Maths Class 10th Introduction to Trigonometry

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

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

Question 2.
In given figure, find tan P – cot R.
 

Solution:

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

Solution:

sin A = 3/4
sin A = BC/AC = sin A = 34
Let BC = 3k and AC = 4k

Question 4.
Given 15 cot A = 8, find sin A and sec A.
 

Solution:

Question 5.
Given sec θ =13/12, calculate all other trigonometric ratios.
 

Solution:

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

Solution:

Question 8.
If 3 cot A = 4, check whether 1−tan2A/1+tan2A= cos² A - sin² A or not.
 

Solution:

Question 9.
In triangle ABC, right angled at B, if tan A =1/√3, find the value of:

• (i) sin A cos C + cos A sin C
• (ii) cos A cos C - sin A sin C

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

Solution:

Question 11.
State whether the following statements 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.
   

Solution:

• (i) tan 60° = √3 , Since √3 > 1.(False)
• (ii) sec A is always ≥ 1.(True)
• (iii) cos A is the abbreviation for cosine A.(False)
• (iv) cot without ∠A is meaningless. (False)
• (v) sin θ can never be greater than 1.
• ∴ sin θ = P/H, hypotenuse is always greater than other two sides.(False)

Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.
Evaluate the following:

Solution:

Question 2.
Choose the correct option and justify your choice:

Solution

Question 3.
If tan (A + B) = √3 and tan (A - B) = 1√3; 0° < A + B ≤ 90°; A > B, find A and B.

Solution:
tan (A + B) = √3
⇒ tan (A + B) = tan 60°
⇒ A + B = 60° ..(i)
tan (A - B) = 1/√3
⇒ tan (A - B) = tan 30°
⇒ A - B = 30° ....(ii)
Adding equation (i) and (ii), we get
2A = 90° ⇒ A = 45°
From (i), 45° + B = 60° ⇒ B = 60° - 45° = 15°
Hence, ∠A = 45°, ∠B = 15°

Question 4.
State whether the following statements are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.

Solution:

Introduction to Trigonometry Ex 8.3

Question 1.

Solution:

Question 2.
Show that:

(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° - sin 38° sin 52° = 0

Solution:
(i) LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 23° tan (90° - 48°) tan (90° - 23°)
= tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° . 1tan48??/1tan23?
= 1 = RHS

(ii) LHS = cos 38° cos 52° - sin 38° sin 52°
= cos 38° cos (90° - 38°) - sin 38° sin (90° - 38°)
= cos 38° sin 38°- sin 38° cos 38° = 0 = RHS

Question 3.
If tan 2A = cot (A - 18°), where 2A is an acute angle, find the value of A.

Solution:
tan 2A = cot (A - 18°)
⇒ cot (90° - 2A) = cot (A - 18°) [∵cot (90° - θ) = tan θ]
⇒90° - 2A = A - 18° ⇒ 3A = 108° ⇒ A = 108°/3
∴ ∠ A = 36°

Question 4.
If tan A = cot B, prove that A + B = 90°.

Solution:
tan A = cot B ⇒ tan A = tan (90° - B) [ ∵ tan (90° - θ) = cot θ]
⇒ A = 90° - B ⇒ A + B = 90° Proved

Question 5.
If sec 4A = cosec (A - 20°), where 4A is an acute angle, find the value of A.

Solution:
sec 4A = cosec (A - 20°)
⇒ cosec (90° - 4A) = cosec (A - 20°) [cosec (90° - θ) = sec θ]
⇒ 90° - 4A = A - 20° ⇒ 5A = 110°
A = 110°/5
A = 22°
∴ ∠ A = 22°

Question 6.
If A, Band Care interior angles of a triangle ABC, then show that: sin (B+C/2) = cos A/2
 

Solution:

Question 7. Express sin 61° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution: sin 67° + cos 75° = sin (90° - 23°) + cos (90° - 15°) = cos 23° + sin 15°

Introduction to Trigonometry Ex 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
 

Solution:
From trigonometric identity, cosec² A – cot² A = 1, we get

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
 

Solution:
Since sin² A + cos² A = 1, therefore

Question 3.
Evaluate:
(i)

(ii) sin 25° cos 65°+ cos 25° sin 65°
Solution:

(ii) sin 25° cos 65° + cos 25° sin 65° = sin 25° cos (90° - 25°) + cos 25° sin (90° - 25°)
= sin 25° sin 25° + cos 25° cos 25°
= sin² 25° + cos² 25° = 1

Question 4.
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) (sec A + tan A) (1 – sin A) =
• (A) sec A
• (B) sin A
• (C) cosec A
• (D) cos A
• (iv) 1+tan2A1+cot2A
• (A) sec² A
• (B) -1
• (C) cot² A
• (D) tan² A

Solution:
(i) 9 sec² A – 9 tan² A = 9(sec² A – tan² A) = 9 x 1 = 9
Correct option is (B)

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Solution: