RBSE Solutions for Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 7 Trigonometric Identities Exercise 7.1.
Rajasthan Board RBSE Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.1
RBSE Solutions For Class 10 Maths Chapter 7 Question 1.
For ∠θ express all the (RBSESolutions.com) Trigonometric ratios in terms of sec θ.
Solution :
(i) sin2 θ + cos2 θ = 1
⇒ sin2 θ = 1 – cos2 θ
(ii) Cos θ = \(\frac { 1 }{ sec\theta }\)
(iii) ∵ 1 + tan2 θ = sec2 θ
or, tan2 θ = sec2 θ – 1
or, tan θ = \(\sqrt { { sec }^{ 2 }\theta -1 }\)
(iv) cot θ = \(\frac { 1 }{ tan\theta }\)
= \(\frac { 1 }{ \sqrt { { sec }^{ 2 }\theta -1 } }\)
(v) cosec θ = \(\frac { 1 }{ sin\theta }\)
RBSE Class 10 Maths Chapter 7 Question 2.
Express (RBSESolutions.com) trigonometric sin θ, sec θ, and tan θ in terms of cot θ.
Solution :
(i) sin θ = \(\frac { 1 }{ cosec\theta }\)
(ii) sec2 θ = 1 + tan2 θ
(iii) tan θ = \(\frac { 1 }{ cot\theta }\)
Exercise 7.1 Class 10 RBSE Question 3.
Verify the following (RBSESolutions.com) with the help of identities.
cos2θ + cos2θ cot2θ = cot2θ
Solution :
L.H.S. = cos2θ + cos2θ cot2θ
= cos2θ(1 + cot2θ)
= cot2θ
= R.H.S.
Ex 7.1 Class 10 RBSE Question 4.
sec θ (1 – sin θ) (sec θ + tan θ) = 1
Solution :
L.H.S. = sec θ(1 – sinθ) (secθ + tanθ)
= 1
= R.H.S
RBSE Solutions For Class 10 Maths Chapter 7.1 Question 5.
cosec2 θ + sec2 θ = cosec2 θ sec2 θ
Solution :
LH.S. = cosec2 θ + sec2 θ
= cosec2 θ sec2 θ
= R.H.S
RBSE Class 10 Maths Chapter 7 Exercise 7.1 Question 6.
\(\sqrt { \frac { 1-sin\theta }{ 1+sin\theta } }\) = sec θ – tan θ
Solution :
L.H.S = \(\sqrt { \frac { 1-sin\theta }{ 1+sin\theta } }\)
Multiply numerator (RBSESolutions.com) and denominator by (1 – sin θ)
= \(\frac { 1 }{ cos\theta } -\frac { sin\theta }{ cos\theta }\)
= sec θ – tan θ
= R.H.S.
Class 10 Maths Solution RBSE Ch 7 Question 7.
\(\sqrt { { sec }^{ 2 }\theta +{ cosec }^{ 2 }\theta } \) = tan θ + cot θ
Solution :
L.H.S. = \(\sqrt { { sec }^{ 2 }\theta +{ cosec }^{ 2 }\theta } \)
= tan θ + cot θ
= R.H.S.
RBSE 10th Maths Solution 7.1 Question 8.
\(\frac { tan\alpha +tan\beta }{ cot\alpha +cot\beta }\) = tan α tan β
Solution :
L.H.S. = \(\frac { tan\alpha +tan\beta }{ cot\alpha +cot\beta }\)
= tan α tan β
= R.H.S.
7.1 Class 10 RBSE Question 9.
\(\frac { 1+sin\theta }{ cos\theta } +\frac { cos\theta }{ 1+sin\theta }\) = 2 sec θ
Solution :
L.H.S = \(\frac { 1+sin\theta }{ cos\theta } +\frac { cos\theta }{ 1+sin\theta }\)
= 2 sec θ
= R.H.S.
RBSE Solutions For Class 10 Maths Chapter 7 Ex 7.1 Question 10.
