Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions Answers

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Extra Questions for Class 10 Maths Introduction to Trigonometry with Answers Solutions

Extra Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Solutions Answers

Introduction to Trigonometry Class 10 Extra Questions Objective Type

Question 1.
The value of \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\) will be:
(i) sin 60°
(ii) cos 60°
(iii) tan 60°
(iv) sin 30°
Answer:
(ii) cos 60°
Solution.

= 2 sin 30o. cos 30°
= sin 60°
Hence, Choice (ii) is correct

Question 2.
If sin θ = 1, the value of sin 20 will be :
(i) -1
(ii) 0
(iii) 1
(iv) 2
Answer:
(ii) 0

Question 3.
The value of cot (- 1470°) will be :
(i) –\(\frac{1}{\sqrt{3}}\)
(ii) – √3
(iii) \(\frac{1}{\sqrt{3}}\)
(iv) √3
Answer:
(ii) – √3

Question 4.
In triangle ABC, the value of sin (B + C) will be :
(i) sin B + sin C
(ii) sin A
(iii) 0
(iv) cos A
Answer:
(ii) sin A

Question 5.
The value of cos 15° is :
(i)
(ii)
(iii)
(iv)
Answer:
(i)

Question 6.
If sin θ = cosec θ and ≤ θ ≤ π then value of 0 will be :
(i) π
(ii) \(\frac{\pi}{2}\)
(iii) \(\frac{\pi}{4}\)
(iv) 0
Answer:
(ii) \(\frac{\pi}{2}\)

Question 7.
The value of cosec 810 will be :
(i) -1
(ii) 0
(iii) 1
(iv) ∞
Answer:
(iii) 1

Question 8.
Find the value of \(\frac{\sin ^{2} 15^{\circ}-\cos ^{2} 15^{\circ}}{\sin ^{2} 15^{\circ}+\cos ^{2} 15^{\circ}}\).
(i) 1
(ii) -1
(iii) \(\frac{2}{\sqrt{3}}\)
(iv) \(\frac{-\sqrt{3}}{2}\)
Answer:
(iv) \(\frac{-\sqrt{3}}{2}\)

Question 9.
The value of will be :
(i) \(\frac{1}{\sqrt{2}}\)
(ii) √2 – 1
(iii) √2
(iv) \(\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
Answer:
(iii) √2

Question 10.
The value of sin 570° will be :
(i) \(\frac{-1}{2}\)
(ii) \(\frac{-\sqrt{3}}{2}\)
(iii) \(\frac{1}{2}\)
(iv) \(\frac{\sqrt{3}}{2}\)
Answer:
(i) \(\frac{-1}{2}\)

Question 11.
If tan A = \(\frac{1}{\sqrt{3}}\) and tan B = √3 then the value of tan (A + B) will be :
(i) 0
(ii) \(\frac{1}{\sqrt{3}}\)
(iii) 1
(iv) ∞
Answer:
(iv) ∞

Question 12.
If sin α = cos α then the value of α will be:
(i) 30°
(ii) 45°
(iii) 60°
(iv) 90°
Answer:
(ii) 45°

Question 13.
The value of sin2 θ +\(\frac{1}{1+\tan ^{2} \theta}\) is :
(i) cos² θ
(ii) sin² θ
(iii) 1
(iv) sec² θ
Answer:
(iii) 1

Question 14.
If sec θ = 2 then the value of θ will be :
(i) \(\frac{\pi}{2}\)
(ii) \(\frac{\pi}{3}\)
(iii) \(\frac{\pi}{4}\)
(iv) \(\frac{\pi}{6}\)
Answer:
(ii) \(\frac{\pi}{3}\)

Question 15.
The value of cos (-1920°) will be :
(i) – 1/2
(ii) 0
(iii) 1/2
(iv) 1
Answer:
(i) – 1/2

Question 16.
If 2 cos 30 = 1. Then the value of a will be :
(i) 30°
(ii) 45°
(iii) 60°
(iv) 20°
Answer:
(iv) 20°

Question 17.
The value of \(\frac{\sin 20^{\circ}}{\cos 70^{\circ}}\) will be :
(i) more than 1
(ii) 1
(iii) 0
(iv) less than 1
Answer:
(ii) 1

Question 18.
The value of sin (- 570°) is :
(i) –\(\frac{\sqrt{3}}{2}\)
(ii) \(\frac{\sqrt{3}}{2}\)
(iii) \(\frac{1}{2}\)
(iv) –\(\frac{1}{2}\)
Answer:
(iii) \(\frac{1}{2}\)

Question 19.
The value of sin 840° will be:
(i) \(\frac{1}{2}\)
(ii) –\(\frac{1}{2}\)
(iii) \(\frac{\sqrt{3}}{2}\)
(iv) –\(\frac{\sqrt{3}}{2}\)
Answer:
(iii) \(\frac{\sqrt{3}}{2}\)

