In this chapter, we provide RD Sharma Solutions for Class 10 Chapter 6 Trigonometric Identities Exercise 6.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 10 Chapter 6 Trigonometric Identities Exercise 6.1 pdf, free RD Sharma Solutions for Class 10 Chapter 6 Trigonometric Identities Exercise 6.1 book pdf download. Now you will get step by step solution to each question.
Textbook | NCERT |
Class | Class 10 |
Subject | Maths |
Chapter | Chapter 6 |
Chapter Name | Trigonometric Identities |
Exercise | 6.1 |
Category | RD Sharma Solutions |
RD Sharma Class 10 Solutions Trigonometric Identities Exercise 6.1
Prove the following trigonometric identities :
Question 1.
(1 – cos2 A) cosec2 A = 1
Solution:
(1 – cos2 A) cosec2 A = 1
L.H.S. = (1 – cos2 A) cosec2 A = sin2 A cosec2 A (∵ 1 – cos2 A = sin2 A)
= (sin A cosec A)2 = (l)2 = 1 = R.H.S. {sin A cosec A = 1 }
Question 2.
(1 + cot2 A) sin2 A = 1
Solution:
(1 + cot2 A) sin2 A = 1
L.H.S. = (1 + cot2 A) (sin2 A)
= cosec2 A sin2 A {1 + cot2 A = cosec2 A}
= [cosec A sin A]2
= (1)2= 1 = R.H.S. (∵ sin A cosec A = 1
Question 3.
tan2 θ cos2 θ = 1- cos2 θ
Solution:
Question 4.
Solution:
Question 5.
(sec2 θ – 1) (cosec2 θ – 1) = 1
Solution:
Question 6.
Solution:
Question 7.
Solution:
Question 8.
Solution:
Question 9.
Solution:
Question 10.
Solution:
Question 11.
Solution:
Question 12.
Solution:
Question 13.
Solution:
Question 14.
Solution:
Question 15.
Solution:
Question 16.
tan2 θ – sin2 θ = tan2 θ sin2 θ
Solution:
Question 17.
(sec θ + cos θ ) (sec θ – cos θ ) = tan2 θ + sin2 θ
Solution:
Question 18.
(cosec θ + sin θ) (cosec θ – sin θ) = cot2 θ + cos2 θ
Solution:
Question 19.
sec A (1 – sin A) (sec A + tan A) = 1 (C.B.S.E. 1993)
Solution:
Question 20.
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Solution:
Question 21.
(1 + tan2 θ) (1 – sin θ) (1 + sin θ) = 1
Solution:
Question 22.
sin2 A cot2 A + cos2 A tan2 A = 1 (C.B.S.E. 1992C)
Solution:
Question 23.
Solution:
Question 24.
Solution:
Question 25.
Solution:
Question 26.
Solution:
Question 27.
Solution:
Question 28.
Solution:
Question 29.
Solution:
Question 30.
Solution:
Question 31.
sec6 θ= tan6 θ + 3 tan2 θ sec2 θ + 1
Solution:
Question 32.
cosec6 θ = cot6 θ+ 3cot2θ cosec2 θ + 1
Solution:
Question 33.
Solution:
Question 34.
Solution:
Question 35.
Solution:
Question 36.
Solution:
Question 37.
Solution:
Question 38.
Solution:
Question 39.
Solution:
Question 40.
Solution:
Question 41.
Solution:
Question 42.
Solution:
Question 43.
Solution:
Question 44.
Solution:
Question 45.
Solution:
Question 46.
Solution:
Question 47.
Solution:
Question 48.
Solution:
Question 49.
tan2 A + cot2 A = sec2 A cosec2 A – 2
Solution:
Question 50.
Solution:
Question 51.
Solution:
Question 52.
Solution:
Question 53.
Solution:
Question 54.
sin2 A cos2 B – cos2 A sin2 B = sin2 A – sin2 B.
Solution:
L.H.S. = sin2 A cos2 B – cos2 A sin2 B
= sin2 A (1 – sin2 B) – (1 – sin2 A) sin2 B
= sin2 A – sin2 A sin2 B – sin2 B + sin2 A sin2 B
= sin2 A – sin2 B
Hence, L.H.S. = R.H.S.
Question 55.
Solution:
Question 56.
cot2 A cosec2 B – cot2 B cosec2 A = cot2 A – cot2 B
Solution:
Question 57.
tan2 A sec2 B – sec2 A tan2 B = tan2 A – tan2 B
Solution:
Prove the following identities: (58-75)
Question 58.
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x1 – y2 = a2 – b1. [C.B.S.E. 2001, 20O2C]
Solution:
x – a sec θ + b tan θ
y = a tan θ + b sec θ
Squaring and subtracting, we get
x2-y2 = {a sec θ + b tan θ)2 – (a tan θ + b sec θ)2
= (a2 sec2 θ + b2 tan2 θ + 2ab sec θ x tan θ) – (a2 tan2 θ + b2 sec2 θ + 2ab tan θ sec θ)
= a2 sec2 θ + b tan2 θ + lab tan θ sec θ – a2 tan2 θ – b2 sec2 θ – 2ab sec θ tan θ
= a2 (sec2 θ – tan2 θ) + b2 (tan2 θ – sec2 θ)
= a2 (sec2 θ – tan2 θ) – b2 (sec2 θ – tan2 θ)
= a2 x 1-b2 x 1 =a2-b2 = R.H.S.
Question 59.
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ±3
Solutioon:
Question 60.
If cosec θ + cot θ = mand cosec θ – cot θ = n,prove that mn= 1
Solution:
Question 61.
Solution:
Question 62.
Solution:
Question 63.
Solution:
Question 64.
Solution:
Question 65.
Solution:
Question 66.
(sec A + tan A – 1) (sec A – tan A + 1) = 2 tan A
Solution:
Question 67.
(1 + cot A – cosec A) (1 + tan A + sec A) = 2
Solution:
Question 68.
(cosec θ – sec θ) (cot θ – tan θ) = (cosec θ + sec θ) (sec θ cosec θ-2)
Solution:
Question 69.
(sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A
Solution:
Question 70.
Solution:
Question 71.
Solution:
Question 72.
Solution:
Question 73.
sec4 A (1 – sin4 A) – 2tan2 A = 1
Solution:
Question 74.
Solution:
Question 75.
Solution:
Question 76.
Solution:
Question 77.
If cosec θ – sin θ = a3, sec θ – cos θ = b3, prove that a2b2 (a2 + b2) = 1
Solution:
Question 78.
Solution:
Question 79.
Solution:
Question 80.
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
Solution:
Question 81.
If cos A + cos2 A = 1, prove that sin2 A + sin4 A = 1
Solution:
cos A + cos2 A = 1
⇒ cos A = 1 – cos2 A
⇒cos A = sin2 A
Now, sin2 A + sin4 A = sin2 A + (sin2 A)2
= cos A + cos2 A = 1 = R.H.S.
Question 82.
If cos θ + cos2 θ = 1, prove that
sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ-2 = 1
Solution:
Question 83.
Given that :
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 – cos α) (l – cos β) (1 – cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
Solution:
Question 84.
Solution:
Question 85.
Solution:
Question 86.
If sin θ + 2cos θ = 1 prove that 2sin θ – cos θ = 2. [NCERT Exemplar]
Solution:
All Chapter RD Sharma Solutions For Class10 Maths
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All Subject NCERT Exemplar Problems Solutions For Class10
All Subject NCERT Solutions For Class 10
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