Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4

Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 त्रिकोणमितीय अनुपात एवं असमिकाएँ

Ex 10.4 Trigonometrical Ratios and Identities अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

प्रश्न 1.
यदि sinθ = cost तब θ का मान ज्ञात कीजिए।
हल:
∵ sinθ = cosθ
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 1

प्रश्न 2.
sin2θ sec2θ का मान ज्ञात कीजिए।
हल:
sin2 θ · sec2θ (UPBoardSolutions.com) = sin2θ · [latex]\frac{1}{\cos ^{2} \theta}[/latex]
= tan2θ

प्रश्न 3.
secθ sinθ का मान ज्ञात कीजिए।
हलः
secθ ·sinθ = [latex]\frac{1}{\cos \theta}[/latex]·sin2
= tanθ

UP Board Solutions

प्रश्न 4.
(secθ – tanθ) (secθ + tanθ) का मान ज्ञात कीजिए।
हलः
(secθ – tanθ)(secθ + tanθ) = sec2θ – tan2θ
=1

प्रश्न 5.
यदि cosθ = [latex]\frac{a}{b}[/latex], तब sine का मान ज्ञात कीजिए।
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 2
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 3

प्रश्न 6.
यदि tanθ = [latex]\frac{a}{b}[/latex], तब [latex]\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}[/latex] का मान ज्ञात कीजिए।
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 4
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 5

UP Board Solutions

प्रश्न 7.
यदि x = a cosθ, y = b sinθ, a (b2x2 + a2y2) का मान ज्ञात कीजिए।
हलः
x = acosθ तथा y = bsinθ
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 6

प्रश्न 8.
(sin4 θ – cos4 θ + 1) cosec2θ का मान ज्ञात कीजिए।
हलः
(sin4 θ – cos4 θ + 1)cosec2θ
= [(sin2θ)2 – (cos2 θ)2 + 1]cosec2θ
= [(sin2θ + cos2 θ) (UPBoardSolutions.com) (sin2 θ – cos2θ) + 1]·cosec2θ
= [1(sin2 θ – cos2θ) + sin2θ + cos2 θ]cosec2θ
= (sin2 θ – cos2θ + sin2 θ + cos2θ).cosec2θ
= 2 sin2θ × [latex]\frac{1}{\sin ^{2} \theta}[/latex] = 2

UP Board Solutions

प्रश्न 9.
[latex]\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}[/latex] का मान ज्ञात कीजिए। (NCERT)
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 7

प्रश्न 10.
[latex]\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}[/latex] का मान ज्ञात कीजिए। (NCERT)
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 8
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 9

प्रश्न 11.
[latex]\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}[/latex] का मान ज्ञात कीजिए। (NCERT)
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 10

प्रश्न 12.
[latex]\frac{\sin \theta}{1+\cos \theta}[/latex] का मान ज्ञात कीजिए। (NCERT)
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 11

UP Board Solutions

प्रश्न 13.
(1 + tanθ + secθ)(1 + cotθ – cosecθ) का मान ज्ञात कीजिए।
हलः
(1 + tanθ + secθ) (UPBoardSolutions.com) (1 + cotθ – cosecθ)
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 12

प्रश्न 14.
sin2 θ + sin4 θ = (यदि cosθ + cos2 θ = 1) का मान ज्ञात कीजिए।
हलः
यदि cosθ + cos2θ = 1
coso = 1 – cos2 θ
cosθ = (UPBoardSolutions.com) sin2θ
या sin2 θ = cos 2
तब sin2θ + sin4 θ = sin2 θ(1 + sin2θ)
= cos θ(1 + cosθ)
= cosθ + cos2 θ = 1

UP Board Solutions

Ex 10.4 Trigonometrical Ratios and Identities लघु उत्तरीय प्रश्न-I (Short Answer Type Questions-I)

निम्न को सिद्ध कीजिए-

प्रश्न 15.
(1 – sin2θ) sec2θ = 1
हलः
L.H.S. = (1 – sin2 θ)sec2θ
= cos2 θ – sec2 θ (UPBoardSolutions.com) (सर्वसम्मिका sin2θ + cos2θ = 1 से)
= cos 2θ·[latex]\frac{1}{\cos ^{2} \theta}[/latex]
= 1 = R.H.S

