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CHAPTER – 13
TRIGONOMETRIC IDENTITIES
Exercise 13A
Page number: 583
Prove each of the following identities:
Question 1:
(i) (1 – cos2θ) cosec2θ = 1
(ii) (1 + cot2θ) sin2θ = 1
Solution:
(i) (1 – cos2θ) cosec2θ = 1
L.H.S. = (1 – cos2θ) cosec2θ
= (sin2θ) × cosec2θ (Using identity sin2θ + cos2 θ = 1)
= 1
cosec2 θ × cosec2θ
= 1
= R.H.S.
Hence Proved.
(ii) (1 + cot2θ) sin2θ = 1
L.H.S. = (1 + cot2θ) × sin2 θ
= (cosec2 θ) × sin2 θ (Using identity 1 + cot2 θ = cosec2 θ)
= 1
sin2 θ × sin2 θ
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= 1
= R.H.S.
Hence Proved.
Question 2:
(i) (sec2θ − 1) cot2θ = 1
(ii) (sec2θ − 1) (cosec2θ − 1) = 1
(iii) (1− cos2θ) sec2θ = tan2θ
Solution:
(i) (sec2θ − 1) cot2θ = 1
L.H.S. = (sec2 θ – 1) × cot2 θ
= (tan2θ) × cot2θ (using identity 1 + tan2 θ = sec2 θ)
= 1
cot2θ × cot2θ
= 1
= R.H.S.
Hence Proved.
(ii) (sec2θ − 1) (cosec2θ − 1) = 1
L.H.S. = (sec2 θ – 1) (cosec2 θ – 1)
= (tan2θ) × cot2θ
(using identity 1 + cot2 θ = cosec2 θ and 1 + tan2 θ = sec2 θ)
= tan2θ ×1
tan2θ
= 1
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= R.H.S.
Hence Proved.
(iii) (1− cos2θ) sec2θ = tan2θ
L.H.S. = (1 – cos2 θ) sec2 θ
= (sin2θ) × (1
cos2θ) (using identity sin2 θ = 1- cos2 θ)
= tan2 θ
= R.H.S.
Hence Proved.
Question 3: Prove
(i) 𝐬𝐢𝐧𝟐 𝜽 +𝟏
𝟏+𝐭𝐚𝐧𝟐 𝜽= 𝟏
(ii) 𝟏
𝟏+𝐭𝐚𝐧𝟐 𝜽+
𝟏
𝟏+𝐜𝐨𝐭𝟐 𝜽= 𝟏
Solution:
(i) L.H.S.
sin2 𝜃 +1
1+tan2 𝜃
= (sin2 𝜃) + (1
sec2 𝜃) (using 1 + tan2 𝜃 = sec2 𝜃)
= (sin2 𝜃) + (cos2 𝜃) We know, sin2 𝜃 + cos2 𝜃 = 1
= 1
= RHS.
Hence proved.
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(ii) LHS.
1
1+tan2 𝜃+
1
1+cot2 𝜃
= (1
sec2 𝜃) + (
1
cosec2 𝜃)
(Using 1 + tan2 𝜃 = sec2 𝜃 and 1 + tan2 𝜃 = sec2 𝜃)
= (cos2 𝜃) + (sin2 𝜃) We know, sin2 𝜃 + cos2 𝜃 = 1
= 1
= RHS.
Hence proved.
Question 4: Prove
(i) (1 + cos θ) (1 − cos θ) (1 + cot2θ) = 1
(ii) cosec θ (1 + cos θ) (cosec θ − cot θ) = 1
Solution:
(i) (1 + cos θ) (1 − cos θ) (1 + cot2θ) = 1
LHS: (1 + cos θ) (1 − cos θ) (1 + cot2θ)
= (1 – cos2 θ) × cosec2 θ (Using sin2 θ + cos2 θ = 1)
= (sin2 θ) × cosec2 θ
= sin2 θ ×1
sin2 θ
= 1
= R.H.S.
Hence Proved
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(ii) cosec θ (1 + cos θ) (cosec θ − cot θ) = 1
L.H.S.
cosec θ (1 + cos θ) (cosec θ − cot θ)
= (cosec θ + cosec θ cos θ) (cosec θ – cot θ)
We know, cosec θ = 1
sin θ and cot θ =
cosθ
sinθ
(sec θ + cot θ) (cosec θ – cotθ)
Apply formula: (a + b) (a – b) = a2 – b2
= cosec2 θ – cot2 θ
= 1
= R.H.S.
Hence proved.
Question 5: Prove
(i) 𝐜𝐨𝐭𝟐 𝛉 −𝟏
𝐬𝐢𝐧𝟐 𝛉= −𝟏
(ii) 𝐭𝐚𝐧𝟐 𝛉 −𝟏
𝐜𝐨𝐬𝟐 𝛉= −𝟏
(i) 𝐜𝐨𝐬𝟐 𝛉 −𝟏
𝟏+𝐜𝐨𝐭𝟐 𝛉= 𝟏
Solution:
(i)
L.H.S.
= cot2 θ – 1
sin2 θ
= cos2θ
sin2θ –
1
sin2 θ
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= (cos2θ−1)
sin2 θ
= − sin2 θ
sin2 θ
= - 1
= R.H.S
(ii)
L.H.S.
= tan2 θ – 1
cos2θ
= sin2θ
cos2θ –
1
cos2θ
= (sin2θ – 1)
cos2θ
= −cos2θ
cos2θ
= -1
= R.H.S
(iii)
L.H.S.
= cos2 θ + 1
(1+cot2θ)
= cos2 θ + 1
cosec2θ
= cos2 θ + sin2θ
= 1
= R.H.S
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Question 6: Prove
𝟏
𝟏+𝐬𝐢𝐧 𝛉+
𝟏
𝟏−𝐬𝐢𝐧 𝛉= 𝟐 𝐬𝐞𝐜𝟐 𝛉
Solution:
L.H.S.
= 1
(1+sin 𝜃)+
1
(1−sin 𝜃)
= (1−sin 𝜃)+(1+sin 𝜃)
(1+sin 𝜃)(1−sin 𝜃)
= 2
1−sin2 θ
= 2
cos2 𝜃
= 2 sec2 𝜃
= R.H.S.
Question 7: Prove
(i) sec θ (1 − sin θ) (sec θ + tan θ) = 1
(ii) sin θ (1 + tan θ) + cos θ (1 + cot θ) = (sec θ + cosec θ)
Solution:
(i) L.H.S.
= sec θ (1 − sin θ) (sec θ + tan θ)
= (1
cos 𝜃) ×(1 − sin θ) × (
1
cos 𝜃+
sin 𝜃
cos 𝜃)
= (1
cos 𝜃) ×(1 − sin θ) × (
1+sin 𝜃
cos 𝜃)
= 1+sin2 𝜃
cos2 𝜃
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= cos2θ
cos2θ
= 1
= R.H.S.
(ii) L.H.S. = sin θ (1 + tan θ) + cos θ (1+ cot θ)
= sin θ (1 +sin θ
cos θ) + cos θ (
1+cos θ
sinθ)
= sin θ{(cosθ +sinθ)
cosθ} + cos θ{
(sinθ+cosθ)
sinθ}
= (cos θ + sin θ) (sinθ
cos θ+
cos θ
sinθ)
= (cos θ+sin θ)
cos θ sin θ
= cosec θ + sec θ
= R.H.S.
Question 8:
(i) 𝟏 +𝐜𝐨𝐭𝟐 𝜽
𝟏+𝐜𝐨𝐬𝐞𝐜 𝜽= 𝐜𝐨𝐬𝐞𝐜 𝜽
(ii) 𝟏 +𝐭𝐚𝐧𝟐 𝜽
𝟏+𝐬𝐞𝐜 𝜽= 𝐬𝐞𝐜 𝜽
Solution:
L.H.S. = 1 +cot2 𝜃
1+cosec 𝜃
= 1 +cos2 𝜃
sin2 𝜃
1+1
sin 𝜃
= 1 +cos2 𝜃
1+sin 𝜃×
sin 𝜃
sin2 𝜃
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= 1 +cos2 𝜃
(1+sin 𝜃) sin 𝜃
= sin 𝜃+sin2 𝜃+cos2 𝜃
sin 𝜃+sin2 𝜃
= sin 𝜃+1
sin 𝜃(+1 sin 𝜃)
= 1
sin 𝜃
= cosec 𝜃
= RHS.
(ii)
L.H.S. = 1 +tan2 𝜃
1+sec 𝜃
= 1 +sin2 𝜃
cos2 𝜃
1+1
cos 𝜃
= 1 +sin2 𝜃
1+cos 𝜃×
cos 𝜃
cos2 𝜃
= 1 +sin2 𝜃
(1+cos 𝜃) cos 𝜃
= cos 𝜃+cos2 𝜃+sin2 𝜃
cos 𝜃+cos2 𝜃
= cos 𝜃+1
cos 𝜃(+1 cos 𝜃)
= 1
cos 𝜃
= sec θ
= R.H.S.
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Question 9: Prove
(𝟏+𝐭𝐚𝐧𝟐 𝜽) 𝐜𝐨𝐭 𝜽
𝐜𝐨𝐬𝐞𝐜𝟐 𝜽= 𝐭𝐚𝐧 𝜽
Solution:
L.H.S. = (1+tan2 𝜃) cot 𝜃
cosec2 𝜃
= (1+
sin2 𝜃
cos2 𝜃)×
cos 𝜃
sin 𝜃1
sin2 𝜃
= cos2 𝜃+sin2 𝜃
cos2 𝜃×
cos 𝜃
sin 𝜃× sin2 𝜃
= 1 ×sin 𝜃
cos 𝜃
= tan 𝜃
= R.H.S.
