NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

Question

Our NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry is prepared by the subject matter expert in a very precise way to make it easier for students to learn and understand all the concepts. Experts have used methods like graphs, diagrams, shortcuts, equations, identities to make it intuitive for the students. An easy-to-understand and intuitive explanation of the solutions will students score good marks in their CBSE board examination.

In NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry we get to learn about many different terms and trigonometric identities. Some of the important topics mentioned here are:

• Sin (Sine), Cos (cosine), tan (tangent), and their uses in a triangle.
• Sec (secant), cosec (cosecant), cot (cotangent) and how are they related to sin, cos, and tan.
• Application of trigonometric ratios such as sin, cos, tan, sec, cosec, cot.
• Trigonometric ratios of some important angles such as 0֯, 30֯, 45֯, 60֯, 90֯.
• Sin θ = Perpendicular/Hypotenuse, cos θ = Base/Hypotenuse, tan θ = perpendicular/base.
• Trigonometric identities such as: Pythagorean Trigonometric Identities.
  • Sin² θ + cos² θ = 1
  • 1 + tan² θ = 1
  • Cot² θ + 1 = cosec² θ
• Reciprocal Trigonometric Identities
  • Sin θ = 1/Cosec θ or Cosec θ = 1/Sin θ
  • Cos θ = 1/Sec θ or Sec θ = 1/Cos θ
  • Tan θ = 1/Cot θ or Cot θ = 1/Tan θ
• Ratio Trigonometric Identities:
  • Tan θ = Sin θ/Cos θ
  • Cot θ = Cos θ/Sin θ
• Trigonometric Identities of Opposite Angles:
  • Sin (-θ) = – Sin θ
  • Cos (-θ) = Cos θ
  • Tan (-θ) = – Tan θ
  • Cot (-θ) = – Cot θ
  • Sec (-θ) = Sec θ
  • Cosec (-θ) = -Cosec θ
• Trigonometric Identities of Supplementary Angles
  • sin (180°- θ) = sinθ
  • cos (180°- θ) = -cos θ
  • cosec (180°- θ) = cosec θ
  • sec (180°- θ)= -sec θ
  • tan (180°- θ) = -tan
  • cot (180°- θ) = -cot θ

Our NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry provides all the required information in a very concise format for easy learning, recalling the concepts. One must opt for our NCERT solutions for last-minute preparation and revision on their examination. Our experts as Sarthaks advise one to completely go through the complete solution while preparing for all kinds of exams including JEE Mains, JEE Advance, NTSE, Olympiad, and other similar examinations.

Ace the exam, get started now.

Answer 1

NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

1. In ∆ABC right angled at B, AB = 24 cm, BC = 7 m. Determine 

(i) sin A, cos A 

(ii) sin C, cos C

Answer:

Applying Pythagoras theorem for ∆ABC, we obtain 

AC² = AB² + BC² 

= (24 cm)² + (7 cm)² 

= (576 + 49) cm² 

= 625 cm² 

∴ AC = \(\sqrt{625}\) cm = 25 cm

(i) 

sin A = \(\frac{Side \,opposite\,to\,\angle A}{Hypotenuse}= \frac{BC}{AC}= \frac{7}{25}\)

cos A = \(\frac{Side \,opposite\,to\,\angle A}{Hypotenuse}= \frac{AB}{AC}= \frac{24}{25}\)

(ii)

sin C = \(\frac{Side \,opposite\,to\,\angle C}{Hypotenuse}= \frac{AB}{AC}= \frac{24}{25}\)

cos C = \(\frac{Side \,opposite\,to\,\angle C}{Hypotenuse}= \frac{BC}{AC}= \frac{7}{25}\)

2. In the given figure find tan P − cot R

Answer:

Applying Pythagoras theorem for ∆PQR, we obtain 

PR² = PQ² + QR² 

(13 cm)² = (12 cm)² + QR² 

169 cm² = 144 cm² + QR² 

25 cm² = QR² 

QR = 5 cm

= \(\frac{5}{12}\)

= \(\frac{5}{12}\)

tan P − cot R = \(\frac{5}{12}\) - \(\frac{5}{12}\) = 0

3. If sin A = \(\frac{3}{4}\), calculate cos A and tan A.

Answer:

Let ∆ABC be a right-angled triangle, right-angled at point B.

Given that,

\(sin A = \frac{3}{4}\)

\(\frac{BC}{AC}= \frac{3}{4}\)

Let BC be 3k. 

Therefore, AC will be 4k, where k is a positive integer. 

Applying Pythagoras theorem in ∆ABC, we obtain 

AC² = AB² + BC² 

(4k)² = AB² + (3k)² 

16k² − 9k² = AB² 

7k² = AB²

AB = \(\sqrt{7k}\)

4. Given 15 cot A = 8. Find sin A and sec A.

Answer:

Consider a right-angled triangle, right-angled at B.

It is given that, Let AB be 8k.

Therefore, BC will be 15k, where k is a positive integer. 

Applying Pythagoras theorem in ∆ABC, we obtain 

AC² = AB² + BC² = (8k)² + (15k)² 

= 64k² + 225k² 

= 289k²

AC = 17k

5. Given sec θ = \(\frac{13}{12}\), calculate all other trigonometric ratios.

Answer:

Consider a right-angle triangle ∆ABC, right-angled at point B.

If AC is 13k, AB will be 12k, where k is a positive integer. 

Applying Pythagoras theorem in ∆ABC, we obtain 

(AC)² = (AB)² + (BC)² 

(13k)² = (12k)² + (BC)² 

169k² = 144k² + BC²

25k² = BC²

BC = 5k

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

Answer:

Let us consider a triangle ABC in which CD ⊥ AB.

It is given that 

cos A = cos B

We have to prove ∠A = ∠B. 

To prove this, let us extend AC to P such that BC = CP.

From equation (1), we obtain

By using the converse of B.P.T, 

CD || BP 

⇒ ∠ACD = ∠CPB (Corresponding angles) … (3) And, 

∠BCD = ∠CBP (Alternate interior angles) … (4) 

By construction, we have BC = CP. 

∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (5) 

From equations (3), (4), and (5), we obtain 

∠ACD = ∠BCD … (6) 

In ∆CAD and ∆CBD, 

∠ACD = ∠BCD [Using equation (6)] 

∠CDA = ∠CDB [Both 90°] 

Therefore, the remaining angles should be equal. 

∴ ∠CAD = ∠CBD 

⇒∠A = ∠B

Answer 2

7. If cot θ = 7/8, evaluate

(i) \(\frac{(1+sinθ)(1-sinθ )}{(1 +cos θ )(1-cosθ )}\)

(ii) cot² θ

Answer:

Let us consider a right triangle ABC, right-angled at point B.

If BC is 7k, then AB will be 8k, where k is a positive integer. 

Applying Pythagoras theorem in ∆ABC, we obtain 

AC² = AB² + BC² 

= (8k)² + (7k)² 

= 64k² + 49k² 

= 113k²

AC = \(\sqrt{113}k\)

(i)

(ii) cot² θ = (cot θ)² = \((\frac{7}{8})^2 =\frac{49}{64}\)

8. If 3 cot A = 4, Check whether \(\frac{1-tan^2 A}{1+tan^2 A}= cos^2 A- sin^2 A\) or not.

Answer:

It is given that 3cot A = 4 

Or, cot A = 4/3

Consider a right triangle ABC, right-angled at point B.

If AB is 4k, then BC will be 3k, where k is a positive integer. 

In ∆ABC, 

(AC)² = (AB)² + (BC)² 

= (4k)² + (3k)² 

= 16k² + 9k² 

= 25k² 

AC = 5k

9. In ∆ABC, right angled at B. If tan A = \(\frac{1}{\sqrt{3}}\), find the value of

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

(ii) cos A cos C − sin A sin C

Answer:

tan A = \(\frac{1}{\sqrt{3}}\) 

\(\frac{BC}{AB}=\frac{1}{\sqrt{3}}\)

If BC is k, then AB will be \(\sqrt{3}k\), where k is a positive integer. 

In ∆ABC, 

AC² = AB² + BC²

\(=(\sqrt{3}k)^2+(k)^2\)

= 3k² + k² = 4k²

∴ AC = 2k

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

(ii) cos A cos C − sin A sin C

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

Answer:

Given that, PR + QR = 25 

PQ = 5 

Let PR be x. 

Therefore, QR = 25 − x

Applying Pythagoras theorem in ∆PQR, we obtain 

PR² = PQ² + QR² 

x² = (5)² + (25 − x)² 

x² = 25 + 625 + x² − 50x 

50x = 650 

x = 13 

Therefore, PR = 13 cm 

QR = (25 − 13) cm = 12 cm

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

(i) The value of 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

Answer:

(i) Consider a ∆ABC, right-angled at B.

But 12/5 > 1 

∴ tan A > 1 

So, tan A < 1 is not always true. 

Hence, the given statement is false.

(ii) sec A = 12/5

Let AC be 12k, AB will be 5k, where k is a positive integer. 

Applying Pythagoras theorem in ∆ABC, we obtain 

AC² = AB² + BC² 

(12k)² = (5k)² + BC² 

144k² = 25k² + BC²

BC² = 119k²

BC = 119k 

It can be observed that for given two sides AC = 12k and AB = 5k, 

BC should be such that, 

AC − AB < BC < AC + AB 

12k − 5k < BC < 12k + 5k 

7k < BC < 17 k 

However, BC = 119k. Clearly, such a triangle is possible and hence, such value of sec A is possible. 

