बोर्ड परीक्षा के पिछले वर्ष के प्रश्न (PYQs)
ये CBSE और राज्य बोर्ड के हाल के प्रश्नपत्रों के आधार पर परीक्षा-शैली के प्रश्न हैं। प्रत्येक को उस तर्क के साथ पूर्णतः हल किया गया है जो एक बोर्ड परीक्षक अपेक्षा करता है।
अंक-युक्ति: आप जो सर्वसमिका या मान उपयोग करते हैं उसे सदैव बताएँ। सर्वसमिका उपपत्तियों में, हर बीजगणितीय चरण दिखाएँ — परीक्षक विधि के अंक उदारता से देते हैं।
सभी 26 प्रश्नों का अभ्यास करें। ये 1-अंक, 2-अंक, 3-अंक और 5-अंक के प्रारूपों को समेटते हैं।
PYQ 1 (1 अंक): यदि sin θ = 3 5 \sin\theta = \tfrac{3}{5} sin θ = 5 3 cos θ \cos\theta cos θ
हल:
cos θ = 1 − 9 25 = 16 25 = 4 5 \cos\theta = \sqrt{1 - \tfrac{9}{25}} = \sqrt{\tfrac{16}{25}} = \tfrac45 cos θ = 1 − 25 9 = 25 16 = 5 4
अंतिम उत्तर: cos θ = 4 5 \cos\theta = \tfrac45 cos θ = 5 4
PYQ 2 (1 अंक): 2 sin 30 ° cos 30 ° 2\sin 30° \cos 30° 2 sin 30° cos 30°
हल:
= 2 ⋅ 1 2 ⋅ 3 2 = 3 2 = 2\cdot\tfrac12\cdot\tfrac{\sqrt3}{2} = \tfrac{\sqrt3}{2} = 2 ⋅ 2 1 ⋅ 2 3 = 2 3
अंतिम उत्तर: 3 2 \tfrac{\sqrt3}{2} 2 3
PYQ 3 (1 अंक): tan 35 ° cot 55 ° \dfrac{\tan 35°}{\cot 55°} cot 55° tan 35°
हल:
cot 55 ° = tan 35 ° \cot 55° = \tan 35° cot 55° = tan 35°
अंतिम उत्तर: 1।
PYQ 4 (1 अंक): यदि sec θ = 17 8 \sec\theta = \tfrac{17}{8} sec θ = 8 17 tan θ \tan\theta tan θ
हल:
tan 2 θ = sec 2 θ − 1 = 289 64 − 1 = 225 64 \tan^2\theta = \sec^2\theta - 1 = \tfrac{289}{64} - 1 = \tfrac{225}{64} tan 2 θ = sec 2 θ − 1 = 64 289 − 1 = 64 225 tan θ = 15 8 \tan\theta = \tfrac{15}{8} tan θ = 8 15
अंतिम उत्तर: tan θ = 15 8 \tan\theta = \tfrac{15}{8} tan θ = 8 15
PYQ 5 (2 अंक): sin 2 30 ° + cos 2 45 ° + tan 2 60 ° \sin^2 30° + \cos^2 45° + \tan^2 60° sin 2 30° + cos 2 45° + tan 2 60°
हल:
sin 2 30 ° = 1 4 \sin^2 30° = \tfrac14 sin 2 30° = 4 1 cos 2 45 ° = 1 2 \cos^2 45° = \tfrac12 cos 2 45° = 2 1 tan 2 60 ° = 3 \tan^2 60° = 3 tan 2 60° = 3 1 4 + 1 2 + 3 = 1 + 2 + 12 4 = 15 4 \tfrac14 + \tfrac12 + 3 = \tfrac{1 + 2 + 12}{4} = \tfrac{15}{4} 4 1 + 2 1 + 3 = 4 1 + 2 + 12 = 4 15
अंतिम उत्तर: 15 4 \tfrac{15}{4} 4 15
PYQ 6 (2 अंक): यदि tan A = cot B \tan A = \cot B tan A = cot B A + B = 90 ° A + B = 90° A + B = 90°
हल:
cot B = tan ( 90 ° − B ) \cot B = \tan(90° - B) cot B = tan ( 90° − B ) tan A = tan ( 90 ° − B ) \tan A = \tan(90° - B) tan A = tan ( 90° − B ) अतः A = 90 ° − B A = 90° - B A = 90° − B A + B = 90 ° A + B = 90° A + B = 90°