Question 10: Prove
(i) 𝐭𝐚𝐧𝟐 𝜽
(𝟏+𝐭𝐚𝐧𝟐 𝜽)+
𝐜𝐨𝐭𝟐 𝜽
(𝟏+𝐜𝐨𝐭𝟐 𝜽)= 𝟏
Solution:
L.H.S. = tan2 𝜃
1+tan2 𝜃+
cot2 𝜃
1+cot2 𝜃
=
sin2 𝜃
cos2 𝜃
1+sin2 𝜃
cos2 𝜃
+cos2 𝜃
sin2 𝜃
1+cos2 𝜃
sin2 𝜃
= sin2 𝜃
1+sin2 𝜃+
cos2 𝜃
1+cos2 𝜃
= sin2 𝜃+sin2 𝜃 cos2 𝜃+cos2 𝜃+cos2 𝜃 sin2 𝜃
(1+sin2 𝜃)(1+cos2 𝜃)
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= sin2 𝜃+sin2 𝜃 cos2 𝜃+cos2 𝜃+cos2 𝜃 sin2 𝜃
sin2 𝜃+sin2 𝜃 cos2 𝜃+cos2 𝜃+cos2 𝜃 sin2 𝜃
= 1
= R.H.S.
Question 11:
𝐬𝐢𝐧𝛉
𝟏+𝐜𝐨𝐬𝛉+
(𝟏+𝐜𝐨𝐬𝛉)
𝐬𝐢𝐧𝛉= 𝟐𝐜𝐨𝐬𝐞𝐜𝛉
Solution:
LHS =sinθ
1+cosθ+
(1+cosθ)
sinθ
= sin2θ+(1+cosθ)2
(1+cosθ)sinθ
= sin2θ+1+cos2θ+2cosθ
(1+cosθ)sinθ
= 1+1+2cosθ
(1+cosθ)sinθ
= 2+2cosθ
(1+cosθ)sinθ
= 2(1+cosθ)
(1+cosθ)sinθ
= 2
sinθ
= 2cosecθ
= RHS
Hence, LHS = RHS
Question 12: 𝐭𝐚𝐧𝛉
(𝟏−𝐜𝐨𝐭𝛉)+
𝐜𝐨𝐭𝛉
(𝟏−𝐭𝐚𝐧𝛉)= (𝟏 + 𝐬𝐞𝐜𝛉 𝐜𝐨𝐬𝐞𝐜𝛉)
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Solution:
LHS = tanθ
(1−cotθ)+
cotθ
(1−tanθ)
= tanθ (1 −cosθ
sinθ) + cotθ (1 −
sinθ
cosθ)
= sinθ tanθ
(sinθ−cosθ)+
cosθ cotθ
(cosθ−sinθ)
= sinθ×
sinθ
cosθ−cosθ×
cosθ
sinθ
(sinθ−cosθ)
= sin2θcosθ −cos2θsinθ
(sinθ−cosθ)
=sin3θ−cos3θ
cosθsinθ(sinθ−cosθ)
=(sinθ−cosθ)(sin2θ+sinθcosθ+cos2θ)
cosθsinθ(sinθ−cosθ)
= 1+sinθ
cosθcosθsinθ
=1
cosθsinθ+
sinθcosθ
cosθsinθ
= secθcosecθ + 1
= 1 + secθcosecθ
=RHS
Question 13: 𝐜𝐨𝐬𝟐𝛉
(𝟏−𝐭𝐚𝐧𝛉)+
𝐬𝐢𝐧𝟑𝛉
(𝐬𝐢𝐧𝛉−𝐜𝐨𝐬𝛉)= (𝟏 + 𝐬𝐢𝐧𝛉 𝐜𝐨𝐬𝛉)
Solution:
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cos2θ
(1−tanθ)+
sin3θ
(sinθ−cosθ)= (1 + sinθ cosθ)
LHS = cos2θ
(1−tanθ)+
sin3θ
(sinθ−cosθ)
=cos2θ
(1+sinθ
cosθ)
+sin3θ
(sinθ−cosθ)
= cos3θ
(cosθ−sinθ)+
sin3θ
(sinθ−cosθ)
= cos3θsin3θ
(cosθ−sinθ)
= (cosθ−sinθ)(cos2θ+cosθsinθ+sin2θ)
(cosθ−sinθ)
= (sin2θ + cos2θ + cosθsinθ)
= (1 + sinθcosθ)
= RHS
Hence, L.H.S. = R.H.S.
Question 14: 𝐜𝐨𝐬𝛉
(𝟏−𝐭𝐚𝐧𝛉)+
𝐬𝐢𝐧𝟐𝛉
(𝐜𝐨𝐬𝛉−𝐬𝐢𝐧𝛉)= (𝐜𝐨𝐬𝛉 + 𝐬𝐢𝐧𝛉)
Solution:
LHS = cosθ
(1−tanθ)+
sin2θ
(cosθ−sinθ)
= cosθ
(1−sinθ
cosθ)
−sin2θ
(cosθ−sinθ)
= cos2θ
(cosθ−sinθ)−
sin2θ
(cosθ−sinθ)
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= cos2θ−sin2θ
(cosθ−sinθ)
= (cosθ+sinθ)(cosθ−sinθ)
(cosθ−sinθ)
= (cosθ + sinθ)
= RHS
Hence, LHS = RHS
Question 15: (𝟏 + 𝐭𝐚𝐧𝟐𝛉)(𝟏 + 𝐜𝐨𝐭𝟐𝛉) =𝟏
(𝐬𝐢𝐧𝟐𝛉−𝐬𝐢𝐧𝟒𝛉)
Solution:
LHS =(1 + tan2θ)(1 + cot2θ)
= sec2θ. cosec2θ
(∵ sec2θ − tan2θ = 1 and cosec2θ − cot2θ = 1)
= 1
cos2θ.sin2θ
= 1
(1−sin2θ)sin2θ
= 1
(sin2θ−sin4θ)
= RHS
Hence, LHS = RHS
Question 16: 𝐭𝐚𝐧𝛉
(𝟏+𝐭𝐚𝐧𝟐𝛉)𝟐 +
𝐜𝐨𝐭𝛉
(𝟏+𝐜𝐨𝐭𝟐𝛉)𝟐 = 𝐬𝐢𝐧𝛉 𝐜𝐨𝐬𝛉
Solution:
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LHS = tanθ
(1+tan2θ)2 +cotθ
(1+cot2θ)2
= tanθ
(sec2θ)2 +cotθ
(cosec2θ)2
= tanθ
sec4θ+
cotθ
cosec4θ
= sinθ
cosθ× cos4θ +
cosθ
sinθ× sin4θ
= sinθcos3θ + cosθsin3θ
= sinθcosθ(cos2θ + sin2θ)
= sinθcosθ
= RHS
Hence, LHS = RHS
Question 17:
(i) sin6θ + cos6θ = 1 – 3 sin2θ cos2θ
(ii) sin2θ + cos4θ = cos2θ + sin4θ
(iii) cosec4θ – cosec2θ = cot4θ + cot2θ
Solution:
(i) LHS = sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ) (sin4θ − sin2θcos2θ + cos4θ)
=1×{(sin2θ)2 + 2sin2θcos2θ + (cos2θ)2 − 3sin2θcos2θ}
= (sin2θ + cos2θ)2 − 3sin2θcos2θ
= (1)2 – 3 sin2θcos2θ
=1 – 3 sin2θcos2θ
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= RHS
Hence, LHS = RHS
(ii) LHS = sin2θ + cos4θ
= sin2θ + (cos2θ)2
= sin2θ + (1 − sin2θ)2
= sin2θ + 1 − 2sin2θ + sin4θ
= 1 − sin2θ + sin4θ
= cos2θ + sin4θ
= RHS
Hence, LHS = RHS
(iii) LHS = cosec4θ − cosec2θ
= cosec2θ (cosec2θ−1)
= cosec2θ × cot2θ (∵ cosec2θ − cot2θ = 1)
= (1 + cot2θ) × cot2θ
= cot2θ + cot4θ
= RHS
Hence, LHS = RHS
Question 18:
(i) 𝟏−𝐭𝐚𝐧𝟐𝛉
𝟏+𝐭𝐚𝐧𝟐𝛉= (𝐜𝐨𝐬𝟐𝛉 − 𝐬𝐢𝐧𝟐𝛉)
(ii) 𝟏−𝐭𝐚𝐧𝟐𝛉
𝐜𝐨𝐭𝟐𝛉−𝟏= 𝐭𝐚𝐧𝟐𝛉
Solution:
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(i) LHS = 1−tan2θ
1+tan2θ
=1−
sin2θ
cos2θ
1+sin2θ
cos2θ
= cos2θ−sin2θ
cos2θ+sin2θ
= cos2θ−sin2θ
1
= cos2θ − sin2θ
= RHS
(ii) LHS = 1−tan2θ
cot2θ−1
= 1−
sin2θ
cos2θ
cos2θ
sin2θ−1
=
cos2θ−sin2θ
cos2θ
cos2θ−sin2θ
sin2θ
= sin2θ
cos2θ
= tan2θ
= RHS
Question 19:
(i) 𝐭𝐚𝐧𝛉
(𝐬𝐞𝐜𝛉−𝟏)+
𝐭𝐚𝐧𝛉
(𝐬𝐞𝐜𝛉+𝟏)= 𝟐𝐜𝐨𝐬𝐞𝐜𝛉
(ii) 𝐜𝐨𝐭𝛉
(𝐜𝐨𝐬𝐞𝐜𝛉+𝟏)+
(𝐜𝐨𝐬𝐞𝐜𝛉+𝟏)
