Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2

Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 त्रिकोणमितीय अनुपात एवं असमिकाएँ

Ex 10.2 Trigonometrical Ratios and Identities अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

प्रश्न 1.
sec 45° का मान ज्ञात कीजिए।
हल:
sec 45° का मान (UPBoardSolutions.com) [latex] \sqrt{{2}} [/latex] होता है।

प्रश्न 2.
tan 60° का मान ज्ञात कीजिए।
हल:
tan60° का मांन [latex] \sqrt{{3}} [/latex] होता है।

UP Board Solutions

प्रश्न 3.
यदि tanθ = [latex]\frac{1}{\sqrt{3}}[/latex], तब θ का मान ज्ञात कीजिए।
हलः
यदि tanθ = [latex]\frac{1}{\sqrt{3}}[/latex] तो tanθ = tan30° अत: θ = 30°

प्रश्न 4.
cos 60° × sin 60° का मान ज्ञात कीजिए।
हलः
cos60° × sin60° (UPBoardSolutions.com) = [latex]\frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}[/latex]

Ex 10.2 Trigonometrical Ratios and Identities लघु उत्तरीय प्रश्न-I (Short Answer Type Questions-I)

प्रश्न 5.
यदि θ = 30°, तब सिद्ध कीजिए कि sin2θ = [latex]\frac{\sqrt{3}}{2}[/latex]
हलः
यदि θ = 30° ∴ sin20 = sin2 × 30° = sin60° = [latex]\frac{\sqrt{3}}{2}[/latex]

प्रश्न 6.
यदि θ = 45°, तब सिद्ध कीजिए कि sin2θ = 1
हलः
यदि θ = 45° ∴ sin20 (UPBoardSolutions.com) = sin2 × 45° = sin90° = 1

UP Board Solutions

प्रश्न 7.
सिद्ध कीजिए कि sin30°.cosec30°.tan30° = [latex]\frac{1}{\sqrt{3}}[/latex]
हल:
sin30° × cosec30° × tan30° = [latex]\frac{1}{2} \times \frac{2}{1} \times \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}[/latex]

Ex 10.2 Trigonometrical Ratios and Identities लघु उत्तरीय प्रश्न-II (Short Answer Type Questions-II)

निम्न की सत्यता की जाँच कीजिए –
प्रश्न 8.
cos60° cos45° – sin60°sin 45° = [latex]\frac{\sqrt{2}-\sqrt{6}}{4}[/latex]
हल:
LHS = cos 60°.cos 45° – sin60°sin 45°
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 1

UP Board Solutions

प्रश्न 9.
cos30°cos45° – sin30° sin 45° = [latex]\frac{\sqrt{3}-1}{2 \sqrt{2}}[/latex]
हलः
LHS = cos 30°cos 45° – sin30°sin 45°
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 2

प्रश्न 10.
cosec230°sin2 45° – sec260° = -2
हलः
LHS = (UPBoardSolutions.com) cosec 230°sin245° – sec260°
= (2)2 · [latex]\left(\frac{1}{\sqrt{2}}\right)^{2}[/latex] – (2)2
= 4·[latex]\frac{1}{2}[/latex] – 4 = – 2 = RHS

प्रश्न 11.
4cos260° + 4sin2 45° – sin2 30° = [latex]\frac{1}{4}[/latex]
हल:
LHS = 4 cos260° + 4sin2 45° – sin230°
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 3

UP Board Solutions

प्रश्न 12.
cos 90° = 1 – 2 sin2 45° = 2 cos2 45° – 1
हल:
cos 90° = 1 – 2 sin245° = 2 cos2 45° – 1
LHS → cos 90 = 0
RHS = 1 – 2 sin2 45°
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 4

प्रश्न 13.
cos60° = 1 – 2 sin2 30° = 2 cos230° – 1
हल:
cos 60° = 1 – 2 sin2 30° = 2 cos2 30° – 1
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 5

UP Board Solutions

प्रश्न 14.
cos2 30° + cos2 45° + cos260° = sin2 30° + sin2 45° + sin2 60°
हल:
LHS = cos230° + (UPBoardSolutions.com) cos2 45° + cos260°
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 6

UP Board Solutions

Ex 10.2 Trigonometrical Ratios and Identities दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)

प्रश्न 15.
सिद्ध कीजिए कि cot2 30° + cot2 45° + cot2 60° = [latex]\frac{13}{3}[/latex]
हल:
LHS = cot2 30° + cot2 45° + cot260°
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 7

प्रश्न 16.
सिद्ध कीजिए कि 2(sin2 45° + cot230°) – 3(cosec260° – sec260°) = 15
हल:
LHS = 2(sin2 45° + cot230°) – 3(cosec260° – sec260°)
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 8

प्रश्न 17.
सिद्ध कीजिए कि 4 cot2 45° – sec260° + sin230° = [latex]\frac{1}{4}[/latex]
हलः
LHS = 4cot2 45° – sec260° + sin230° (UPBoardSolutions.com)
= 4 x (1)2 – (2)2 + [latex]\left(\frac{1}{2}\right)^{2}[/latex]
= 4 – 4 + [latex]\frac{1}{4}[/latex] = [latex]\frac{1}{4}[/latex] = RHS

प्रश्न 18.
यदि tan(A – B) = [latex]\frac{1}{\sqrt{3}}[/latex], व tan(A + B) = [latex] \sqrt{{3}} [/latex], तब A व B के मान ज्ञात कीजिए।
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 9
A का मान समीकरण (2) में रखने पर 45° + B = 60°
B = 60° – 45° = 150

UP Board Solutions

प्रश्न 19.
यदि sin(A – B) = cos(A + B) = [latex]\frac{1}{2}[/latex], तब A व B के मान ज्ञात कीजिए।
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 10
समी० (1) में A (UPBoardSolutions.com) का मान रखने पर
45° – B = 30°
– B = 30° – 450
– B = -15° ⇒ B = 150

प्रश्न 20.
यदि A = 30%, तो सिद्ध कीजिए कि –
(i) sin 3A = 3 sin A – 4 sin3 A
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 11
हलः
यदि A = 30°
sin3A = 3sinA – 4sin3 A
LHS → sin3 × 30° = sin90° = 1
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 12
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 13

UP Board Solutions

प्रश्न 21.
यदि A = 45° तथा B = 30°, तो सिद्ध कीजिए कि –
(i) tan(A + B) = [latex]\frac{\tan A+\tan B}{1-\tan A \tan B}[/latex]
(ii) sin(A + B) = sin (UPBoardSolutions.com) A cos B + cos A sin B
(iii) cos(A – B) = cos A cos B + sin A sin B
हलः
यदि A = 45° और B = 30°
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 14
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 15

UP Board Solutions

प्रश्न 22.
यदि θ = 30°, तब सिद्ध कीजिए कि cosθ = [latex]\sqrt{\frac{1+\cos 2 \theta}{2}}[/latex]
हलः
Balaji Class 10 Maths Solutions Chapter 10 Trigonometrical Ratios and Identities Ex 10.2 16

Balaji Publications Mathematics Class 10 Solutions

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