Question 20.
The value of sin 3270°
(i) \(\frac{1}{\sqrt{2}}\)
(ii) \(\frac{1}{3}\)
(iii) \(\frac{1}{2}\)
(iv) \(\frac{\sqrt{3}}{2}\)
Answer:
(iii) \(\frac{1}{2}\)

Question 21.
The value of cos 2940° is :
(i) \(\frac{1}{\sqrt{3}}\)
(ii) \(\frac{1}{2}\)
(iii) \(\frac{1}{3}\)
(iv) \(\frac{\sqrt{3}}{2}\)
Answer:
(ii) \(\frac{1}{2}\)

Question 22.
The value of cos² 61° + cos² 29° will
(i) 0
(ii) 1
(iii) -1
(iv) 2
Answer:
(ii) 1

Question 23.
If sin θ = \(\frac{\sqrt{3}}{2}\) and 0° < θ < 90°, then the value of tan 2θ will be :
(i) -√3
(ii) –\(\frac{1}{\sqrt{3}}\)
(iii) \(\frac{1}{\sqrt{3}}\)
(iv) √3
Answer:
(i) -√3

Question 24.
The value of cos 2A is :
(i) cos² A – sin²A
(ii) 1 – 2 cos²A
(iii) cos²A + sin²A
(iv) 2 sin² A – 1.
Answer:
(i) cos² A – sin²A

Question 25.
The palue of sin 2A is :
(i) \(\frac{1+\tan ^{2} \mathrm{A}}{1-\tan ^{2} \mathrm{A}}\)
(ii) \(\frac{1-\tan ^{2} \mathrm{A}}{1+\tan ^{2} \mathrm{A}}\)
(iii) \(\frac{2 \tan \mathrm{A}}{1-\tan ^{2} \mathrm{A}}\)
(iv) \(\frac{2 \tan \mathrm{A}}{1+\tan ^{2} \mathrm{A}}\)
Answer:
(iv) \(\frac{2 \tan \mathrm{A}}{1+\tan ^{2} \mathrm{A}}\)

Question 26.
The value of cos (-405°) is :
(i) –\(\frac{1}{\sqrt{2}}\)
(ii) \(\frac{1}{\sqrt{2}}\)
(iii) \(\frac{1}{2}\)
(iv) –\(\frac{1}{2}\)
Answer:
(ii) \(\frac{1}{\sqrt{2}}\)

Question 27.
If sin θ = \(\frac{\sqrt{3}}{2}\) and 0° < θ < 90°, the value of cot 2θ will be :
(i) –\(\frac{1}{\sqrt{3}}\)
(ii) -√3
(iii) \(\frac{1}{\sqrt{3}}\)
(iv) √3
Answer:
(i) –\(\frac{1}{\sqrt{3}}\)

Question 28.
The value will be :
(i) -1
(ii) 0
(iii) 1
(iv) 3
Answer:
(i) -1

Question 29.
The value will be :
(i) -∞
(ii) -1
(iii) +1
(iv) ∞
Answer:
(ii) -1

Question 30.
The value of sec 70° sin 20° – cos 20° . cosec 70° will be :
(i) -1
(ii) 0
(iii) 1
(iv) infinity.
Answer:
(ii) 0

Question 31.
The value of sin 12° cos 78° + sin 78° cos 12° will be:
(i) 2
(ii) 1
(iii) 0
(iv) -1.
Answer:
(ii) 1

Question 32.
The maximum value of sin x for the value of x is :
(i) x = \(\frac{\pi}{4}\)
(ii) x = \(\frac{\pi}{2}\)
(iii) x = π
(iv) x = \(\frac{3 \pi}{2}\)
Answer:
(ii) x = \(\frac{\pi}{2}\)

Question 33.
Value of \(\frac{\sin 31^{\circ}}{\cos 59^{\circ}}\) will be :
(i) -1
(ii) 0
(iii) 1
(iv) 2.
Answer:
(iii) 1

Question 34.
If cosθ = \(\frac{1}{2}\), the value of cosec²θ is :
(i) \(\frac{1}{2}\)
(ii) \(\frac{\sqrt{3}}{2}\)
(iii) \(\frac{3}{4}\)
(iv) \(\frac{4}{3}\)
Answer:
(iv) \(\frac{4}{3}\)

Question 35.
The value of sec θ cosec θ tan θ is :
(i) sec²θ
(ii) cosec²θ
(iii) cos²θ
(iv) cosθ.
Answer:
(i) sec²θ

Question 36.
The value of cos 240° is :
(i) –\(\frac{\sqrt{3}}{2}\)
(ii) –\(\frac{1}{2}\)
(iii) \(\frac{1}{2}\)
(iv) \(\frac{\sqrt{3}}{2}\)
Answer:
(ii) –\(\frac{1}{2}\)