𝐜𝐨𝐭𝛉= 𝟐𝐬𝐞𝐜𝛉
Solution:
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(i) LHS = tanθ
(secθ−1)+
tanθ
(secθ+1)
= tanθ {secθ+1+secθ−1
(secθ−1)(secθ+1)}
= tanθ {2secθ
(sec2θ−1)}
= tanθ ×2secθ
tan2θ
= 2secθ
tanθ
= 21
cosθsinθ
cosθ
= 21
sinθ
= 2 cosecθ
= RHS
Hence, LHS = RHS
(ii) LHS = cotθ
(cosecθ+1)+
(cosecθ+1)
cotθ
= cot2θ+(cosecθ+1)2
(cosecθ+1)cotθ
= cot2θ+cosec2θ+2cosecθ+1
(cosecθ+1)cotθ
= cot2θ+cosec2θ+2cosecθ+cosec2θ−cot2θ
(cosecθ+1)cotθ
= 2cosec2θ+2cosecθ
(cosecθ+1)cotθ
= 2cosecθ(cosecθ+1)
(cosecθ+1)cotθ
= 2cosecθ
cotθ
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= 2 ×1
sinθ×
sinθ
cosθ
= 2secθ
= RHS
Hence, LHS = RHS
Question 20:
(i) 𝐬𝐞𝐜𝛉−𝟏
𝐬𝐞𝐜𝛉+𝟏=
𝐬𝐢𝐧𝟐𝛉
(𝟏+𝐜𝐨𝐬𝛉)𝟐
(ii) 𝐬𝐞𝐜𝛉−𝐭𝐚𝐧𝛉
𝐬𝐞𝐜𝛉+𝐭𝐚𝐧𝛉=
𝐜𝐨𝐬𝟐𝛉
(𝟏+𝐬𝐢𝐧𝛉)𝟐
Solution:
(i) LHS = secθ−1
secθ+1
=
1
cosθ−1
1
cosθ+1
=
1−cosθ
cosθ1+cosθ
cosθ
= 1−cosθ
1+cosθ
= (1−cosθ)(1+cosθ)
(1+cosθ)(1+cosθ)
{Dividing the numerator and denominator by (1 + cosθ)}
= 1 −cos2θ
(1+cosθ)2
= sin2θ
(1+cosθ)2
= RHS
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(ii) LHS = secθ−tanθ
secθ+tanθ
=
1
cosθ−
sinθ
cosθ1
cosθ+
sinθ
cosθ
=
1−sinθ
cosθ1+sinθ
cosθ
= 1−sinθ
1+sinθ
= (1−sinθ)(1+sinθ)
(1+sinθ)(1+sinθ)
{Dividing the numerator and denominator by (1+sinθ)}
= (1−sin2θ)
(1+sinθ)2
= cos2θ
(1+sinθ)2
= RHS
Question 21: Prove each of the following identities:
(i) √𝟏+𝐬𝐢𝐧𝛉
𝟏−𝐬𝐢𝐧𝛉= (𝐬𝐞𝐜𝛉 + 𝐭𝐚𝐧𝛉)
(ii) √𝟏−𝐜𝐨𝐬𝛉
𝟏+𝐜𝐨𝐬𝛉= (𝐜𝐨𝐬𝐞𝐜𝛉 − 𝐜𝐨𝐭𝛉)
(iii) √𝟏+𝐜𝐨𝐬𝛉
𝟏−𝐜𝐨𝐬𝛉+ √
𝟏−𝐜𝐨𝐬𝛉
𝟏+𝐜𝐨𝐬𝛉= 𝟐𝐜𝐨𝐬𝐞𝐜𝛉
Solution:
(i) LHS = √1+sinθ
1−sinθ
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= √(1 + sinθ)(1 − sinθ) × (1 + sinθ)(1 + sinθ)
= √(1+sinθ)2
1−sin2θ
= √(1+sinθ)2
cos2θ
= 1+sinθ
cosθ
= 1
cosθ+
sinθ
cosθ
= (secθ + tanθ)
= RHS
(ii) LHS
= √1−cosθ
1+cosθ
= √(1−cosθ)
(1+cosθ)×
(1−cosθ)
(1−cosθ)
= √(1−cosθ)2
1−cos2θ
= √(1−cosθ)2
sin2θ
= 1−cosθ
sinθ
= 1
sinθ−
cosθ
sinθ
= (cosecθ − cotθ)
= RHS
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(iii) LHS = √1+cosθ
1−cosθ+ √
1−cosθ
1+cosθ
= √(1+cosθ)2
(1−cosθ)(1+cosθ)+ √
(1−cosθ)2
(1+cosθ)(1−cosθ)
= √(1+cosθ)2
(1−cos2θ)+ √
(1−cosθ)2
(1−cos2θ)
= √(1+cosθ)2
sin2θ+ √
(1−cosθ)2
sin2θ
= (1+cosθ)
sinθ+
1−cosθ
sinθ
= 1+cosθ+1−cosθ
sinθ
= 2sinθ
= 2cosecθ
= RHS
Question 22: 𝐜𝐨𝐬𝟑𝛉+𝐬𝐢𝐧𝟑𝛉
𝐜𝐨𝐬𝛉+𝐬𝐢𝐧𝛉+
𝐜𝐨𝐬𝟑𝛉−𝐬𝐢𝐧𝟑𝛉
𝐜𝐨𝐬𝛉−𝐬𝐢𝐧𝛉= 𝟐
Solution:
LHS = cos3θ+sin3θ
cosθ+sinθ+
cos3θ−sin3θ
cosθ−sinθ
= (cosθ+sinθ)(cos2θ−cosθsinθ+sin2θ)
(cosθ+sinθ)+
(cosθ−sinθ)(cos2θ+cosθsinθ+sin2θ)
(cosθ−sinθ)
= (cos2θ + sin2θ − cosθsinθ) + (cos2θ + sin2θ + cosθsinθ)
= (1 − cosθsinθ) + (1 + cosθsinθ)
= 2
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= RHS
Hence, LHS = RHS
Question 23: 𝐬𝐢𝐧𝛉
(𝐜𝐨𝐭𝛉+𝐜𝐨𝐬𝐞𝐜𝛉)−
𝐬𝐢𝐧𝛉
(𝐜𝐨𝐭𝛉−𝐜𝐨𝐬𝐞𝐜𝛉)= 𝟐
Solution:
LHS = sinθ
(cotθ+cosecθ)−
sinθ
(cotθ−cosecθ)
= sinθ {(cotθ−cosecθ)−(cotθ+cosecθ)
(cotθ+cosecθ)(cotθ−cosecθ)}
= sinθ {−2cosecθ
(cot2θ−cosec2θ)}
= sinθ (−2cosecθ
−1) (∵ cosec2θ − cot2θ = 1)
= sinθ. 2cosecθ
= sinθ × 2 ×1
sinθ
= 2
= RHS
Question 24:
(i) 𝐬𝐢𝐧𝛉−𝐜𝐨𝐬𝛉
𝐬𝐢𝐧𝛉+𝐜𝐨𝐬𝛉+
𝐬𝐢𝐧𝛉+𝐜𝐨𝐬𝛉
𝐬𝐢𝐧𝛉−𝐜𝐨𝐬𝛉=
𝟐
(𝟐𝐬𝐢𝐧𝟐𝛉−𝟏)
(ii) 𝐬𝐢𝐧𝛉+𝐜𝐨𝐬𝛉
𝐬𝐢𝐧𝛉−𝐜𝐨𝐬𝛉+
𝐬𝐢𝐧𝛉−𝐜𝐨𝐬𝛉
𝐬𝐢𝐧𝛉+𝐜𝐨𝐬𝛉=
𝟐
(𝟏−𝟐𝐜𝐨𝐬𝟐𝛉)
Solution:
(i) LHS = sinθ−cosθ
sinθ+cosθ+
sinθ+cosθ
sinθ−cosθ
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= (sinθ−cosθ)2+(sinθ+cosθ)2
(sinθ+cosθ)(sinθ−cosθ)
= sin2θ+cos2θ−2sinθcosθ+sin2θ+cos2θ+2sinθcosθ
sin2θ−cos2θ
= 1+1
sin2 θ−(1−sin2θ) (∵ sin2θ + cos2θ = 1)
= 2
sin2θ−1+sin2θ
= 2
2sin2θ−1
= RHS
(ii) LHS
= sinθ+cosθ
sinθ−cosθ+
sinθ−cosθ
sinθ+cosθ
= (sinθ+cosθ)2+(sinθ−cosθ)2
(sinθ−cosθ)(sinθ+cosθ)
= sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ−2sinθcosθ
(sin2θ−cos2θ)
= 1+1
(1−cos2θ)−cos2θ (∵ sin2θ + cos2θ = 1)
= 2
1−2cos2θ
= RHS
Question 25:
𝟏+𝐜𝐨𝐬𝛉−𝐬𝐢𝐧𝟐𝛉
𝐬𝐢𝐧𝛉(𝟏+𝐜𝐨𝐬𝛉)= 𝐜𝐨𝐭𝛉
Solution:
LHS = 1+cosθ−sin2θ
sinθ(1+cosθ)
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= (1+cosθ)−(1−cos2θ)
sinθ(1+cosθ)
= cosθ+cos2θ
sinθ(1+cosθ)
= cosθ(1+cosθ)
sinθ(1+cosθ)
= cosθ
sinθ
= cot θ
= RHS
Hence, L.H.S. = R.H.S.