प्रश्न 16.
(secθ + tanθ)2 = [latex]\frac{1+\sin \theta}{1-\sin \theta}[/latex]
हलः
L.H.S. = (secθ + tanθ)2
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 13

UP Board Solutions

प्रश्न 17.
(sin θ – cosθ)(cotθ + tanθ) = secθ – cosec θ
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 14

प्रश्न 18.
[latex]\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}[/latex] = 1 – 2 secθ tan θ + 2 tan2θ
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 15

UP Board Solutions

प्रश्न 19.
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 16
हलः
अंश का परिमेयकरण (UPBoardSolutions.com) करने पर,
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 17

प्रश्न 20.
(1 – sinθ) (1 + sin θ) (1 + tan2θ) = 1
हलः
L.H.S. = (1 – sinθ)(1 + sinθ)(1 + tan2θ)
= (1 – sin2θ)sec2θ
= cos2θ · (UPBoardSolutions.com) sec2θ
= cos2θ [latex]\frac{1}{\cos ^{2} \theta}[/latex] = 1 = R.H.S.

प्रश्न 21.
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 18
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 39

प्रश्न 22.
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 18
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 19

प्रश्न 23.
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 20
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 21

UP Board Solutions

Ex 10.4 Trigonometrical Ratios and Identities लघु उत्तरीय प्रश्न-II (Short Answer Type Questions-II)

प्रश्न 24.
यदि tan θ + sin θ = a, tanθ – sin θ = b, तब सिद्ध कीजिए a2 – b2 = 4[latex]\sqrt{a b}[/latex]
हलः
L.H.S. = a2 – b2
= (tan θ + sinθ)2 – (UPBoardSolutions.com) (tan θ – sinθ)2
= tan2 θ + sin2 θ + 2 tanθsinθ – tan2 θ – sin2θ + 2 tanθ sin2
= 4 tanθ sinθ
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 22
∴ L.H.S. = R.H.S.

प्रश्न 25.
यदि asin 3 θ + b cos3 θ = sin θ cosθ in a sinθ – b cos θ = 0, तब सिद्ध कीजिए कि a2 + b2 = 1
हलः
asin3 θ + bcos 3 θ = sinθ cosθ … (1)
asinθ – bcos θ = 0 … (2)
समीकरण (2) से,
asinθ = bcos θ …(3)
समीकरण (1) से, asinθ, का मान रखने पर
(asinθ).sin2θ + bcos3 θ = sinθ cosθ
bcosθsin2θ + bcos3θ = (UPBoardSolutions.com) sinθ cosθ
bcos θ(sin2θ + cos2θ) = sinθ cosθ
bcosθ (1) = sin θcosθ
∴ b = sin θ …(4)
समीकरण (3) से, b का मान रखने पर
asinθ = sin θcos θ
∴ a = cos θ
L.H.S. = a2 + b2 …(5)
= cos 2θ + sin2θ
= 1 = R.H.S.

सिद्ध कीजिए कि –

UP Board Solutions
प्रश्न 26.
tanθ – cotθ = [latex]\frac{2 \sin ^{2} \theta-1}{\sin \theta \cos \theta}=\frac{1-2 \cos ^{2} \theta}{\sin \theta \cos \theta}[/latex]
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 23

प्रश्न 27.
sec2 θ + cosec2θ = sec2 θ cosec2θ (UP 2000, 02)
हलः
L. H. S. = sec2θ + (UPBoardSolutions.com) cosec2θ
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 24

प्रश्न 28.
(cosecθ – cotθ)2 = [latex]\frac{1-\cos \theta}{1+\cos \theta}[/latex] (NCERT)
हलः
L. H. S. = (cosecθ – cotθ)2
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 25

प्रश्न 29.
[latex]\sqrt{\sec ^{2} \theta+\csc ^{2} \theta}[/latex] = tanθ + cotθ
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 26

UP Board Solutions

प्रश्न 30.
[latex]\frac{1+\cos \theta}{1-\cos \theta}[/latex] = (cosecθ + cotθ)2
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 27

UP Board Solutions

Ex 10.4 Trigonometrical Ratios and Identities दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)