Question 37.
The value of \(\frac{2 \tan 15^{\circ}}{1+\tan ^{2} 15^{\circ}}\) is :
(i) \(\frac{\sqrt{3}}{2}\)
(ii) \(\frac{1}{\sqrt{2}}\)
(iii) \(\frac{1}{2}\)
(iv) \(\frac{1}{\sqrt{3}}\)
Answer:
(iii) \(\frac{1}{2}\)

Question 38.
The value of  is :
(i) \(\frac{1}{2}\)
(ii) \(\frac{\sqrt{3}}{2}\)
(iii) \(\frac{2}{\sqrt{3}}\)
(iv) \(\frac{\sqrt{2}}{1}\)
Answer:
(iii) \(\frac{2}{\sqrt{3}}\)

Question 39.
The value of sin (-300°) will be :
(i) \(\frac{\sqrt{3}}{2}\)
(ii) –\(\frac{\sqrt{3}}{2}\)
(iii) \(\frac{1}{2}\)
(iv) –\(\frac{1}{2}\)
Answer:
(i) \(\frac{\sqrt{3}}{2}\)

Question 40.
The value of sin (-1080°) will be:
(i) -1
(ii) 0
(iii) 1
(iv) ∞
Answer:
(ii) 0

Question 41.
The measure of the angle subtended at the centre of a circle of radius 10 min radian by the arc of measure 5π m will be :
(i) \(\frac{\pi}{2}\)
(ii) \(\frac{\pi}{4}\)
(iii) \(\frac{\pi}{5}\)
(iv) \(\frac{\pi}{10}\)
Answer:
(i) \(\frac{\pi}{2}\)

Question 42.
If tan α = sin α, then the value of α will be:
(i) 90°
(ii) 60°
(iii) 45°
(iv) 0°
Answer:
(iv) 0°

Question 43.
The value of 9 sec² θ – 9 tan² θ is :
(i) 1
(ii) 8
(iii) 9
(iv) 10
Answer:
(iii) 9
Solution.
9 sec² θ – 9 tan² θ = 9 ( sec² θ – tan² θ ) = 9 × 1 = 9
Hence, choice (iii) is correct

Question 44.
It tan θ = \(\frac{a}{b}\), then the value of \(\frac{b \sin \theta-a \cos \theta}{b \sin \theta+a \cos \theta}\) will be :
(i) 1
(ii) \(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\)
(iii) \(\frac{b^{2}-a^{2}}{b^{2}+a^{2}}\)
(iv) 0
Answer:
(iv) 0

Question 45.
The value of \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30}\) will be
(i) cos 60°
(ii) sin 60°
(iii) tan 60°
(iv) cot 60°
Answer:
(iii) tan 60°
Solution.

= √3 = tan 60°
Hence choice (iii) is correct. (iii)

Introduction to Trigonometry Class 10 Extra Questions Very Short Answer Type

Question 1.
Find the value of tan \(\left(\frac{11 \pi}{6}\right)\) :
Solution :
tan \(\left(\frac{11 \times 180}{6}\right)\)
= tan (11 × 30)
= tan 330°
= tan (360° – 30°)
= tan 30°
= – \(\frac{1}{\sqrt{3}}\)

Question 2.
If 3x₁ = cosec θ and \(\frac{3}{x_{2}}\) = cot θ find the value of
Solution.
Given, 3x₁ = cosec θ
∴ x₁ = \(\frac{1}{3}\) cosec θ

Question 3.
Prove that

Solution.

Question 4.
Find the value of \(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\).
Solution.

Question 5.
If cos A = \(\frac{\sqrt{3}}{2}\) , then find the value of sin 2A.
Solution.
Given, cos A = \(\frac{\sqrt{3}}{2}\) = cos 30°
∴ A = 30°
sin 2A = sin 2 × 30° = sin 60° = \(\frac{\sqrt{3}}{2}\)

Question 6.
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°) [as cot (90 – θ) = tan θ]
⇒ 90° – 2A = A – 18°
⇒ 3A = 90° + 18° = 108°
∴ A = \(\frac{108}{3}\) = 36°

Question 7.
Solve the equation

Solution.

∴ θ = 60°

Question 8.
Prove that sin⁴ θ – cos⁴ θ = 2 sin² θ – 1.
Solution.
L.H.S. = sin⁴ θ – cos⁴ θ
= (sin² θ)² – (cos² θ)²
= (sin² θ + cos² θ) (sin² – cos² θ)
= 1[sin² – (1 – sin a )]
= 2 sin² θ – 1 = R.H.S.