अंतिम उत्तर: सिद्ध हुआ।
PYQ 7 (2 अंक): यदि sin ( A − B ) = 1 2 \sin(A - B) = \tfrac12 sin ( A − B ) = 2 1 cos ( A + B ) = 1 2 \cos(A + B) = \tfrac12 cos ( A + B ) = 2 1 A > B A > B A > B A A A B B B
हल:
sin ( A − B ) = 1 2 ⇒ A − B = 30 ° \sin(A - B) = \tfrac12 \Rightarrow A - B = 30° sin ( A − B ) = 2 1 ⇒ A − B = 30° cos ( A + B ) = 1 2 ⇒ A + B = 60 ° \cos(A + B) = \tfrac12 \Rightarrow A + B = 60° cos ( A + B ) = 2 1 ⇒ A + B = 60° जोड़ने पर: 2 A = 90 ° ⇒ A = 45 ° 2A = 90° \Rightarrow A = 45° 2 A = 90° ⇒ A = 45° B = 15 ° B = 15° B = 15°
अंतिम उत्तर: A = 45 ° A = 45° A = 45° B = 15 ° B = 15° B = 15°
PYQ 8 (2 अंक): सिद्ध कीजिए cos θ 1 + sin θ + 1 + sin θ cos θ = 2 sec θ \dfrac{\cos\theta}{1 + \sin\theta} + \dfrac{1 + \sin\theta}{\cos\theta} = 2\sec\theta 1 + sin θ cos θ + cos θ 1 + sin θ = 2 sec θ
हल:
LHS = cos 2 θ + ( 1 + sin θ ) 2 cos θ ( 1 + sin θ ) = \dfrac{\cos^2\theta + (1+\sin\theta)^2}{\cos\theta(1+\sin\theta)} = cos θ ( 1 + sin θ ) cos 2 θ + ( 1 + sin θ ) 2
अंश = cos 2 θ + 1 + 2 sin θ + sin 2 θ = 2 + 2 sin θ = 2 ( 1 + sin θ ) = \cos^2\theta + 1 + 2\sin\theta + \sin^2\theta = 2 + 2\sin\theta = 2(1+\sin\theta) = cos 2 θ + 1 + 2 sin θ + sin 2 θ = 2 + 2 sin θ = 2 ( 1 + sin θ )
LHS = 2 ( 1 + sin θ ) cos θ ( 1 + sin θ ) = 2 cos θ = 2 sec θ = \dfrac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)} = \dfrac{2}{\cos\theta} = 2\sec\theta = cos θ ( 1 + sin θ ) 2 ( 1 + sin θ ) = cos θ 2 = 2 sec θ
अंतिम उत्तर: सिद्ध हुआ।
PYQ 9 (2 अंक): यदि 3 cot A = 4 3\cot A = 4 3 cot A = 4 1 − tan 2 A 1 + tan 2 A \dfrac{1 - \tan^2 A}{1 + \tan^2 A} 1 + tan 2 A 1 − tan 2 A
हल:
cot A = 4 3 \cot A = \tfrac43 cot A = 3 4 tan A = 3 4 \tan A = \tfrac34 tan A = 4 3 tan 2 A = 9 16 \tan^2 A = \tfrac{9}{16} tan 2 A = 16 9 1 − 9 / 16 1 + 9 / 16 = 7 / 16 25 / 16 = 7 25 \dfrac{1 - 9/16}{1 + 9/16} = \dfrac{7/16}{25/16} = \tfrac{7}{25} 1 + 9/16 1 − 9/16 = 25/16 7/16 = 25 7
अंतिम उत्तर: 7 25 \tfrac{7}{25} 25 7
PYQ 10 (3 अंक): सिद्ध कीजिए ( sin θ + cos θ ) ( tan θ + cot θ ) = sec θ + csc θ (\sin\theta + \cos\theta)(\tan\theta + \cot\theta) = \sec\theta + \csc\theta ( sin θ + cos θ ) ( tan θ + cot θ ) = sec θ + csc θ
हल:
tan θ + cot θ = 1 sin θ cos θ \tan\theta + \cot\theta = \dfrac{1}{\sin\theta\cos\theta} tan θ + cot θ = sin θ cos θ 1 LHS = ( sin θ + cos θ ) ⋅ 1 sin θ cos θ = sin θ sin θ cos θ + cos θ sin θ cos θ = (\sin\theta + \cos\theta)\cdot\dfrac{1}{\sin\theta\cos\theta} = \dfrac{\sin\theta}{\sin\theta\cos\theta} + \dfrac{\cos\theta}{\sin\theta\cos\theta} = ( sin θ + cos θ ) ⋅ sin θ cos θ 1 = sin θ cos θ sin θ + sin θ cos θ cos θ
= 1 cos θ + 1 sin θ = sec θ + csc θ = \dfrac{1}{\cos\theta} + \dfrac{1}{\sin\theta} = \sec\theta + \csc\theta = cos θ 1 + sin θ 1 = sec θ + csc θ
अंतिम उत्तर: सिद्ध हुआ।
PYQ 11 (3 अंक): सिद्ध कीजिए tan θ 1 − cot θ + cot θ 1 − tan θ = 1 + tan θ + cot θ \dfrac{\tan\theta}{1 - \cot\theta} + \dfrac{\cot\theta}{1 - \tan\theta} = 1 + \tan\theta + \cot\theta 1 − cot θ tan θ + 1 − tan θ cot θ = 1 + tan θ + cot θ
हल:
खंड 6 के हल किए उदाहरण से, LHS = 1 + sec θ csc θ = 1 + 1 sin θ cos θ = 1 + \sec\theta\csc\theta = 1 + \dfrac{1}{\sin\theta\cos\theta} = 1 + sec θ csc θ = 1 + sin θ cos θ 1
tan θ + cot θ = 1 sin θ cos θ \tan\theta + \cot\theta = \dfrac{1}{\sin\theta\cos\theta} tan θ + cot θ = sin θ cos θ 1 = 1 + 1 sin θ cos θ = 1 + \dfrac{1}{\sin\theta\cos\theta} = 1 + sin θ cos θ 1 LHS = RHS।
अंतिम उत्तर: सिद्ध हुआ।
PYQ 12 (3 अंक): सिद्ध कीजिए sin θ − 2 sin 3 θ 2 cos 3 θ − cos θ = tan θ \dfrac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta} = \tan\theta 2 cos 3 θ − cos θ sin θ − 2 sin 3 θ = tan θ
हल:
अंश = sin θ ( 1 − 2 sin 2 θ ) = \sin\theta(1 - 2\sin^2\theta) = sin θ ( 1 − 2 sin 2 θ )
हर = cos θ ( 2 cos 2 θ − 1 ) = \cos\theta(2\cos^2\theta - 1) = cos θ ( 2 cos 2 θ − 1 )
ध्यान दें 1 − 2 sin 2 θ = 2 cos 2 θ − 1 1 - 2\sin^2\theta = 2\cos^2\theta - 1 1 − 2 sin 2 θ = 2 cos 2 θ − 1 cos 2 θ \cos 2\theta cos 2 θ
अतः भिन्न = sin θ cos θ = tan θ = \dfrac{\sin\theta}{\cos\theta} = \tan\theta = cos θ sin θ = tan θ
अंतिम उत्तर: सिद्ध हुआ।
PYQ 13 (3 अंक): cos 45 ° sec 30 ° + csc 30 ° \dfrac{\cos 45°}{\sec 30° + \csc 30°} sec 30° + csc 30° cos 45°
हल:
cos 45 ° = 1 2 \cos 45° = \tfrac{1}{\sqrt2} cos 45° = 2 1 sec 30 ° = 2 3 \sec 30° = \tfrac{2}{\sqrt3} sec 30° = 3 2 csc 30 ° = 2 \csc 30° = 2 csc 30° = 2 हर = 2 3 + 2 = 2 + 2 3 3 = \tfrac{2}{\sqrt3} + 2 = \tfrac{2 + 2\sqrt3}{\sqrt3} = 3 2 + 2 = 3 2 + 2 3