Hence, the given statement is true.

(iii) Abbreviation used for cosecant of angle A is cosec A. 

And cos A is the abbreviation used for cosine of angle A.

Hence, the given statement is false.

(iv) cot A is not the product of cot and A. It is the cotangent of ∠A. 

Hence, the given statement is false.

(v) sin θ = 4/3

We know that in a right-angled triangle,

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. 

Therefore, such value of sin θ is not possible. 

Hence, the given statement is false.

Answer 3

12. Evaluate the following 

(i) sin60° cos30° + sin30° cos 60° 

(ii) 2tan²45° + cos²30° − sin²60°

(iii) \(\cfrac{cos\,45°}{sec\,30°+ \,cosec\,30°}\)

(iv) \(\cfrac{sin\,30°+\,tan\,45°-\,cosec\,60°}{sec\,30°+\,cos\,60°+\,cot\,45°}\)

(v) \(\cfrac{5 \,cos^2\,60°+\,4\,sec^2\,30°-tan^2\,45°}{sin^2\,30°+\,cos^2\,30°}\) 

Answer:

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan²45° + cos²30° − sin²60°

 (iii) \(\cfrac{cos\,45°}{sec\,30°+ \,cosec\,30°}\)

(iv) \(\cfrac{sin\,30°+\,tan\,45°-\,cosec\,60°}{sec\,30°+\,cos\,60°+\,cot\,45°}\)

(v) \(\cfrac{5 \,cos^2\,60°+\,4\,sec^2\,30°-tan^2\,45°}{sin^2\,30°+\,cos^2\,30°}\) 

13. Choose the correct option and justify your choice.

(i) \(\frac{2\,tan\,30°}{1+\,tan^2\,30°}=\)

(A). sin60° 

(B). cos60° 

(C). tan60° 

(D). sin30°

(ii) \(\frac{1-tan^2\,45°}{1 \,+\,tan^2\,45°}=\)

(A). tan90° 

(B). 1 

(C). sin45° 

(D). 0

(iii) sin2A = 2sinA is true when A = 

(A). 0° 

(B). 30° 

(C). 45° 

(D). 60°

(iv) \(\frac{2\,tan\,30°}{1\,-\,tan^2\,30°}\)

(A). cos60° 

(B). sin60° 

(C). tan60° 

(D). sin30°

Answer:

 (i) \(\frac{2\,tan\,30°}{1+\,tan^2\,30°}\)

Out of the given alternatives, only sin60° = \(\frac{\sqrt{3}}{2}\)

Hence, (A) is correct.

Hence, (D) is correct. 

(iii) Out of the given alternatives, only A = 0° is correct.

As sin 2A = sin 0° = 0 2 

sinA = 2sin 0° = 2(0) = 0 

Hence, (A) is correct.

(iv) \(\frac{2\,tan\,30°}{1\,-\,tan^2\,30°}\)

Out of the given alternatives, only tan 60° = \(\sqrt{3}\)

Hence, (C) is correct.

14. If tan(A + B) = \(\sqrt{3}\) and tan(A - B) = \(\frac{1}{\sqrt{3}}\)

 0° < A + B ≤ 90°, A > B find A and B.

Answer:

 tan(A + B) = \(\sqrt{3}\) 

tan(A + B) = tan 60

⇒ A + B = 60 …………… (1)

tan (A - B) = \(\frac{1}{\sqrt{3}}\)

⇒tan (A − B) = tan30 

⇒A − B = 30 …………….. (2)

On adding both equations, we obtain 

2A = 90

⇒ A = 45

From equation (1), we obtain

45 + B = 60

B = 15

Therefore, ∠A = 45° and ∠B = 15°

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

(i) sin (A + B) = sin A + sin B 

(ii) The value of sinθ increases as θ increases 

(iii) The value of cos θ increases as θ increases 

(iv) sinθ = cos θ for all values of θ (v) cot A is not defined for A = 0°

Answer:

(i) sin (A + B) = sin A + sin B 

Let A = 30° and B = 60° 

sin (A + B) = sin (30° + 60°) 

= sin 90° = 1 

And sin A + sin B = sin 30° + sin 60°

Clearly, sin (A + B) ≠ sin A + sin B

Hence, the given statement is false.

(ii) The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as sin 0° = 0

sin 90° = 1 

Hence, the given statement is true.

(iii) cos 0° = 1

cos90° = 0 

It can be observed that the value of cos θ does not increase in the interval of 0°<θ<90°. 

Hence, the given statement is false.

(iv) sin θ = cos θ for all values of θ. 

This is true when θ = 45°

It is not true for all other values of θ.

Hence, the given statement is false.