Question 9.
If cosec A = 2, Find the value of
Solution.
Given, cosec A = 2 = cosec 30°
∴ A = 30° sin A


= √3 + 2 – √3
= 2

Question 10.
In ∆ABC, Prove :

Solution.
In ∆ABC
A + B + C = 180°(angle sum property of ∆)
= A + B = 180° – C

[∴ cos (90° – θ) = sin θ]

Question 11.
In a right ∆ABC, right angled at point C. If tan A = 1 prove that 2 sin A cos A = 1.

Solution.
tan A = 1 = \(\frac{BC}{CA}\)
∴ CA = BC = x (let)
By pythagoras theorem
BA² = BC² + CA²
= x² + x² = 2x²
∴ BA = x √2

∴ L.H.S. = 2. sin A cos A
= 2.\(\frac{1}{\sqrt{2}}\) × \(\frac{1}{\sqrt{2}}\)
= 1 = R.H.S

Question 12.
Prove


Solution.
(i) R.H.S. = (sec θ – tan θ)²

(ii) R.H.S. (cosec θ + cot θ)²

Question 13.
Prove (1 – sin θ) (1 + sin θ) (1 + tan² θ) = 1
Solution.
L.H.S. = (1 – sin θ) (1 + sin θ) (1 + tan² θ)
= (1 – sin² θ) (sec² θ)
[∵ 1 – sin² = cos² θ, 1 + tan² θ = sec² θ]
= cos² θ × sec² θ
= cos²θ × \(\frac{1}{\cos ^{2} \theta}\)
= 1 = R.H.S.

Question 14.
Prove : (1 + cot θ + tan θ) (sin θ – cos θ)
=
Solution.
L.H.S. = (1 + cot θ +, tan θ) (sin θ – cos θ)

Question 15.
Prove that:
cot A – cot 2A = cosec 2A.
Solution :
L.H.S. = cot A – cot 2A

= cosec 2A = R.H.S.

Question 16.
Find the value of :
(i) Find the value of top \(\frac{13 \pi}{6}\)
(ii) Find the value of sec \(\frac{23 \pi}{4}\)
(iii) Find the value of cot \(\frac{13 \pi}{3}\)
(iv) Find the value of (cos 75° + cos 15°).
Solution.

Question 17.
Prove that :

Solution :
L. H. S.

= R.H.S.

Question 18.
If cos (-840°) = –\(\frac{1}{2}\) then find the value of sin (-840°).
Solution.
cos (- 840°) = cos (840°)
= cos (120°) = –\(\frac{1}{2}\)
Hence
sin (-840°) = – sin 120° = \(\frac{-\sqrt{3}}{2}\)

Question 19.
If sin θ = \(\frac{4}{5}\), then find the value of cos 2θ.
Solution.
sin θ = \(\frac{4}{5}\)
cos 2θ = 1 – 2 sin²θ
= 1 – 2(\(\frac{4}{5}\))² = –\(\frac{7}{25}\)

Question 20.
Find the value of tan 35° tan 40° tan 45° tan 50° tan 55°
Solution.
tan 35o tan 40° tan 45o tan 50° tan 55°
= tan (90° – 55°) tan (90° – 50°) .1.tan 50° tan 55°
= cot 55° cot 50°.1.tan 50° tan 55°
= 1.

Question 21.
Prove that :

Solution :
R. H. S.

Question 22.
Prove.

Solution.
L. H. S.

Question 23.
Prove that : (cos A + cos B)² + (sin A + sin B)²

Solution :
L.H.S.
cos² A + cos² B + 2 cos A cos B
+ sin² A + sin² B + 2 sinA sinB
⇒ 1 + 1 + 2 (cos A cos B + sin A sin B)
⇒ 2 + 2 cos (A – B)
⇒ 2 [1 + cos (A – B)]

Question 24.
Prove the following:
(i)
(ii)
(iii)
(iv) Find the value of

(v) Find the value of

(vi) Find the value of sin 3A in terms of sin A. Hence find the value of sin 135°. If A = 45°.
(vii) If sin θ = \(\frac{3}{25}\), find the value of sin 2θ
Solution.
(i)

(ii)

(iii)

(iv)

(v)

(vi) sin 3A = sin(2A + A)
= sin 2A. cos A + cos 2A.sin A
= 2sin A.cos A. cos A + (1 – 2 sin²A).sin A
= 2 sin A(1 – sin²A) + sin A – 2 sin³A
= 2 sin A – 2 sin³A + sin A – 2 sin³A
= 3 sin A – 4 sin³A.
∴ sin 135° = sin(3 × 45°)
= 3 sin 45° – 4 sin³ 45°

(vii) As sin 2 θ = 2 sin θ . cos θ

Question 25.
Prove that : sec θ (1 – sin θ) (sec θ + tan θ) = 1
Solution :
L.H.S.