\(\frac { { sin }^{ 4 }\theta -{ cos }^{ 4 }\theta }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta }\) = 1
Solution :
L.H.S. = \(\frac { { sin }^{ 4 }\theta -{ cos }^{ 4 }\theta }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta }\)
= 1
= R.H.S.
RBSE Class 10 Maths Chapter 7.1 Question 11.
cot θ – tan θ = \(\frac { { 1-2sin }^{ 2 }\theta }{ sin\theta cos\theta }\)
Solution :
L.H.S. = cot θ – tan θ
= \(\frac { { 1-2sin }^{ 2 }\theta }{ sin\theta cos\theta }\)
= R.H.S.
Class 10 Maths RBSE Solution Chapter 7 Question 12.
cos4 θ + sin4 θ = 1 – 2cos2 θ sin2 θ
Solution :
L.H.S. = cos4 θ + sin4 θ
= (cos2 θ)2 + (sin2 θ)2 + 2sin2 θ. cos2 θ – 2 sin2 θ cos2 θ
= (sin2 θ + cos2 θ)2 – 2sin2 θ cos2 θ
= 1 – 2sin2 θ cos2 θ [∵ sin2 θ + cos2 θ = 1]
= R.H.S
Ch 7 Class 10 Maths RBSE Question 13.
(sec θ – cos θ) (cot θ + tan θ) = tan θ sec θ
solution :
L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= tan θ sec θ
= R.H.S.
Chapter 7 Maths Class 10 RBSE Question 14.
\(\frac { { 1-tan }^{ 2 }\alpha }{ { cot }^{ 2 }\alpha -1 }\) = tan2 α
L.H.S. = \(\frac { { 1-tan }^{ 2 }\alpha }{ { cot }^{ 2 }\alpha -1 }\)
= R.H.S.
Trigonometric Identities Class 10 RBSE Question 15.
\(\frac { sin\theta }{ 1-cos\theta }\) = \(\frac { 1+cos\theta }{ sin\theta }\)
Solution :
L.H.S. = \(\frac { sin\theta }{ 1-cos\theta }\)
Multiply numerator (RBSESolutions.com) and denominator by (1 + cos θ)
= \(\frac { sin\theta }{ 1-cos\theta } \times \frac { 1+cos\theta }{ 1+cos\theta }\)
= \(\frac { 1+cos\theta }{ sin\theta }\)
= R.H.S.
RBSE 10th Class Trigonometry Question 16.
sin6 θ + cos6 θ = 1 – 3sin2 θ cos2 θ
Solution :
L.H.S. = (sin2 θ)3 + (cos2 θ)2
= (sin2 θ + cos2 θ)
(sin2 θ + cos2 θ – sin2 θ cos2 θ) [∵ a3 + b3 = (a + b) (a2 + b2 – ab)]
= sin4 θ + cos4 θ – sin2 θ cos2 θ
= (sin2 θ)2 + (cos2 θ)2 + 2sin2 θ cos2 θ
= 2 sin2 θ cos2 θ – sin2 θ cos2 θ
= (sin2 θ + cos2 θ)2 – 3sin2 θ cos2 θ
= 1 – 3sin2 θ cos2 θ
= R.H.S.
Class 10 RBSE Maths Chapter 7 Question 17.
\(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } \) = 1 + tan θ + cot θ
Solution :
L.H.S. = \(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } \)
= 1 + tan θ + cot θ
= R.H.S.
Alternate :
L.H.S. = \(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } \)
= tan θ + 1 + cot θ
= 1 + tan θ + cot θ
= R.H.S.
RBSE Solutions For Class 10 Maths Ex 7.1 Question 18.
sin θ(1 + tan θ) + cos θ(1 + cot θ) = cosec θ + sec θ
Solution :
L.H.S. = sin θ(1 + tan θ) + cos θ(1 + cot θ)
= cosec θ + sec θ
= R.H.S.
RBSE 10th Math Ex 7.1 Question 19.
sin2 θ cos θ + tan θ sin θ + cos3 θ = sec θ.