Question 26:
(i) 𝐜𝐨𝐬𝐞𝐜𝛉+𝐜𝐨𝐭𝛉
𝐜𝐨𝐬𝐞𝐜𝛉−𝐜𝐨𝐭𝛉= (𝐜𝐨𝐬𝐞𝐜𝛉 + 𝐜𝐨𝐭𝛉)𝟐 = 𝟏 + 𝟐𝐜𝐨𝐭𝟐 + 𝟐𝐜𝐨𝐬𝐞𝐜𝛉 𝐜𝐨𝐭𝛉
(ii) 𝐬𝐞𝐜𝛉+𝐭𝐚𝐧𝛉
𝐬𝐞𝐜𝛉−𝐭𝐚𝐧𝛉= (𝐬𝐞𝐜𝛉 + 𝐭𝐚𝐧𝛉)𝟐 = 𝟏 + 𝟐𝐭𝐚𝐧𝟐𝛉 + 𝟐𝐬𝐞𝐜𝛉 𝐭𝐚𝐧𝛉
Solution:
(i) Here, cosecθ+cotθ
cosecθ−cotθ
= (cosecθ+cotθ)(cosecθ+cotθ)
(cosecθ−cotθ)(cosecθ+cotθ)
= (cosecθ+cotθ)2
(cosec2θ−cot2θ)
= (cosecθ+cotθ)2
1
= (cosecθ + cotθ)2
Again, (cosecθ + cotθ)2
= cosec2θ + cot2θ + 2cosecθcotθ
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= 1 + cot2θ + cot2θ + 2cosecθcotθ (∵ cosec2θ − cot2θ = 1)
= 1 + 2cot2θ + 2cosecθcotθ
(ii) Here, secθ+tanθ
secθ−tanθ
= (secθ+tanθ)(secθ+tanθ)
(secθ−tanθ)(secθ+tanθ)
= (secθ+tanθ)2
sec2θ−tan2θ
=(secθ+tanθ)2
1
= (secθ + tanθ)2
Again, (secθ + tanθ)2
= sec2θ + tan2θ + 2secθtanθ
= 1 + tan2θ + tan2θ + 2secθtanθ
= 1 + 2tan2θ + 2secθtanθ
Question 27:
(i) 𝟏+𝐜𝐨𝐬𝛉+𝐬𝐢𝐧𝛉
𝟏+𝐜𝐨𝐬𝛉−𝐬𝐢𝐧𝛉=
𝟏+𝐬𝐢𝐧𝛉
𝐜𝐨𝐬𝛉
(ii) 𝐬𝐢𝐧𝛉+𝟏−𝐜𝐨𝐬𝛉
𝐜𝐨𝐬𝛉−𝟏+𝐬𝐢𝐧𝛉=
𝟏+𝐬𝐢𝐧𝛉
𝐜𝐨𝐬𝛉
Solution:
(i) LHS = 1+cosθ+sinθ
1+cosθ−sinθ
= {(1+cosθ)+sinθ}{(1+cosθ)+sinθ}
{(1+cosθ)−sinθ}{(1+cosθ)+sinθ}
{Multiplying the numerator and denominator by (1 + cosθ + sinθ)}
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= {(1+cosθ)+sinθ}2
{(1+cosθ)2−sin2θ}
= 1+cos2θ+2cosθ+sin2θ+2sinθ(1+cosθ)
1+cos2θ+2cosθ−sin2θ
= 2+2cosθ+2sinθ(1+cosθ)
1+cos2θ+2cosθ−(1−cos2θ)
= 2(1+cosθ)+2sinθ(1+cosθ)
2cos2θ+2cosθ
= 2(1+cosθ)(1+sinθ)
2cosθ(1+cosθ)
= 1+sinθ
cosθ
= RHS
(ii) LHS = sinθ+1−cosθ
cosθ−1+sinθ
= (sinθ+1−cosθ)(sinθ+cosθ+1)
(cosθ−1+sinθ)(sinθ+cosθ+1)
{Multiplying and dividing by 1+ cos θ + sin θ}
= (sinθ+1)2−cos2θ
(sinθ+cosθ)2−12
= sin2θ+1+2sinθ−cos2θ
sin2θ+cos2θ+2sinθcosθ−1
= sin2θ+sin2θ+cos2θ+2sinθ−cos2θ
2sinθcosθ
= 2sin2θ+2sinθ
2sinθcosθ
= 2sinθ(1+sinθ)
2sinθcosθ
= 1+sinθ
cosθ
= RHS
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Question 28: 𝐬𝐢𝐧𝛉
(𝐬𝐞𝐜𝛉+𝐭𝐚𝐧𝛉−𝟏)+
𝐜𝐨𝐬𝛉
(𝐜𝐨𝐬𝐞𝐜𝛉+𝐜𝐨𝐭𝛉−𝟏)= 𝟏
Solution:
LHS = sinθ
(secθ+tanθ−1)+
cosθ
(cosecθ+cotθ−1)
= sinθcosθ
1+sinθ−cosθ+
cosθsinθ
1+cosθ−sinθ
= sinθcosθ [1
1+(sinθ−cosθ)+
1
1−(sinθ−cosθ)]
= sinθcosθ [1−(sinθ−cosθ)+1+(sinθ−cosθ)
{1+(sinθ−cosθ)}{1−(sinθ−cosθ)}]
= sinθcosθ [1−sinθ+cosθ+1+sinθ−cosθ
1−(sinθ−cosθ)2 ]
= 2sinθcosθ
1−(sin2θ+cos2θ−2sinθcosθ)
= 2sinθcosθ
2sinθcosθ
= 1
= RHS
Hence, LHS = RHS
Question 29: 𝐬𝐢𝐧𝛉+𝐜𝐨𝐬𝛉
𝐬𝐢𝐧𝛉−𝐜𝐨𝐬𝛉+
𝐬𝐢𝐧𝛉−𝐜𝐨𝐬𝛉
𝐬𝐢𝐧𝛉+𝐜𝐨𝐬𝛉=
𝟐
(𝐬𝐢𝐧𝟐𝛉−𝐜𝐨𝐬𝟐𝛉)=
𝟐
(𝟐𝐬𝐢𝐧𝟐𝛉−𝟏)
Solution:
We have sinθ+cosθ
sinθ−cosθ+
sinθ−cosθ
sinθ+cosθ
= (sinθ+cosθ)2+(sinθ−cosθ)2
(sinθ−cosθ)(sinθ+cosθ)
= sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ−2sinθcosθ
sin2θ−cos2θ
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= 1+1
sin2θ−cos2θ
= 2
sin2θ−cos2θ
Again, 2
sin2θ−cos2θ
= 2
sin2θ−(1−sin2θ)
= 2
2sin2θ−1
Question 30: 𝐜𝐨𝐬𝛉 𝐜𝐨𝐬𝐞𝐜𝛉−𝐬𝐢𝐧𝛉 𝐬𝐞𝐜𝛉
𝐜𝐨𝐬𝛉+𝐬𝐢𝐧𝛉= 𝐜𝐨𝐬𝐞𝐜𝛉 − 𝐬𝐞𝐜𝛉
Solution:
LHS = cosθ cosecθ−sinθ secθ
cosθ+sinθ
=
cosθ
sinθ−
sinθ
cosθ
cosθ+sinθ
= cos2θ−sin2θ
cosθsinθ(cosθ+sinθ)
= (cosθ+sinθ)(cosθ−sinθ)
cosθsinθ(cosθ+sinθ)
= (cosθ−sinθ)
cosθsinθ
= 1
sinθ−
1
cosθ
= cosecθ − secθ
= RHS
Hence, LHS = RHS
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Question 31: (𝟏 + 𝐭𝐚𝐧𝛉 + 𝐜𝐨𝐭𝛉)(𝐬𝐢𝐧𝛉 − 𝐜𝐨𝐬𝛉) = (𝐬𝐞𝐜𝛉
𝐜𝐨𝐬𝐞𝐜𝟐𝛉−
𝐜𝐨𝐬𝐞𝐜𝛉
𝐬𝐞𝐜𝟐𝛉)
Solution:
LHS = (1 + tanθ + cotθ)(sinθ − cosθ)
= sinθ + tanθsinθ + cotθsinθ − cosθ − tanθcosθ − cotθcosθ
= sinθ + tanθsinθ +cosθ
sinθ× sinθ − cosθ −
sinθ
cosθ× cosθ − cotθcosθ
= sinθ + tanθsinθ + cosθ − cosθ − sinθ − cotθcosθ
= tanθsinθ − cotθcosθ
= sinθ
cosθ×
1
cosecθ−
cosθ
sinθ×
1
secθ
= 1
cosecθ×
1
cosecθ× secθ −
1
secθ×
1
secθ× cosecθ
= secθ
cosec2θ−
cosecθ
sec2θ
= RHS
Hence, LHS = RHS
Question 32: Prove that 𝐜𝐨𝐭𝟐𝛉(𝐬𝐞𝐜𝛉−𝟏)
(𝟏+𝐬𝐢𝐧𝛉)+
𝐬𝐞𝐜𝟐𝛉(𝐬𝐢𝐧𝛉−𝟏)
(𝟏+𝐬𝐞𝐜𝛉)= 𝟎
Solution:
LHS = cot2θ(secθ−1)
(1+sinθ)+
sec2θ(sinθ−1)
(1+secθ)
=
cos2θ
sin2θ(
1
cosθ−1)
(1+sinθ)+
1
cos2θ(sinθ−1)
(1+1
cosθ)
=
cos2θ
sin2θ(1−
cosθ
cosθ)
(1+sinθ)+
(sinθ−1)
cos2θ
(cosθ+1
cosθ)
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= cos2θ(1−cosθ)
sin2θcosθ(1+sinθ)+
(sinθ−1)cosθ
(cosθ+1)cos2 θ
= cosθ(1−cosθ)
(1−cos2θ)(1+sinθ)+
(sinθ−1)cosθ
(cosθ+1)(1−sin2θ)
= cosθ(1−cosθ)
(1−cosθ)(1+cosθ)(1+sinθ)+
−(1−sinθ)cosθ
(cosθ+1)(1−sinθ)(1+sinθ)
= cosθ
(1+cosθ)(1+sinθ)−
cosθ
(cosθ+1)(1+sinθ)
= 0
= RHS
Question 33:
{𝟏
(𝐬𝐞𝐜𝟐𝛉−𝐜𝐨𝐬𝟐𝛉)+
𝟏
(𝐜𝐨𝐬𝐞𝐜𝟐𝛉−𝐬𝐢𝐧𝟐𝛉)} (𝐬𝐢𝐧𝟐𝛉 𝐜𝐨𝐬𝟐𝛉) =
𝟏−𝐬𝐢𝐧𝟐𝛉𝐜𝐨𝐬𝟐𝛉
𝟐+𝐬𝐢𝐧𝟐𝛉 𝐜𝐨𝐬𝟐𝛉
Solution:
LHS = {1
(sec2θ−cos2θ)+
1
(cosec2θ−sin2θ)} (sin2θ cos2θ)
= {cos2θ
1−cos4θ+
sin2θ
1−sin4θ} (sin2θ cos2θ)
= {cos2θ
(1−cos2θ)(1+cos2θ)+