Question 26.
Prove that :
cos 33°cos 27 – cos 57°cos 63° = \(\frac{1}{2}\)
Solution :
L.H.S.
cos 33° cos 27° – cos (90°-33°) cos (90° – 27°)
∴ cos 33° cos 27° – sin 33° sin27°
∴ cos (33° + 27°) = cos (60°)
= \(\frac{1}{2}\)

Question 27.
Find the value of :
cos 80o.cos 70° – cos 10° cos 20°.
Solution.
cos 80° cos 70° – cos 10° cos 20°
= cos(90° – 10°) cos(90° – 20°) – cos 10° cos 20°
= sin 10° sin 20° – cos 10° cos 20°
= – (cos 10° cos 20° – sin 10° sin 20°)
= -cos(10° + 20°)
= -cos 30° = –\(\frac{-\sqrt{3}}{2}\).

Question 28.
Prove :

Solution.

L.H.S. sin 5A – 2 sin 3A + sin A cos 5A – 2 cos 3A + cos A
sin 5A +sin A – 2 sin 3A , cos 5A + cos A – 2 cos 3A
2sin 3A cos 2A – 2sin 3A 2 cos 3A cos 2A – 2 cos 3A 2sin 3A(cos 2A – 1)
2cos 3A (cos 2A – 1) = tan 3A = R.H.S.

Question 29.
Prove cos² π/8 + cos²3π + cos² 5π/8 + cos² 7π/8 = 2.
Solution :

1 + 1 = 2 = R.H.S.

Question 30.
Prove :
cos 4A = 1 – 8 sin²A + 8 sin⁴A
Solution :
L. H. S.
cos 4A = cos 2(2A)
= 1 – 2 sinθ 2A
= 1 – 2 (2 sin A cosA)²
= 1 – 8 sin² A cos²A
= 1 – 8 sin – A (1 – sin² A)
= 1 – 8 sin² A + 8 sin⁴A
= R.H.S.

Introduction to Trigonometry Class 10 Extra Questions Short Answer Type

Question 1.
Prove that :

Solution :
L.H.S.

= 1 + 1 = 2 = R.H.S.

Question 2.
Prove that:

Solution :
L. H. S.

= 2 × cosec A = R.H.S.

Question 3.
Prove that :
\(\frac{\sin 8 A}{\sin A}\) = 8 cos A cos 2A cos 4A
Solution :

= 8 cos A cos 2A cos 4A = R.H.S

Question 4.
Prove that : (1 – sin θ)(1 + sin θ)( 1+ tan2θ)= 1
Solution :
L.H.S.
= (1- sin A)(1 + sin )( 1+ tan2θ)
= (1 – sin² θ) × sec²θ
= cos²θ × \(\frac{1}{\cos ^{2} \theta}\) = 1
= R.H.S.

Question 5.
Prove that : sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A
Solution.

= sec A + cosec A R.H.S.

Question 6.
Find the value of

Solution :
(cos 15° + sin 15°)

Question 7.
Prove that : 16 cos 20° cos 40° cos 60° cos 80° = 1
Solution:
L.H.S.= 16 cos 20° cos 40° cos 60° cos 80°
= 8 × \(\frac{1}{2}\) (2 cos 80° cos 40°) cos 20°
= 4 (cos 120° cos 40°) cos 200 = 2. 2. – cos 20° + cos 40ocos 2007
= 2 [-cos 20° + cos 60° + cos 20°]
= 2 × \(\frac{1}{2}\) = 1 = R.H.S.

Question 8.
Prove that:

Solution :
L.H.S.
= sec²A – sec²A – tan²A × tan²A
= sec⁴ A – tan⁴A
= (sec²A -tan²A)(sec²A + tan²A)
= 1 (1 + tan²A + tan²A)
= 1 + 2 tan2A = R.H.S.

Question 9.
Prove that :

Solution :
L.H.S. =

= 2 cot θ = R.H.S.

Question 10.
Prove that :
(i) If 2 tan P= 3 tan Q, then prove that:
tan(P – Q) = \(\frac{\sin 2 Q}{5-\cos 2 Q}\)
(ii) If cos A + sin A = √2 cos A, prove that cos A – sin A = √2 sin A . sin 63° + cos 630

Solution.

(ii) Given cos A + sin A = √2cos A
∴ sin A = (√2 – 1) cos A Now, multiply both sides by (√2 +1)
∴ (√2 + 1)sin A = (√2 + 1) (√2 – 1) cosA
= √2 sin A + sin A = cos A
Hence, √2sin A = cos A – sin A.

(iii)

(iv) L.H.S.

Question 11.
If cosec A = \(\frac{17}{15}\), then find the value of sec A.
Solution.
cosec A = \(\frac{17}{15}\)
sin A = \(\frac{15}{17}\)
cos A = \(\frac{8}{17}\)
∵ sec A = \(\frac{17}{8}\)

Question 12.
Prove that

(i) θ, (ii) cos² θ + cos² θ.cot² θ = cot² θ
Solution.