Solution :
L.H.S. = sin2 θ cos θ + tan θ sin θ + cos3 θ
= (sin2 θ cos θ + cos3 θ) + tan θ sin θ
= cos θ(sin2 θ + cos2 θ) + tan θ sin θ
= sec θ
= R.H.S.
Trigonometry Class 10 RBSE Question 20.
\(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } \) = 1 + sec θ cosec θ
Solution :
L.H.S. = \(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } \)
= 1 + sec θ cosec θ
= R.H.S.
RBSE Class 10 Chapter 7 Question 21.
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A
Solution :
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
= 5 + sec2 A + cosec2 A
= 5 + (1 + tan2 A) + (1 + cot2 A)
= 7 + tan2 A + cot2 A
R.H.S.
Exercise 7.1 Class 10 Trigonometry Question 22.
sin8θ – cos8θ = (sin2 θ – cos2 θ) (1 – 2sin2 θ cos2 θ)
Solution :
L.H.S. = sin8 θ – cos2 θ
(sin4 θ)2 – (cos4 θ)2
= (sin4 θ + cos4 θ) (sin4 θ – cos4 θ)
= [(sin2 θ)2 + (cos2 θ)2 + 2cos2 θ sin2 θ – 2sin2 θ cos2 θ] [(sin2 θ)2 – (cos2 θ)2]
= [(sin2 θ + cos2 θ) – 2sin2 θ cos2 θ] [(sin2 θ + cos2 θ) (sin2 θ – cos2 θ)]
= (1 – 2Sin2 θ cos2 θ) (sin2 θ – cos2 θ)
= (sin2 θ – cos2 θ) (1 – 2sin2 θ cos2 θ)
= R.H.S.
Trigonometric Identities Class 10 RBSE Formula Question 23.
\(\sqrt { \frac { sec\theta +1 }{ sec\theta -1 } }\) = cot θ + cosec θ
Solution :
L.H.S. = \(\sqrt { \frac { sec\theta +1 }{ sec\theta -1 } }\)
Multiply numerator (RBSESolutions.com) and denominator by (1 + cos θ)
= cosec θ + cot θ or cot θ + cosec θ
= R.H.S.
RBSE Class 10th Maths Chapter 7 Question 24.
= sin2 θ cos2 θ
Solution :
L.H.S.
= sin2 θ cos2 θ
= R.H.S.
Trigonometry Questions For Class 10 RBSE Question 25.
Solution :
L.H.S.
= R.H.S.
Again by equation (i)
= R.H.S.
Www.RBSEsolutions.Com Class 10 Question 26.
\(\frac { cosA }{ 1-tanA } +\frac { SinA }{ 1-cotA }\) = sin A + cos A.
Solution :
L.H.S. = \(\frac { cosA }{ 1-tanA } +\frac { SinA }{ 1-cotA }\)
= sin A + cos A
= R.H.S
RBSE Solutions For Class 10 Maths ex 7.1 Question 27.
(cosec A – sin A) (sec A – cos A) = \(\frac { 1 }{ tanA+cotA }\)
Solution :
L.H.S. = (cosec A – sin A) (sec A – cos A)
= \(\frac { 1 }{ tanA+cotA }\)
= R.H.S.
Class 10th Math 7.1 Question 28.
= 1 + sin θ cos θ
Solution :
L.H.S.
= 1 + sin θ cos θ [∵ sin2 + cos2 = 1]
= R.H.S.
RBSE Class 10 Maths Solutions ex 7.1 Question 29.
If sec θ + tan θ = p then (RBSESolutions.com) prove that \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }\) = sin θ
Solution :
Given, sec θ + tan θ = p
L.H.S. = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }\)
= sin θ
= R.H.S.
Exercise 7.1 Class 10 Question 30.
If \(\frac { cosA }{ cosB }\) = m and \(\frac { cosA }{ sinB }\) = n then prove that (m2 + n2) cos2 B = n2
Solution :
Given m = \(\frac { cosA }{ cosB }\) and n = \(\frac { cosA }{ sinB }\)
L.H.S. = (m2 + n2) cos2B
= n2
= R.H.S.
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