ii) 0

Q10. In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

  • The Logic: This is a “Variable” problem. Let QR = x, then PR = 25 – x. Use the Pythagorean theorem to solve for x.

  • The Steps:

    1. PQ^2 + QR^2 = PR^2

    2. 5^2 + x^2 = (25 – x)^2

    3. 25 + x^2 = 625 + x^2 – 50x

    4. 50x = 600 => x = 12.

    5. So, QR = 12 cm, PR = 25 – 12 = 13 cm, and PQ = 5 cm.

    6. sin P = 12/13, cos P = 5/13, tan P = 12/5.

  • The Result: sin P = 12/13, cos P = 5/13, tan P = 12/5

Q11. State whether the following are true or false. Justify your answer.

  • The Logic: Trigonometry values are strictly bound by the rules of right-angled triangles.

  • The Steps:

    1. The value of tan A is always less than 1: False. tan A = Opposite/Adjacent. The opposite side can definitely be longer than the adjacent side (e.g., tan 60 = 1.732).

    2. sec A = 12/5 for some value of angle A: True. sec = Hypotenuse/Adjacent. Since the hypotenuse (12) is longer than the adjacent side (5), this is possible.

    3. cos A is the abbreviation used for the cosecant of angle A: False. cos is for cosine; cosec is for cosecant.

    4. cot A is the product of cot and A: False. cot A is a single functional value; cot without an angle has no meaning.

    5. sin theta = 4/3 for some angle theta: False. sin = Opposite/Hypotenuse. The hypotenuse MUST be the longest side, so sin can never be greater than 1 (4/3 is 1.33).

Student Summary Table (Advanced Trig Tips)

Checkpoint Pro-Tip Why?
sin & cos Never > 1 Hypotenuse is the longest side.
sec & cosec Always >= 1 Hypotenuse is the numerator.
tan & cot Can be anything Opposite and Adjacent can be any length.
Reciprocals tan = 1/cot They are upside-down versions of each other.

Chapter 8: Introduction to Trigonometry — Exercise 8.2

Welcome to the “Value Table” masterclass! This exercise is a scoring goldmine because once you memorize the specific values for 0, 30, 45, 60, and 90 degrees, you can solve these problems like a human calculator.

Q1. Evaluate the following:

i) sin 60 cos 30 + sin 30 cos 60

  • The Logic: This is a direct substitution problem. We take the fixed values from our trigonometric table and plug them in. Pro-Tip: Notice how sin 60 and cos 30 are the same? That’s because they are “complementary”!

  • The Steps:

    • sin 60 = (root 3)/2

    • cos 30 = (root 3)/2

    • sin 30 = 1/2

    • cos 60 = 1/2

    • Putting it together: ((root 3)/2 * (root 3)/2) + (1/2 * 1/2)

    • = 3/4 + 1/4 = 4/4 = 1

  • The Result: 1

ii) 2 tan^2 45 + cos^2 30 – sin^2 60

  • The Logic: Square the values before multiplying. Human Tip: Since cos 30 and sin 60 are equal, their squares will cancel each other out if one is subtracted from the other!

  • The Steps:

    • tan 45 = 1, so tan^2 45 = 1^2 = 1

    • cos 30 = (root 3)/2, so cos^2 30 = 3/4

    • sin 60 = (root 3)/2, so sin^2 60 = 3/4

    • Calculation: 2(1) + 3/4 – 3/4 = 2

  • The Result: 2

Q2. Choose the correct option and justify your choice:

i) (2 tan 30) / (1 + tan^2 30)

  • The Logic: We simplify the expression to see which standard trigonometric value it matches.

  • The Steps:

    • tan 30 = 1 / (root 3)

    • Numerator: 2 * (1 / root 3) = 2 / (root 3)

    • Denominator: 1 + (1 / root 3)^2 = 1 + 1/3 = 4/3

    • Final: (2 / root 3) / (4/3) = (2 / root 3) * (3 / 4) = 3 / (2 root 3)

    • Rationalizing: (3 * root 3) / (2 * 3) = (root 3) / 2

  • The Result: sin 60 (Option A)

Q3. If tan (A + B) = root 3 and tan (A – B) = 1/root 3; 0 < A + B <= 90; A > B, find A and B.

  • The Logic: We convert the numbers back into angles. If tan (Something) = root 3, then that “Something” must be 60 degrees. This gives us two simple equations to solve.

  • The Steps:

    1. tan (A + B) = root 3 => A + B = 60 (Equation 1)

    2. tan (A – B) = 1/root 3 => A – B = 30 (Equation 2)

    3. Adding both equations: (A + B) + (A – B) = 60 + 30

    4. 2A = 90 => A = 45

    5. Substitute A in Eq 1: 45 + B = 60 => B = 15

  • The Result: A = 45 and B = 15

Q4. State whether the following are true or false. Justify your answer.

  • The Logic: Use the values from your table to test these statements.

  • The Steps:

    1. sin (A + B) = sin A + sin B: False. (Try A=30, B=60. sin 90 is 1, but sin 30 + sin 60 = 0.5 + 0.866 = 1.366).

    2. The value of sin theta increases as theta increases: True. (It goes from 0 at 0 degrees to 1 at 90 degrees).

    3. The value of cos theta increases as theta increases: False. (It actually decreases from 1 to 0).

    4. sin theta = cos theta for all values of theta: False. (This is only true at 45 degrees).

    5. cot A is not defined for A = 0: True. (cot 0 = 1 / tan 0 = 1 / 0, which is undefined).

Student Summary Table (Trigonometry Value Cheat Sheet)

Angle (theta) 0 30 45 60 90
sin theta 0 1/2 1/root 2 root 3/2 1
cos theta 1 root 3/2 1/root 2 1/2 0
tan theta 0 1/root 3 1 root 3 Undefined

Chapter 8: Introduction to Trigonometry — Exercise 8.3

Welcome to the “Grand Finale” of Trigonometry! This exercise is all about Trigonometric Identities—the powerful formulas that allow you to transform complex equations into simple ones, just like a math magician.

Q1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

  • The Logic: Think of this as a “translation” task. We use our three main identities to find a path from sin, sec, and tan until they all look like cot A.

  • The Steps:

    • For sin A: We know that cosec^2 A = 1 + cot^2 A. Since sin is the reciprocal of cosec, we can say sin^2 A = 1 / (1 + cot^2 A).

      Taking the square root: sin A = 1 / root(1 + cot^2 A).

    • For tan A: This is the easiest one! Tan and cot are direct reciprocals.

      tan A = 1 / cot A.

    • For sec A: We use the identity sec^2 A = 1 + tan^2 A.

      Substitute tan A with 1/cot A: sec^2 A = 1 + (1 / cot^2 A).

      Take the LCM: sec^2 A = (cot^2 A + 1) / cot^2 A.

      Taking the square root: sec A = root(cot^2 A + 1) / cot A.

  • The Result: sin A = 1/root(1+cot^2 A), tan A = 1/cot A, sec A = root(cot^2 A+1)/cot A

Q2. Write all the other trigonometric ratios of Angle A in terms of sec A.

  • The Logic: Similar to Q1, we want everything (sin, cos, tan, cosec, cot) to be expressed using only sec A.

  • The Steps:

    • cos A: Directly the reciprocal. cos A = 1 / sec A.

    • tan A: Use sec^2 A – tan^2 A = 1. So, tan^2 A = sec^2 A – 1.

      tan A = root(sec^2 A – 1).

    • sin A: Use sin A = tan A x cos A.

      sin A = root(sec^2 A – 1) / sec A.

    • cot A: Reciprocal of tan. cot A = 1 / root(sec^2 A – 1).

    • cosec A: Reciprocal of sin. cosec A = sec A / root(sec^2 A – 1).

  • The Result: All ratios are now expressed in terms of sec A.

Q3. Choose the correct option. Justify your choice.

(i) 9 sec^2 A – 9 tan^2 A =

  • The Logic: Look for a common factor and a familiar identity.

  • The Steps: Take 9 common: 9(sec^2 A – tan^2 A). We know sec^2 A – tan^2 A = 1.

    9(1) = 9.

  • The Result: (B) 9

(ii) (1 + tan theta + sec theta)(1 + cot theta – cosec theta) =

  • Pro-Tip: When in doubt, convert everything to sin and cos. It usually clears the path!

  • The Steps: After converting to (sin/cos) and taking LCM, the terms simplify beautifully using the (a+b)(a-b) identity and sin^2 + cos^2 = 1.

  • The Result: (C) 2

Q4. Prove the following identities…

(i) (cosec theta – cot theta)^2 = (1 – cos theta) / (1 + cos theta)

  • The Logic: Work on the Left Hand Side (LHS). Convert cosec and cot into sin and cos.

  • The Steps:

    1. LHS = (1/sin – cos/sin)^2 = [(1 – cos) / sin]^2

    2. LHS = (1 – cos)^2 / sin^2

    3. We know sin^2 = 1 – cos^2. So: (1 – cos)^2 / (1 – cos^2)

    4. The denominator (1 – cos^2) is (1 – cos)(1 + cos).

    5. One (1 – cos) cancels out.

  • The Result: (1 – cos theta) / (1 + cos theta). (Proved)

(ii) cos A / (1 + sin A) + (1 + sin A) / cos A = 2 sec A

  • The Logic: Take the LCM of the two fractions and simplify the numerator.

  • The Steps:

    1. Numerator becomes: cos^2 A + (1 + sin A)^2

    2. Expand: cos^2 A + 1 + sin^2 A + 2 sin A

    3. Since cos^2 + sin^2 = 1, we get: 1 + 1 + 2 sin A = 2 + 2 sin A

    4. Take 2 common: 2(1 + sin A).

    5. The (1 + sin A) in the numerator and denominator cancels out, leaving 2 / cos A.

  • The Result: 2 sec A. (Proved)

Student Summary Table (The Big Three Identities)

Standard Identity Variations for Solving Human Tip
sin^2 + cos^2 = 1 sin^2 = 1 – cos^2 Use this to swap between sin and cos.
1 + tan^2 = sec^2 sec^2 – tan^2 = 1 Great for “Choice” questions.
1 + cot^2 = cosec^2 cosec^2 – cot^2 = 1 Use when dealing with reciprocals.
The “Secret” Rule tan = sin / cos Convert to sin/cos if you get stuck!

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