(ii) L.H.S. = cos² θ + cos² x cot² θ
= cos² θ[1 + cot² θ]
= cos² θ × cosec²θ
= cos θ × \(\frac{1}{\sin ^{2} \theta}\) = cot² θ = R.H.S.

Question 13.
Solve the equation :

Solution.

Question 14.
If (A + B) = 45°, then prove that (1 + tan A) (1 + tan B) = 2
Solution.
A + B = 45°
⇒ tan(A + B) = tan 45o
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = 1
⇒ tan A + tan B = 1 – tan A tan B
⇒ tan A + tan B + tan A tan B = 1
⇒ 1 + tan A + tan B + tan A tan B = 1 + 1
(1 + tan A) + tan B(1 + tan A) = 2
(1 + tan A)(1+tan B) = 2

Question 15.
Prove that tan 75° = 2 + √3.
Solution.
tan 75°
= tan (45° + 30°)

Question 16.
Prove that

Solution.

Question 17.
Find the value of \(\frac{\sin 65^{\circ}}{\sin 115^{\circ}}\)
Solution.
We have,

Question 18.
Prove that

Solution.

Question 19.
Prove that sec 70° sin 20° + cos 20° cosec 70° = 2
Solution.
L. H. S.
sec 70° sin 20° + cos 20° cosec 70°

Question 20.
Prove that :
(i) cos 2A = \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\)
(ii)
Solution.
(i) R.H.S

(ii)

Question 21.
If sec θ + tan θ = p, prove that
\(\frac{p^{2}-1}{p^{2}+1}\) = sin θ.
Solution.

Question 22.
If A = 30° and B = 60°, then prove that sin(A + B) = sin A cos B + cos A sin B.
Solution.
L.H.S. = sin (A + B)
= sin (30° + 60°)
= sin 90° = 1
R.H.S. = sin A cos B + cos A sin B
= sin 30°. cos 60° + cos 30°.sin 60°

L.H.S. = R.H.S.

Question 23.
Find the value of sin 65o cos 35° – cos 65° sin 35°.
Solution.
sin 65° cos 35° – cos 65° sin 35°
= sin(65° – 35°)
= sin 30° = \(\frac{1}{2}\)

Question 24.
If sinθ = \(\frac{9}{41}\), then find the value of tan θ – cosec θ.
Solution.

Question 25.
Find the value of sin A cos(90° – A) + cos A sin(90° – A).
Solution.
sin A cos (90° – A) + cos A sin (90° – A)
= sin A sin A + cos A cos A
= sin²A + cos²A = 1

Question 26.
Prove that :
\(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\) = cosec θ – cot θ
Solution.

cosec θ – cot θ = R.H.S.

Question 27.
(a) Prove that :

Or
sin 4θ – cos 4θ = 2 tan²θ – 1.
(b) Prove that :
cos⁴θ – sin²θ = cos 2θ.
(c) Find the value of cos 375°.
Solution.
(a) Do yourself.
(b) Do yourself.
(c) cos 375° = cos (360° + 15°)
= cos 15o = cos (45° – 30°)
= cos 45°.cos 30° + sin 45o.sin 30°

Question 28.
Prove that: sin 3A cos3A + cos 3A. sinA = \(\frac{3}{4}\) sin 4A.
Solution.
Do yourself.

Question 29.
Prove that :

Solution.
Do yourself.

Question 30.
(a) Prove that:

Solution.

Question 31.
Prove :
\(\frac{\sec A-\tan A}{\sec A+\tan A}\)
= 1 – 2 sec A tan A + 2 tan²A
Solution.
L.H.S. – \(\frac{\sec A-\tan A}{\sec A+\tan A}\)

= 1 + tan²A + tan² – A – 2sec A tan A
= 1 – 2 sec A. tan A + 2 tan²A
= R.H.S.

Question 32.
Prove :
2cos A + cos 3A + cos 5A
= 4 cos A cos² 2A.
Solution.
L.H.S. = 2 cos A + cos 3A + cos 5A
= 2 cos A + 2 cos 4A cos (-A)
= 2 cos A + 2 cos 4A cos A
= 2 cos A(1 + cos 4A)
= 2 cos A(1 + 2 cos²2A – 1)
= 4 cos A cos²2A = R.H.S.

Question 33.
Prove : sin²A – sin²B = sin(A + B) sin(A – B).
Solution.
L.H.S. = sin²A – sin²B
= (sin A – sin B) (sin A + sin B)

= sin(A + B).sin(A – B)
= R.H.S.

Question 34.
Prove :

Solution.
L.H.S.


L.H.S. = R.H.S.

Question 35.
Prove :
Proved.
sec A – tan A = \(\frac{1}{\sec A+\tan A}\)
Solution.
L.H.S. = sec A – tan A

So, L.H.S. = R.H.S.