sin2θ
(1−sin2θ)(1+sin2θ)} (sin2θcos2θ)
= [cot2θ
1+cos2θ+
tan2θ
1+sin2θ] sin2θcos2θ
= cos4θ
1+cos2θ+
sin4θ
1+sin2θ
= (cos2θ)
2
1+cos2θ+
(sin2θ)2
1+sin2θ
= (1−sin2θ)
1+cos2θ+
(1−cos2θ)2
1+sin2θ
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= (1−sin2θ)
2(1+sin2)+(1−cos2θ)
2(1+cos2θ)
(1+sin2θ)(1+cos2θ)
= cos4θ(1+sin2θ)+sin4θ(1+cos2θ)
1+sin2θ+cos2 θ+sin2θcos2θ
= cos4θ+cos4θsin2θ+sin4θ+sin4θcos2θ
1+1+sin2θcos2θ
= cos2θ+sin4θ+sin2θcos2θ(sin2θ+cos2θ)
2+sin2θcos2θ
= (cos2θ)
2+(sin2θ)
2+sin2θcos2θ(1)
2+sin2θcos2θ
= (cos2θ+sin2θ)
2−2sin2θcos2θ+sin2θcos2 θ(1)
2+sin2θcos2θ
= 12+cos2θsin2θ−2cos2θsin2θ
2+sin2θcos2θ
= 1−cos2θsin2θ
2+sin2θcos2θ
= RHS
Question 34: (𝐬𝐢𝐧 𝐀−𝐬𝐢𝐧 𝐁)
(𝐜𝐨𝐬 𝐀+𝐜𝐨𝐬 𝐁)+
(𝐜𝐨𝐬 𝐀−𝐜𝐨𝐬 𝐁)
(𝐬𝐢𝐧 𝐀+𝐬𝐢𝐧 𝐁)= 𝟎
Solution:
LHS = (sin A−sin B)
(cos A+cos B)+
(cos A−cos B)
(sin A+sin B)
= (sinA−sinB)(sinA+sinB)+(cosA−cosB)(cosA−cosB)
(cosA+cosB)(sinA+sinB)
= sin2A−sin2B+cos2A−cos2B
(cosA+cosB)(sinA+sinB)
= 0
(cosA+cosB)(sinA+sinB)
= 0
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= RHS
Question 35: 𝐭𝐚𝐧 𝐀+𝐭𝐚𝐧 𝐁
𝐜𝐨𝐭 𝐀+𝐜𝐨𝐭 𝐁= 𝐭𝐚𝐧 𝐀 𝐭𝐚𝐧 𝐁
Solution:
LHS = tan A+tan B
cot A+cot B
= tanA+tanB
1
tanA+
1
tanB
= tanA+tanBtanA+tanB
tanAtanB
= tanAtanB(tanA+tanB)
(tanA+tanB)
= tanAtanB
= RHS
Hence, LHS = RHS
Question 36: Show that none of the following is an identity:
(i) cos2θ + cos θ = 1
(ii) sin2θ + sin θ = 2
(iii) tan2θ + sin θ = cos2θ
Solution:
(i) cos2θ + cos θ = 1
LHS = cos²θ + cosθ
=1 − sin2θ + cosθ
=1 − (sin2θ − cosθ)
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Since LHS ≠ RHS, this is not an identity.
(ii) sin2θ + sinθ = 1
LHS = sin2θ + sinθ
= 1 − cos2θ + sinθ
= 1 − (cos2θ − sinθ)
Since LHS ≠ RHS, this is not an identity.
(iii) tan2θ + sinθ = cos2θ
LHS = tan2θ + sinθ
= sin2θ
cos2θ+ sinθ
= 1−cos2θ
cos2θ+ sinθ
= sec2θ − 1 + sinθ
Since LHS ≠ RHS, this is not an identity.
Question 37: Prove that (𝐬𝐢𝐧𝛉 − 𝟐𝐬𝐢𝐧𝟑𝛉) = (𝟐𝐜𝐨𝐬𝟑𝛉 − 𝐜𝐨𝐬𝛉)𝐭𝐚𝐧𝛉
Solution:
RHS = (2cos3θ − cosθ)tanθ
= (2cos2θ − 1)cosθ ×sinθ
cosθ
= [2(1 − sin2θ) − 1]sinθ
= (2 − 2sin2θ − 1)sinθ
= (1 − 2sin2θ)sinθ
= (sinθ − 2sin3θ)
=LHS
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Question 38: If 1 + sin2θ = 3 sin θ cos θ then prove that tan θ = 1 or 𝟏
𝟐.
Solution:
1 + sin2(θ) = 3sin(θ) cos(θ)
Divide by cos2(θ) throughout to get: sec2(θ) + tan2(θ) = 3 tan(θ)
⇒ (1 + tan2(θ)) + tan2(θ) = 3 tan(θ)
⇒ 2tan2(θ) − 3 tan(θ) + 1 = 0, which is quadratic in tan(θ).
Solving using Quadratic formula, we get:
tan(θ) = +3 ± √(−3)2−4(2)(1)
2(2)= √
+3±9−8
4=
+3±1
4
⇒ tan(θ) = +4
4, +
2
4 = 1 or
1
2.
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Exercise 13.2
Page number: 594
Question 1: If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove
that (m2 + n2) = (a2 + b2).
Solution:
a cos θ + b sin θ = m
Squaring equation, we get
a2 cos2 θ + b2 sin2 θ + 2ab cos θ sin θ = m2 …… (1)
Again Square equation, a sin θ – b cos θ = n
a2 sin2 θ + b2 cos2 θ – 2ab cos θ sin θ = n2 …… (2)
Add (1) and (2)
a2 cos2 θ + b2 sin2 θ + 2ab cos θ sin θ + a2 sin2 θ + b2 cos2 θ – 2ab cos θ
sin θ = m2 +n2
a2 (cos2 θ + sin2θ) + b2 (cos2 θ + sin2 θ) = m2 + n2
(Using cos2 θ + sin2θ = 1)
a2 + b2 = m2 + n2
Hence Proved.
Question 2: If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove
that (x2 − y2) = (a2 − b2).
Solution:
a sec θ + b tan θ = x
a tan θ + b sec θ = y
Squaring above equations:
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a2 sec2 θ + b2 tan2 θ + 2ab sec θ tan θ = x2 …. (1)
a2 tan2 θ + b2 sec2 θ + 2ab sec θ tan θ = y2 …. (2)
Subtract equation (2) from (1):
a2 (sec2 θ – tan2 θ) + b2 (tan2 θ – sec2 θ) = x2 – y2
(using sec2 θ = 1 + tan2 θ)
or a2 – b2 = x2 – y2
Hence proved.
Question 3: If (𝒙
𝒂𝐬𝐢𝐧 𝜽 −
𝒚
𝒃𝐜𝐨𝐬 𝜽) = 𝟏 and (
𝒙
𝒂𝐬𝐢𝐧 𝜽 +
𝒚
𝒃𝐜𝐨𝐬 𝜽) = 𝟏,
Prove that 𝒙𝟐
𝒂𝟐 +𝒚𝟐
𝒃𝟐 = 𝟐
Solution:
𝑥
𝑎 sin θ –
𝑦
𝑏 cos θ = 1
𝑥
𝑎 cos θ +
𝑦
𝑏 sin θ = 1
Squaring both the equations, we have
𝑥2
𝑎2 sin2 θ + 𝑦2
𝑏2 cos2 θ – 2 𝑥𝑦
𝑎𝑏 cos θ sin θ = 1 ….... (1)
𝑥2
𝑎2 cos2 θ + 𝑦2
𝑏2 sin2 θ + 2 𝑥𝑦
𝑎𝑏 cos θ sin θ = 1 …… (2)
Add (1) and (2), we get
𝑥2
𝑎2 (sin2 θ + cos2 θ) + 𝑦2
𝑏2 (sin2 θ + cos2 θ) = 1+1
(Using cos2 θ + sin2θ = 1)
𝑥2
𝑎2 + 𝑦2
𝑏2 = 2
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Question 4: If (sec θ + tan θ) = m and (sec θ − tan θ) = n, show that
mn = 1.