(v) cot A is not defined for A = 0°

Hence, the given statement is true.

Answer 4

16. Evaluate 

(I) \(\frac{sin\,18°}{cos\,72°}\)

(II) \(\frac{tan\,26°}{cot\,64°}\)

(III) cos 48° − sin 42°

(IV) cosec 31° − sec 59°

Answer:

 (I) \(\frac{sin\,18°}{cos\,72°}\)

 (II) \(\frac{tan\,26°}{cot\,64°}\)

(III) cos 48° − sin 42° = cos (90°− 42°) − sin 42°

= sin 42° − sin 42° 

= 0 

(iv) cosec 31° − sec 59° = cosec (90° − 59°) − sec 59°

= sec 59° − sec 59° 

= 0

17. Show that 

(I) tan 48° tan 23° tan 42° tan 67° = 1 

(II) cos 38° cos 52° − sin 38° sin 52° = 0

Answer:

(I) tan 48° tan 23° tan 42° tan 67°

= tan (90° − 42°) tan (90° − 67°) tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67° 

= (cot 42° tan 42°) (cot 67° tan 67°) 

= (1) (1) 

= 1 

(II) cos 38° cos 52° − sin 38° sin 52°

= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52° 

= sin 52° sin 38° − sin 38° sin 52° 

= 0

18. If tan 2A = cot (A− 18°), where 2A is an acute angle, find the value of A.

Answer:

Given that, tan 2A = cot (A− 18°) 

cot (90° − 2A) = cot (A −18°) 

90° − 2A = A− 18° 

108° = 3A 

A = 36°

19. If tan A = cot B, prove that A + B = 90°

Answer:

Given that, tan A = cot B

tan A = tan (90° − B) 

A = 90° − B 

A + B = 90°

20. If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.

Answer:

Given that, sec 4A = cosec (A − 20°) 

cosec (90° − 4A) = cosec (A − 20°) 

90° − 4A= A− 20° 

110° = 5A

 A = 22°

21. If A, Band C are interior angles of a triangle ABC then show that \(sin(\frac{B+C}{2} )= cos\frac{A}{2}\)

Answer:

We know that for a triangle ABC,

∠ A + ∠B + ∠C = 180° 

∠B + ∠C= 180° − ∠A

22. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Answer:

sin 67° + cos 75° 

= sin (90° − 23°) + cos (90° − 15°) 

= cos 23° + sin 15°'

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

Answer:

We know that,

Therefore,

We know that, tan A = \(\frac{sin\,A}{cos\,A}\)

However, cot A = \(\frac{cos\,A}{sin\,A}\)

Therefore, tan A = \(\frac{1}{cot\,A}\)

Also, sec² A = 1 + tan² A

24. Write all the other trigonometric ratios of ∠A in terms of sec A.

Answer:

We know that,

cos A = \(\frac{1}{sec\,A}\)

Also, sin² A + cos² A = 1 

sin² A = 1 − cos² A

tan²A + 1 = sec²A 

tan²A = sec²A − 1

25. Evaluate

(i) \(\cfrac{sin^2\,63° +\,sin^2\,27°}{cos^2\,17°+\,cos^2\,73°}\)

(ii) sin25° cos65° + cos25° sin65°

Answer:

 (i) \(\cfrac{sin^2\,63° +\,sin^2\,27°}{cos^2\,17°+\,cos^2\,73°}\)

(ii) sin25° cos65° + cos25° sin65°

= sin²25° + cos²25° 

= 1 (As sin²A + cos²A = 1)

Answer 5

26. Choose the correct option. Justify your choice. 

(i) 9 sec² A − 9 tan² A = 

(A) 1 

(B) 9 

(C) 8 

(D) 0 

(ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ) 

(A) 0 

(B)1 

(C) 2 

(D) −1 

(iii) (secA + tanA) (1 − sinA) = 

(A) secA 

(B) sinA 

(C) cosecA 

(D) cosA 

(iv) \(\frac{1 \,+\,tan^2 A}{1\,+\,cot^2 A}\)

(A) sec² A 

(B) −1 

(C) cot2 A 

(D) tan2 A

Answer:

(i) 9 sec²A − 9 tan²A 

= 9 (sec²A − tan²A) 

= 9 (1) [As sec² A − tan² A = 1] 

= 9 

Hence, alternative (B) is correct.

(ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ)

Hence, alternative (C) is correct.

(iii) (secA + tanA) (1 − sinA)

= cosA 

Hence, alternative (D) is correct.

 (iv) \(\frac{1 \,+\,tan^2 A}{1\,+\,cot^2 A}\)

Hence, alternative (D) is correct.

27. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer:

= secθ cosec θ + 1 = R.H.S.

= cosec A + cot A 

= R.H.S

Hence, L.H.S = R.H.S