Question 36.
Prove that :

Solution.
L.H.S.

Introduction to Trigonometry Class 10 Extra Questions Long Answer Type

Question 1.
Prove that:
(cos A + cos B)² + (sin A – sin B)²
= 4 cos² \(\left(\frac{A+B}{2}\right)\)
Solution.
L.H.S.
cos²A + cos²B + 2cos A cos B + sin²A
+ sin²B – 2 sin A sinB
⇒ (cosA + sinoA) + (cos’B + sino B)
+ 2 (cosA cos B – sin A sin B)
⇒ 1 + 1 + 2 cos (A – B)
⇒ 2[1 + cos (A – B)]

= 4cos² \(\frac{A-B}{2}\)
= R.H.S.

Question 2.
Prove that:
\(\sqrt{\frac{1-\sin A}{1+\sin A}}\)= sec A – tan A
Solution.

= sec A – tan A
= R.H.S.

Question 3.
Prove that :

Solution :
L.H.S.


= sec A- sec A + tan A
= tan A ……(ii)
From (i) and (ii)
L.H.S. = R.H.S.

Question 4.
Prove that : cos³A cos3A + sin³A sin3A = cos³2A.
Solution :

= cos³ 2A = R.H.S.

Question 5.
If \(\frac{\cos \alpha}{\cos \beta}\) = n, \(\frac{\sin \alpha}{\sin \beta}\) = m prove that
(m² – n²) sin²β = 1 – n².
Solution.
cos α = n cos β and sin α = m sin β
Squaring and adding we get
cos²α + sin² α = n² cos²β + m² sin² β.
⇒ 1 = n² (1 – sin²β) + m² sin² β
⇒ 1 = n² – n² sin² β + m2 sino β
⇒ 1 – n² = sin²β (m² – n²)

Question 6.
Prove that :
sin 20° sin 40° sin 80° = \(\sqrt{\frac{3}{8}}\)
Solution :

Question 7.
Prove :

Solution :

Question 8.
Prove :

Solution.
L.H.S.

= (cosec θ + cot θ)² = R.H.S. Proved

Question 9.
Prove:
tan² θ + cot² θ + 2 = sec² θ.cosec² θ
Solution.
tan²θ + cot² θ + 2
= (1 + tan²θ) + (1 + cot²θ)
= sec² θ + cosec² θ

= sec² θ cosec² θ = R.H.S.

Question 10.
If A + B + C = 180°, Prove that :
(i) sin A + sin B + sin C = 4 cos\(\frac{A}{2}\) . cos\(\frac{B}{2}\) . cos\(\frac{C}{2}\)
(ii) cot A . cot B + cot B . cot C + cot C . cot A = 1
(iii) sin 2A + sin 2B + sin 2C = 4 sin A sin B . sin C
(iv) cos A + cos B + cos C = 1 + 4 sin\(\frac{A}{2}\) . sin\(\frac{B}{2}\) . sin\(\frac{C}{2}\)
(v) sin A + sin B – sin C = 4 sin\(\frac{A}{2}\) . sin\(\frac{B}{2}\) . cos\(\frac{C}{2}\)
(vi) Prove :

Solution.
(i) L.H.S.
= sin A + sin B + sin C

(ii) We have,
A + B + C = 180°
⇒ A + B = 180°- C
∴ cot(A + B) = cot(180° – C)
⇒ \(\frac{\cot A \cdot \cot B-1}{\cot B+\cot A}\) = -cot C
⇒ cot A . cot B – 1 = -cot B . cot C – cot A . cot C
⇒ cot A . cot B + cot B . cot C + cot C . cot A = 1
= R.H.S.

(iii) L.H.S. = sin 2A + sin 2B + sin 2C

= 2 sin(A + B) . cos(A – B) + 2 sin C.cos C…(i)
∵ A+B+C = 180°
∴ A + B = 180° – C
⇒ sin(A + B) = sin(180° – C)
= sin C
and cos (A + B) = cos (180° – C)
= -cos C
Putting value in (i),
= 2 sin C.cos(A – B) – 2 sin C. cos(A + B)
= 2 sin C[cos (A – B) – cos(A + B)]

= 2 sin C(2 sin A.sin B]
= 4 sin A . sin B . sin C
= R.H.S.

(iv) L.H.S. = cos A + cos B + cos C

= R.H.S.

(v) Do yourself.

(vi) L.H.S.

Question 11.
Prove that:

= cos 2θ – tan 3θ . sin 2θ.
(ii) sin 10° . sin 30° . sin 50° . sin 70° = \(\frac{1}{16}\)
(iii) cos 20° . cos 40° cos 60°. cos 80° = \(\frac{1}{16}\)
(iv)
(v) sin 20°.sin 40° sin 60°.sin 80° = \(\frac{3}{16}\)
(vi) tan 70° = tan 20° + 2 tan 50°.
(vii) tan 3A – tan 2A – tan A = tan 3A . tan 2A tan A.
(viii) If A + B + C = 90°, prove that
tan A tan B + tan B tan C + tan C tan A = 1
(ix) Prove that :

(x) Prove :

+ (sin A + sin B)²
Solution.
(i) L.H.S.