Solution:
(sec θ + tan θ) = m … (1) and
(sec θ − tan θ) = n …. (2)
Multiply (1) and (2), we have
(sec θ + tan θ) (sec θ – tan θ) = mn
(sec2 θ – tan2 θ) = mn (Because sec2 θ – tan2 θ = 1)
1 = mn
Or mn = 1
Hence Proved
Question 5: If (cosec θ + cot θ) = m and (cosec θ − cot θ) = n, show
that mn = 1.
Solution:
(cosec θ + cot θ) = m … (1) and
(cosec θ − cot θ) = n … (2)
Multiply (1) and (2)
(cosec2 θ – cot2 θ) = mn (Because cosec2 θ – cot2 θ = 1)
1 = mn
Or mn = 1
Hence Proved
Question 6: If x = a cos3 θ and y = b sin3 θ, prove that
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(𝑥
𝑎)
2
3+ (
𝑦
𝑏)
2
3= 1
Solution:
x = a cos3 θ
y = b sin3 θ
L.H.S.
(𝑥
𝑎)
2
3+ (
𝑦
𝑏)
2
3
= (𝑎 cos3 𝜃
𝑎)
2
3+ (
𝑏 sin3 𝜃
𝑏)
2
3
= (cos3 𝜃)2
3 + (sin3 𝜃)2
3
= cos2 θ + sin2 θ
= 1
= R.H.S.
Question 7: If (tan θ + sin θ) = m and (tan θ − sin θ) = n, prove that
(m2 − n2)2 = 16mn.
Solution:
(tan θ + sin θ) = m and (tan θ − sin θ) = n
To Prove: (m2 − n2)2 = 16mn
L.H.S. = (m2 − n2)2
= [(tan θ + sin θ)2 – (tan θ − sin θ)2]2
= (4tan θ sin θ)2
= 16 tan2 θ sin2 θ … (1)
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R.H.S. = 16mn
= 16(tan θ + sin θ) (tan θ − sin θ)
= 16(tan2 θ − sin2 θ)
= 16 [sin2 θ(1 – cos2θ)
cos2θ]
= 16 × sin2θ
cos2θ × (1 – cos2 θ)
= 16 tan2 θ sin2 θ … (2)
From (1) and (2)
L.H.S. = R.H.S.
Question 8: If (cot θ + tan θ) = m and (sec θ − cos θ) = n, prove that
(𝐦𝟐𝐧)(𝟐
𝟑) – (𝐦𝐧𝟐)(
𝟐
𝟑) = 1.
Solution:
(cot θ + tan θ) = m and (sec θ − cos θ) = n
m = 1
tanθ + tan θ =
(1+tan2 θ)
tan θ=
(sec2θ)
tan θ
= 1
sinθcosθ
or m = 1
sinθcosθ
Again, n = sec θ − cos θ
= 1
cosθ− cosθ
= (1 – cos2 θ)
cos θ
= sin2 θ
cosθ
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or n = sin2 θ
cosθ
To prove: (m2n)(2
3) – (mn2)(
2
3) = 1
L.H.S.
Substituting the values of m and n, we have
= (m2n)(2
3) – (mn2)(
2
3)
= [(1
sinθcosθ)
2×
sin2 θ
cosθ]
2
3
– [(1
sinθcosθ) × (
sin2 θ
cosθ)
2
]
2
3
= [sin2 θ
sin2 θ cos3θ]
2
3 – [(
sin4 θ
sinθcos3θ)]
2
3
= [1
cos3θ]
2
3 – [(
sin3 θ
cos3θ)]
2
3
= 1
cos2θ−
sin2 θ
cos2 θ
= (1 – sin2 θ) cos2 θ (We know, 1 – sin2 θ = cos2 θ)
= cos2 θ
cos2 θ
= 1
=R.H.S.
Hence proved.
Question 9: If (cosec θ − sin θ) = a3 and (sec θ − cos θ) = b3, prove that
a2b2(a2 + b2) = 1.
Solution:
(cosec θ − sin θ) = a3 and (sec θ − cos θ) = b3
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(cosec θ − sin θ) = a3
(1
sinθ− sin θ) = a3
cos2θ
sinθ = a3
And a2 = (a3)(2
3)
= (cos2θ
sinθ)
(2
3)
…..(1)
Again
(sec θ − cos θ) = b3
(1
cosθ− cos θ) = b3
= sin2 θ
cos θ = b3
And, b2 = (b3)(2
3)
= (sin2 θ
cos θ)
(2
3)
To Prove: a2b2(a2 + b2) = 1
L.H.S. a2b2(a2 + b2)
= [(cos2θ
sinθ)]
2
3× [(
sin2 θ
cos θ)]
2
3× ([(
cos2θ
sinθ)]
2
3+ [(
sin2 θ
cos θ)]
2
3)
= (cos2θsin2 θ
cos θ sinθ)
2
3× ([(
cos2θ
sinθ)]
2
3+ [(
sin2 θ
cos θ)]
2
3)
= (cos3θ)2
3 × (sin3θ)2
3
= sin2 θ + cos2 θ
= 1
=R.H.S.
Hence proved.
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Question 10: If (2 sin θ + 3 cos θ) = 2, show that (3 sin θ − 2 cos θ) = ±
3.
Solution:
(2 sin θ + 3 cos θ) = 2 … (1)
(2 sin θ + 3 cos θ)2 + (3 sin θ – 2 cos θ)2
= 4sin2 θ + 9 cos2 θ + 12sin θ cos θ + 9 sin2 θ + 4 cos2 θ – 12 sin θ cos θ
= 13sin2 θ + 13 cos2 θ
= 13(sin2 θ + cos2 θ)
= 13 (Because (sin2 θ + cos2 θ) = 1)
⇒ (2 sin θ + 3 cos θ)2 + (3 sin θ – 2 cos θ)2 = 13
Using equation (1)
⇒ (2)2 + (3 sin θ – 2 cos θ)2 = 13
⇒ (3 sin θ – 2 cos θ)2 = 9
or (3 sin θ – 2 cos θ) = ± 3
Hence Proved.
Question 11: If (𝐬𝐢𝐧𝛉 + 𝐜𝐨𝐬𝛉) = √𝟐𝐜𝐨𝐬𝛉, show that 𝐜𝐨𝐭𝛉 = (√𝟐 +
𝟏).
Solution:
We have, (sinθ + cosθ) = √2cosθ
Dividing both sides by sinθ we get
sinθ
sinθ+
cosθ
sinθ=
√2cosθ
sinθ
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⇒ 1 + cotθ = √2cotθ
⇒ √2cotθ − cotθ = 1
⇒ (√2 − 1)cotθ = 1
⇒ cotθ =1
(√2−1)
⇒ cotθ =1
(√2−1)×
(√2+1)
(√2+1)
⇒ cotθ =(√2+1)
2−1
⇒ cotθ =(√2+1)
1
∴ cotθ = (√2 + 1)
Question 12: If cos θ + sin θ = √𝟐 sin θ, show that sin θ − cos θ
= √𝟐 cos θ.
Solution:
Given: cos θ + sin θ = √2 sin θ
We have (sinθ + cosθ)2 + (sinθ − cosθ)2 = 2(sin2θ + cos2θ)
⇒ (√2sinθ)² + (sinθ − cosθ)² = 2
⇒ 2sin2θ + (sinθ − cosθ)2 = 2
⇒ (sinθ − cosθ)2 = 2 − 2sin2θ
⇒ (sinθ − cosθ)2 = 2(1 − sin2θ)
⇒ (sinθ − cosθ)2 = 2cos2θ
⇒ (sinθ − cosθ) = √2cosθ
Hence proved.
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Question 13: If 𝐬𝐞𝐜𝛉 + 𝐭𝐚𝐧𝛉 = 𝒑, prove that
(i) 𝐬𝐞𝐜 𝜽 =𝟏
𝟐(𝒑 +
𝟏
𝒑) (ii) 𝐭𝐚𝐧𝛉 =
𝟏
𝟐(𝒑 −
𝟏
𝒑) (iii) 𝐬𝐢𝐧𝛉 =
𝒑𝟐−𝟏
𝒑𝟐+𝟏
Solution:
(i)We have, secθ + tanθ = 𝑝 .....(1)
⇒secθ+tanθ
1×
secθ−tanθ
secθ−tanθ= 𝑝
⇒sec2θ−tan2θ
secθ−tanθ= 𝑝
⇒1
secθ−tanθ= 𝑝
⇒ secθ − tanθ =1
𝑝 ..... (2)
Adding (1) and (2), we get
2secθ = 𝑝 +1
𝑝
⇒ secθ =1
2(𝑝 +
1
𝑝)
(ii) Subtracting (2) from (1), we get
2tanθ = (𝑝 −1
𝑝)
⇒ tanθ =1
2(𝑝 −
1
𝑝)
(iii) Using (i) and (ii), we get
sinθ =tanθ
secθ
=
1
2(𝑝−
1
𝑝)
1
2(𝑝+
1
𝑝)
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=(
𝑝2−1
𝑝)
(𝑝2+1
𝑝)
∴ sinθ =𝑝2−1
𝑝2+1
Question 14: If tan A = n tan B and sin A = m sin B, prove that
cos2A = (𝒎𝟐−𝟏)
𝒏𝟐−𝟏.