प्रश्न 31.
सिद्ध कीजिए कि tan2 A – tan2 B = [latex]\frac{\sin ^{2} A-\sin ^{2} B}{\cos ^{2} A \cos ^{2} B}=\frac{\cos ^{2} B-\cos ^{2} A}{\cos ^{2} B \cos ^{2} A}[/latex]
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 28

प्रश्न 32.
सिद्ध कीजिए [latex s = 2]\frac{\cot \theta+\csc \theta-1}{\cot \theta-\csc \theta+1}=\frac{1+\cos \theta}{\sin \theta}[/latex]
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 29
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 30

UP Board Solutions

प्रश्न 33.
सिद्ध कीजिए [latex s=2]\frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta[/latex] (UP 2012)
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 31

प्रश्न 34.
सिद्ध कीजिए [latex s=2]\frac{\csc \theta+\cot \theta}{\csc \theta-\cot \theta}[/latex] = (cosecθ + cotθ)2 = 1 + 2 cot2 θ + 2 cosecθ cotθ.
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 32

UP Board Solutions

प्रश्न 35.
सिद्ध कीजिए
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 33
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 34
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 35
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 36

UP Board Solutions

प्रश्न 36.
यदि sinθ + cosθ = p a secθ + cosecθ = q, तब सिद्ध कीजिए कि
q(p2 – 1) = 2p
हलः
L. H. S. = 9(p2 – 1)- 2p
= (sec θ + cosec θ) (UPBoardSolutions.com) [(sin θ + cos θ)2 – 1]
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 37
∵ (sinθ + cosθ) = P
= P × 2 = 2P = R.H.S.

UP Board Solutions

प्रश्न 37.
यदि cosθ + sinθ = [latex] \sqrt{{2}} [/latex] · cosθ, तब सिद्ध कीजिए कि (cosθ – sinθ) = [latex] \sqrt{{2}} [/latex] · sinθ
हलः
यदि cosθ + sinθ = [latex] \sqrt{{2}} [/latex]cosθ
दोनों पक्षों का वर्ग करने पर,
(cosθ + sinθ)2 = ([latex] \sqrt{{2}} [/latex] cosθ)2
cos2θ + sin2θ + 2cosθ sinθ = 2cos2θ
sin2θ = 2cos2θ – cos2θ (UPBoardSolutions.com) = 2sinθ · cosθ
cos2θ – 2cosθ·sinθ = sin2θ
cos2θ + sin2θ – 2cosθ·sinθ = sin2θ + sin2θ
(cosθ – sinθ)2 = 2sin2θ
cosθ – sinθ = [latex] \sqrt{{2}} [/latex] sinθ इति सिद्धम्।

प्रश्न 38.
यदि sinθ – cosθ = 0 तब सिद्ध कौजिए कि sin4θ + cos4θ = [latex]\frac{1}{2}[/latex]
हलः
दिया है, sinθ – cosθ = 0
दोनों और वर्ग करने पर,
(sinθ – cosθ)2 = 0
sin2θ + cos2θ – 2sinθ cosθ = 0
1 – 2 sinθ cosθ = 0
1 = 2sinθ cosθ
sinθ·cosθ = [latex]\frac{1}{2}[/latex]
L. H. S. = sin4θ + cos4θ
= (sin2θ)2 + (cos2θ)2 + 2sin2θcos2θ – 2sin2θcos2θ
= (sin2θ + cos2θ)2 – 2(sinθ·cosθ)2
= (1)2 – 2[latex]\left(\frac{1}{2}\right)^{2}[/latex] (UPBoardSolutions.com)
= 1 – 2 × [latex]\frac{1}{4}[/latex] = 1 – [latex]\frac{1}{2}=\frac{1}{2}[/latex] = R.H.S

UP Board Solutions

प्रश्न 39.
यदि [latex]\frac{x}{a}[/latex]cosθ + [latex]\frac{y}{b}[/latex] sinθ = 1 व [latex]\frac{x}{a}[/latex]sinθ – [latex]\frac{y}{b}[/latex] cosθ = 1, तब सिद्ध कीजिए कि [latex]\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}[/latex] = 2
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.4 38

Balaji Publications Mathematics Class 10 Solutions

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