= cos 2θ – tan 3θ.sin 2θ
= R.H.S.

(ii) L.H.S. = sin 10° sin 50° sin 70° sin 30°

(iii) L.H.S. = cos 20° cos 40°.cos 80°.cos 60°

(iv)

(v) L.H.S.
= sin 20° sin 40° sin 60° sin 80°
= sin 60° sin 20° sin 40° sin 80°

L.H.S. = R.H.S.

(vi) tan 70° = tan(50° + 20°) .

⇒ tan 70° – tan 70° tan 50° tan 20°
= tan 50° + tan 20°
⇒ tan 70° – tan(90° – 20°) tan 50°
tan 20° = tan 50° + tan 20°
⇒ tan 70° – cot 20° tan 50° tan 20°
= tan 50° + tan 20°
⇒ tan 70° – \(\frac{1}{\tan 20^{\circ}}\) tan 50° tan 20°
= tan 50° + tan 20°
⇒ tan 70° – tan 50° = tan 50° + tan 20°
⇒ tan 70° = tan 50° + tan 20°+ tan 50°
⇒ tan 70°= tan 20° + 2tan 50°

(vii) We have
tan 3A = tan(A + 2A)
tan A + tan 2A
⇒ tan 3A = \(\frac{\tan A+\tan 2 A}{1-\tan A \tan 2 A}\)
⇒ tan 3A – tan 3A tan A tan 2A = tan A + tan 2A
⇒ tan 3A – tan 2A – tan A
= tan 3A tan 2A tan A.

(viii) Given
A + B + C = 90°
⇒ A + B = 90° – C
⇒ tan(A + B) = tan(90° – C)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = cot C = \(\frac{1}{\tan C}\)
⇒ tan =tan A . tan C + tan B . tan C
= 1 – tan A tan B
⇒ tan A tan B + tan B . tan C + tan C . tan A = 1

(ix)


= sec A + tan A
= R.H.S.

(x) R.H.S. = (cos A-cos B)² + (sin A + sin B)²
= cos²A + cos²B – 2cos A cos B
+ sin²A + sin²B + 2 sin A sin B
= (cos² A +sin² A) + (cos²B + sin²B)
– 2{cos A cos B – sin A sin B]
= 1 + 1 – 2[cos A.cos B – sin A.sin B]
= 2[1 – cos(A + B)]

= L.H.S.

Question 12.
Solve :

Solution.

⇒ cos θ = cos 60°
∴ θ = 60°

Question 13.
Prove : (cos A + sec A)² + (sin A + cosec A)² 7 + tan² A + cot² A.
Solution.
L.H.S.
= (cos A + sec A)² + (sin A + cosec A)²
= cos²A + sec²A + 2.cos A sec A + sin² A + cosec²A + 2sin A cosec A
= 1 + 2 + 2 + sec²A + cosec² A
= 5 + 1 + tan²A + 1 + cot² A
= 7 + tan²A + cot²A
= R.H.S.

Question 14.
Prove :

Solution :

Question 15.
Prove that:
tan (A + B) tan (A – B)

Solution.
tan (A + B)

Question 16.

Solution.
L. H. S.


= tan 5 A = R.H.S.

Question 17.
Prove :
cos A cos (60° + A). Cos (60° – A) = \(\frac{1}{4}\) cos 3A
Solution.
L.H.S.
= cos A. cos (60° + A). cos (60° – A)
= cos A (cos 60°. cos A-sin 60° sin A)
(cos 60°. cos A + sin 60°sin A)

Question 18.
Prove :

Solution.
L.H.S.


= tan (A – B)
= R.H.S.

Question 19.
Prove :
sin A sin 2 A + sin 3A sin 6A = sin 4A sin 5A.
Solution.
L.H.S.
= sin A sin 2A + sin3 A sin 6A
= \(\frac{1}{2}\) [2 sin A sin 2A + 2 sin3A sin 6A]
= \(\frac{1}{2}\) [cos A – cos 3A + cos 3 A – cos 9A]
= \(\frac{1}{2}\) [cos A – cos 9A]
= \(\frac{1}{2}\) × 2 sin \(\frac{9 \mathrm{A}+\mathrm{A}}{2}\) sin \(\frac{9 \mathrm{A}-\mathrm{A}}{2}\)
= sin 5A sin 4A
= sin 4A sin 5A = R.H.S.

Question 20.
Prove :

Solution.
L.H.S.

= sec θ + tan θ = R.H.S.