Solution:
We have tanA = 𝑛 tanB
⇒ cotB =𝑛
tanA ...(i)
Again, sinA = 𝑚 sinB
⇒ cosecB =𝑚
sinA ...(ii)
Squaring (i) and (ii) and subtracting (ii) from (i), we get
𝑚2
sin2A−
𝑛2
tan2A= 𝐜𝐨𝐬𝐞𝐜2𝐁 − cot2B
⇒𝑚2
sin2A−
n2cosA
sin²A= 1
⇒ 𝑚² − 𝑛2cos2A = sin2A
⇒ 𝑚² − 𝑛2 cos2A = 1 − cos2A
⇒ 𝑛2cos2A − cos2A = 𝑚2 − 1
⇒ cos2A(𝑛2 − 1) = (𝑚2 − 1)
⇒ cos2A = (𝑚2 − 1)(𝑛2 − 1)
∴ cos2A = (𝑚2 − 1)(𝑛2 − 1)
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Question 15: If 𝒎 = (𝐜𝐨𝐬𝛉 − 𝐬𝐢𝐧𝛉) and 𝒏 = (𝐜𝐨𝐬𝛉 + 𝐬𝐢𝐧𝛉), then
show that √𝒎
𝒏+ √
𝒏
𝒎=
𝟐
√𝟏−𝐭𝐚𝐧𝟐𝛉.
Solution:
LHS = √𝑚
𝑛+ √
𝑛
𝑚
= √𝑚
√𝑛+
√𝑛
√𝑚
= 𝑚+𝑛
√𝑚𝑛
= (cosθ−sinθ)+cosθ+sinθ
√(cosθ−sinθ)(cosθ+sinθ)
= 2cosθ
√cos2θ−sin2θ
= (
2cosθ
cosθ)
(√cos2θ−sin2θ
cosθ)
= 2
√cos2θ
cos2θ−
sin2θ
cos2θ
= 2
√1−tan2θ
= RHS
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Exercise 13.3
Page number: 596
Very Short and Short Answer Questions:
Question 1: Write the value of (1 – sin2 θ) sec2 θ.
Solution:
(1 – sin2θ) sec2 θ = (cos2 θ) × 1
cos2𝜃
= 1
Question 2: Write the value of (1-cos2θ) cosec2θ.
Solution:
(1-cos2θ) cosec2θ = sin2θ × 1
sin2θ
= 1
Question 3: Write the value of (1 + tan2θ) cos2θ.
Solution:
(1 + tan2θ) cos2θ = sec2 θ × 1
sec2θ
= 1
Question 4: Write the value of (1 + cot2θ) sin2θ.
Solution:
(1 + cot2θ) sin2θ = cose2θ × 1
cosec2𝜃
= 1
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Question 5: Write the value of sin2θ + 𝟏
(𝟏+𝐭𝐚𝐧𝟐 𝜽)
Solution:
sin2θ + 1
(1+tan2 𝜃)
= sin2θ + 1
(sec2 𝜃)
= sin2θ + cos2θ
= 1
Question 6: Write the value of (𝐜𝐨𝐭𝟐𝛉 −𝟏
𝐬𝐢𝐧𝟐𝛉)
Solution:
(cot2θ −1
sin2θ) = (
cos2θ
sin2θ –
1
sin2θ)
= (cos2θ−1)
sin2θ
= −sin2θ
sin2θ
= −1
Question 7: Write the value of sin θ cos(90°-θ) + cos θ sin(90°-θ).
Solution:
sin θ cos(90°- θ) + cos θ sin(90°- θ) = sinθ × sinθ + cosθ × cosθ
= sin2θ + cos2θ
= 1
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Question 8: Write the value of cosec2(90°-θ) – tan2θ.
Solution:
cosec2(90° - θ) – tan2θ = sec2θ – tan2θ
= 1
Question 9: Write the value of sec2 θ (1 + sin θ) (1 – sin θ).
Solution:
sec2θ (1 + sin θ) (1 – sin θ) = sec2θ (1 – sin2 θ)
= sec2θ × cos2θ
= 1
cos2θ × cos2θ
= 1
Question 10: Write the value of cosec2θ (1 + cos θ) (1 – cos θ).
Solution:
cosec2θ (1 + cos θ) (1 – cos θ) = cosec2θ (1 – cos2 θ)
= cosec2θ × sin2θ
= cosec2θ × 1
cosec2𝜃
= 1
Question 11: Write the value of sin2θ cos2θ (1 + tan2θ) (1 + cot2θ).
Solution:
sin2θ cos2θ (1 + tan2θ) (1 + cot2θ) = sin2 θ × cos2 θ × sec2 θ × cosec2 θ
= sin2θ × cos2θ × 1
cos2θ×
1
sin2θ =1
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Question 12: Write the value of (1 + tan2θ) (1 + sin θ) (1 – sin θ).
Solution:
(1 + tan2θ) (1 + sin θ) (1 – sin θ) = sec2θ (1 – sin2 θ)
= sec2θ × cos2θ
= 1
cos2θ× cos2θ
= 1
Question 13: Write the value of 3 cot2θ – 3 cosec2θ.
Solution:
3 cot2θ – 3 cosec2θ = 3(cot2θ – cosec2θ)
= 3 × -1
= -3
Question 14: Write the value of 4tan2θ – 𝟒
𝐜𝐨𝐬𝟐𝛉
Solution:
4tan2θ – 4
cos2θ = 4 ×
sin2θ
cos2θ–
4
cos2θ
= (4(sin2θ – 1))
cos2θ
= 4(−cos2θ)
cos2θ
= -4
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Question 15: Write the value of (𝐭𝐚𝐧2𝛉 – 𝐬𝐞𝐜2𝛉)
(𝐜𝐨𝐭2𝛉 – 𝐜𝐨𝐬𝐞𝐜2𝛉)
Solution:
(tan2θ – sec2θ)
(cot2θ – cosec2θ) =
−1
−1
= 1 (Using 1 + cot2θ = cose2θ and 1 + tan2θ = sec2θ)
Question 16: If 𝐬𝐢𝐧𝛉 =𝟏
𝟐, then write the value of (𝟑𝐜𝐨𝐭𝟐𝛉 + 𝟑).
Solution:
As, sinθ =1
2
So, cosec θ =1
sin θ= 2 .....(i)
Now,
3cot2θ + 3
= 3(cot2θ + 1)
= 3cosec2θ
= 3(2)2 [Using (i)]
= 3(4)
= 12
Question 17: If 𝐜𝐨𝐬𝛉 =𝟐
𝟑, the write the value of (𝟒 + 𝟒 𝐭𝐚𝐧𝟐𝛉).
Solution:
4 + 4 tan2θ
= 4(1 + tan2θ)
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= 4sec2θ
=4
cos2θ
=4
(2
3)
2
=4
(4
9)
=4×9
4
= 9
Question 18: If 𝐜𝐨𝐬𝛉 =𝟕
𝟐𝟓, then write the value of (𝐭𝐚𝐧𝛉 + 𝐜𝐨𝐭𝛉).
Solution:
As sin2θ = 1 − cos2θ
= 1 − (7
25)
2
= 1 −49
625
=625−49
625
⇒ sin2θ =576
625
⇒ sinθ = √576
625
⇒ sinθ =24
25
Now,
tanθ + cotθ
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=sinθ
cosθ+
cosθ
sinθ
=sin2θ+cos2θ
cosθ sinθ
=1
(7
25×
24
25)
=1
(168
625)
=625
168
Question 19: If 𝐜𝐨𝐬𝛉 =𝟐
𝟑, then write the value of
(𝐬𝐞𝐜𝛉−𝟏)
(𝐬𝐞𝐜𝛉+𝟏).
Solution:
sec θ−1
sec θ+1
= (1
cos θ−
1
1) (
1
cos θ+
1
1)
= (1−cos θ
cos θ) (
1+cos θ
cos θ)
=1−cosθ
1+cosθ
=(
1
1−
2
3)
(1
1+
2
3)
=(
1
3)
(5
3)
=1
5
Question 20: If 𝟓 𝐭𝐚𝐧𝛉 = 𝟒, then write the value of (𝐜𝐨𝐬𝛉−𝐬𝐢𝐧𝛉)
(𝐜𝐨𝐬𝛉+𝐬𝐢𝐧𝛉).
Solution:
We have,
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5 tanθ = 4
⇒ tanθ =4
5
Now,
(cosθ−sinθ)
(cosθ+sinθ)
=(
cosθ
cosθ−
sinθ
cosθ)
(cosθ
cosθ+
sinθ
cosθ) (Dividing numerator and denominator by cosθ)
= (1−tanθ)
(1+tanθ)
=(
1
1−
4
5)
(1
1+
4
5)
=(
1
5)
(9
5)
=1
9
Question 21: If 𝟑 𝐜𝐨𝐭𝛉 = 𝟒, then write the value of (𝟐 𝐜𝐨𝐬𝛉+𝐬𝐢𝐧𝛉)
(𝟒 𝐜𝐨𝐬𝛉−𝐬𝐢𝐧𝛉).
Solution:
We have,
3 cotθ = 4
⇒ cotθ =4
3
Now, (2 cosθ+sinθ)
(4 cosθ−sinθ)
=(
2cosθ
sinθ+
sinθ
sinθ)
(4cosθ
sinθ−
sinθ
sinθ) (Dividing numerator and denominator by sinθ)
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= (2cotθ+1)
(4cotθ−1)
=(2×
4
3+1)
(4×4
3−1)
=(
8
3+
1
1)
(16
3−
1
1)
=(
8+3
3)
(16−3
3)
=(
11
3)
(13
3)
=11
13
Question 22: If 𝐜𝐨𝐭𝛉 =𝟏
√𝟑, then write the value of
(𝟏−𝐜𝐨𝐬𝟐𝛉)
(𝟐−𝐬𝐢𝐧𝟐𝛉).
Solution:
We have,
cotθ =1
√3
⇒ cotθ = cot (π
3)
⇒ θ =π
3
Now, (1−cos2θ)
(2−sin2θ)
=1−cos2(
𝜋
3)
2−sin2(𝜋
3)
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=1−(
1
2)
2
2−(√3
2)
2
=(
1
1−
1
4)
(2
1−
3
4)
=(
3
4)
(5
4)
=3
5
Question 23: If 𝐭𝐚𝐧𝛉 =𝟏
√𝟓, then write the value of
(𝐜𝐨𝐬𝐞𝐜𝟐𝛉−𝐬𝐞𝐜𝟐𝛉)
(𝐜𝐨𝐬𝐞𝐜𝟐𝛉+𝐬𝐞𝐜𝟐𝛉).
Solution:
(cosec2θ−sec2θ)
(cosec2θ+sec2θ)
= (1+cot2θ)−(1+tan2θ)
(1+cot2θ)+(1+tan2θ)
= (1+
1
tan2θ)−(1+tan2θ)
(1+1
tan2θ)+(1+tan2θ)
= (1+
1
tan2θ−1−tan2θ)
(1+1
tan2θ+1+tan2θ)
= (
1
tan2θ−tan2θ)
(1
tan2θ+tan2θ+2)
= (
√5
1)
2
−(1
√5)
2
(√5
1)
2
+(1
√5)
2+2
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= (
5
1−
1
5)
(5
1+
1
5+
2
1)
= (
24
5)
(36
5)
= 24
36
= 2
3
Question 24: If 𝐜𝐨𝐭 𝑨 =𝟒
𝟑 and (𝐀 + 𝐁) = 𝟗𝟎°, then what is the value
of tan B?
Solution:
We have,
cotA =4
3
⇒ cot(90° − B) =4
3 (As, A + B = 90°)
∴ tanB =4
3
Question 25: If 𝐜𝐨𝐬𝐁 =𝟑
𝟓 and (𝐀 + 𝐁) = 𝟗𝟎°, then find the value of
𝐬𝐢𝐧𝐀.
Solution:
We have,
cosB =3
5
⇒ cos(90° − A) =3
5 (As, A + B = 90°)
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∴ sinA =3
5
Question 26: If √𝟑 𝐬𝐢𝐧𝛉 = 𝐜𝐨𝐬𝛉 and θ is an acute angle, then find the
value of θ.
Solution:
We have,
√3 sinθ = cosθ
⇒sinθ
cosθ=
1
√3
⇒ tanθ =1
√3
⇒ tanθ = tan30°
∴ θ = 30°
Question 27: Write the value of tan10° tan20° tan70° tan80°.
Solution:
tan10° tan20° tan70° tan80°
= cot (90° − 10°) cot (90° − 20°) tan 70° tan 80°
= cot 80° cot 70° tan 70° tan 80°
= 1
tan 80°×
1
tan 70°× tan70° × tan80°
= 1
Question 28: Write the value of tan1° tan2° ... tan89°.
Solution:
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tan1° tan2° ... tan89°=tan1° tan2° tan3° ... tan45° ...tan87° tan88° tan89°
= tan1° tan2° tan3° ... tan45° ...cot (90°−87°) cot (90°−88°) cot (90°−89°)
= tan1° tan2° tan3° ... tan45° ...cot3° cot2° cot1°
= tan1° × tan2° × tan3° × ... × 1 × ... × 1
tan 3° ×
1
tan 2° ×
1
tan 1°
= 1
Question 29: Write the value of cos1° cos2° ... cos180°.
Solution:
cos1° cos2° ... cos180°
= cos1° cos2° ... cos90° ... cos180°
= cos1° cos2° ... 0 ... cos180°
= 0
Question 30: If 𝐭𝐚𝐧𝐀 =𝟓
𝟏𝟐, then find the value of (sinA+cosA)secA.
Solution:
(sin A + cos A) sec A
= (sinA + cosA)1
cosA
=sinA
cosA+
cosA
cosA
= tanA + 1
=5
12+
1
1
=5+12
12
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=17
12
Question 31: If 𝐬𝐢𝐧𝛉 = 𝐜𝐨𝐬(𝛉 − 𝟒𝟓°), where θ is acute, then find the
value of θ.
Solution:
We have,
sinθ = cos(θ − 45°)
⇒ cos(90° − θ) = cos(θ − 45°)
Comparing both sides, we get
90° − θ = θ − 45°
⇒ θ + θ = 90° + 45°
⇒ 2θ = 135°
⇒ θ = (135
2)
o
∴ θ = 67.5°
Question 32: Find the value of 𝐬𝐢𝐧𝟓𝟎°
𝐜𝐨𝐬𝟒𝟎°+
𝐜𝐨𝐬𝐞𝐜𝟒𝟎°
𝐬𝐞𝐜𝟓𝟎°− 𝟒𝐜𝐨𝐬𝟓𝟎° 𝐜𝐨𝐬𝐞𝐜𝟒𝟎°.
Solution:
sin50°
cos40°+
cosec40°
sec50°− 4cos50° cosec40°
= cos(90°−50°)
cos40°+
sec(90°−40°)
sec50°− 4sin(90° − 50°) cosec40°
= cos40°
cos40°+
sec50°
sec50°− 4 sin40° ×
1
sin40°
= 1 + 1 − 4 = −2
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Question 33: Find the value of sin 48° sec 42° + cos 48° cosec 42°.
Solution:
sin 48° sec 42° + cos 48° cosec 42°
= sin 48° cosec (90° − 42°) + cos 48° sec (90° − 42°)
= sin48° cosec 48° + cos 48° sec 48°
= sin48° × 1
sin48° + cos48° ×
1
cos48°
= 1 + 1
= 2
Question 34: If 𝒙 = 𝒂 𝐬𝐢𝐧𝛉 and 𝒚 = 𝒃 𝐜𝐨𝐬 𝜽, then write the value of
(𝒃𝟐𝒙𝟐 + 𝒂𝟐𝒚𝟐).
Solution:
(𝑏2𝑥2 + 𝑎2𝑦2)
= 𝑏2(𝑎 sinθ)2 + 𝑎2(𝑏 cosθ)2
= 𝑏2𝑎2 sin2θ + 𝑎2𝑏2cos2θ
= 𝑎2𝑏2(sin2θ + cos2θ)
= 𝑎2𝑏2 (1)
= 𝑎2𝑏2
Question 35: If 𝟓𝒙 = 𝐬𝐞𝐜𝛉 and 𝟓𝒙 = 𝐭𝐚𝐧𝛉, then find the value of
𝟓 (𝒙𝟐 −𝟏
𝒙𝟐).
Solution:
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5 (𝑥2 −1
𝑥2)
= 25
5(𝑥2 −
1
𝑥2)
=1
5(25𝑥2 −
25
𝑥2)
=1
5[(5𝑥)2 − (5𝑥)2]
=1
5[(secθ)2 − (tanθ)2]
=1
5(sec2θ − tan2θ)
=1
5(1)
=1
5
Question 36: If 𝐜𝐨𝐬𝐞𝐜𝛉 = 𝟐𝒙 and 𝐜𝐨𝐭𝛉 = 𝟐𝒙, then find the value of
𝟐 (𝒙𝟐 −𝟏
𝒙𝟐).
Solution:
2 (𝑥2 −1
𝑥2)
= 4
2(𝑥2 −
1
𝑥2)
= 1
2(4𝑥2 − 4𝑥2)
= 1
2[(2𝑥)2 − (2𝑥)2]
= 1
2[(cosecθ)2 − (secθ)2]
= 1
2(cosec2θ − sec2θ)
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= 1
2(1)
= 1
2
Question 37: If 𝐬𝐞𝐜𝛉 + 𝐭𝐚𝐧𝛉 = 𝒙, then find the value of sec θ.
Solution:
We have,
secθ + tanθ = 𝑥 ..... (i)
⇒ secθ+tanθ
1×
secθ−tanθ
secθ−tanθ= 𝑥
⇒ sec2θ−tan2θ
secθ−tanθ= 𝑥
⇒ 1
secθ−tanθ=
𝑥
1
⇒ secθ − tanθ =1
𝑥 ..... (ii)
Adding (i) and (ii), we get
2 secθ = 𝑥 +1
𝑥
⇒2 secθ =𝑥2+1
𝑥
∴ secθ =𝑥2+1
2𝑥
Question 38: Find the value of 𝐜𝐨𝐬𝟑𝟖° 𝐜𝐨𝐬𝐞𝐜𝟓𝟐°
𝐭𝐚𝐧𝟏𝟖° 𝐭𝐚𝐧𝟑𝟓° 𝐭𝐚𝐧𝟔𝟎° 𝐭𝐚𝐧𝟕𝟐° 𝐭𝐚𝐧𝟓𝟓°.
Solution:
cos38° cosec52°
tan18° tan35° tan60° tan72° tan55°
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= cos38° sec(90°−52°)
cot(90°−18°) cot(90°−35°)tan60° tan72° tan55°
= cos38° sec38°
cot72° cot55° tan60° tan72° tan55°
=
cos38°×1
cos38°1
tan72°×
1
tan55°×√3×tan72°×tan55°
=1
√3
Question 39: If 𝐬𝐢𝐧 𝜽 = 𝒙, then write the value of cotθ.
Solution:
cotθ =cosθ
sinθ
= √1−sin2θ
sinθ
= √1−𝑥2
2
Question 40: If 𝐬𝐞𝐜𝛉 = 𝒙, then write the value of tan θ.
Solution:
As, tan2θ = sec2θ − 1
So, tanθ = √sec2θ − 1 = √𝑥2 − 1
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