1 / 2 ( 2 + 2 3 ) / 3 = 3 2 2 ( 1 + 3 ) = 3 ( 3 − 1 ) 2 2 ( 2 ) = 3 − 3 4 2 \dfrac{1/\sqrt2}{(2+2\sqrt3)/\sqrt3} = \dfrac{\sqrt3}{2\sqrt2(1+\sqrt3)} = \dfrac{\sqrt3(\sqrt3-1)}{2\sqrt2(2)} = \dfrac{3 - \sqrt3}{4\sqrt2} ( 2 + 2 3 ) / 3 1/ 2 = 2 2 ( 1 + 3 ) 3 = 2 2 ( 2 ) 3 ( 3 − 1 ) = 4 2 3 − 3
अंतिम उत्तर: 3 − 3 4 2 \dfrac{3 - \sqrt3}{4\sqrt2} 4 2 3 − 3
PYQ 14 (3 अंक): यदि sin θ + cos θ = 3 \sin\theta + \cos\theta = \sqrt3 sin θ + cos θ = 3 tan θ + cot θ = 1 \tan\theta + \cot\theta = 1 tan θ + cot θ = 1
हल:
वर्ग करें: 1 + 2 sin θ cos θ = 3 ⇒ sin θ cos θ = 1 1 + 2\sin\theta\cos\theta = 3 \Rightarrow \sin\theta\cos\theta = 1 1 + 2 sin θ cos θ = 3 ⇒ sin θ cos θ = 1
tan θ + cot θ = 1 sin θ cos θ = 1 1 = 1 \tan\theta + \cot\theta = \dfrac{1}{\sin\theta\cos\theta} = \dfrac{1}{1} = 1 tan θ + cot θ = sin θ cos θ 1 = 1 1 = 1
अंतिम उत्तर: सिद्ध हुआ।
PYQ 15 (3 अंक): यदि tan θ = 1 7 \tan\theta = \tfrac{1}{\sqrt7} tan θ = 7 1 csc 2 θ − sec 2 θ csc 2 θ + sec 2 θ \dfrac{\csc^2\theta - \sec^2\theta}{\csc^2\theta + \sec^2\theta} csc 2 θ + sec 2 θ csc 2 θ − sec 2 θ
हल:
csc 2 θ = 1 + cot 2 θ = 1 + 7 = 8 \csc^2\theta = 1 + \cot^2\theta = 1 + 7 = 8 csc 2 θ = 1 + cot 2 θ = 1 + 7 = 8 sec 2 θ = 1 + tan 2 θ = 1 + 1 7 = 8 7 \sec^2\theta = 1 + \tan^2\theta = 1 + \tfrac17 = \tfrac87 sec 2 θ = 1 + tan 2 θ = 1 + 7 1 = 7 8 8 − 8 / 7 8 + 8 / 7 = 48 / 7 64 / 7 = 48 64 = 3 4 \dfrac{8 - 8/7}{8 + 8/7} = \dfrac{48/7}{64/7} = \dfrac{48}{64} = \tfrac34 8 + 8/7 8 − 8/7 = 64/7 48/7 = 64 48 = 4 3
अंतिम उत्तर: 3 4 \tfrac34 4 3
PYQ 16 (3 अंक): सिद्ध कीजिए sec θ ( 1 − sin θ ) ( sec θ + tan θ ) = 1 \sec\theta(1 - \sin\theta)(\sec\theta + \tan\theta) = 1 sec θ ( 1 − sin θ ) ( sec θ + tan θ ) = 1
हल:
sec θ + tan θ = 1 + sin θ cos θ \sec\theta + \tan\theta = \dfrac{1 + \sin\theta}{\cos\theta} sec θ + tan θ = cos θ 1 + sin θ LHS = 1 cos θ ( 1 − sin θ ) ⋅ 1 + sin θ cos θ = ( 1 − sin θ ) ( 1 + sin θ ) cos 2 θ = \dfrac{1}{\cos\theta}(1 - \sin\theta)\cdot\dfrac{1 + \sin\theta}{\cos\theta} = \dfrac{(1-\sin\theta)(1+\sin\theta)}{\cos^2\theta} = cos θ 1 ( 1 − sin θ ) ⋅ cos θ 1 + sin θ = cos 2 θ ( 1 − sin θ ) ( 1 + sin θ )
= 1 − sin 2 θ cos 2 θ = cos 2 θ cos 2 θ = 1 = \dfrac{1 - \sin^2\theta}{\cos^2\theta} = \dfrac{\cos^2\theta}{\cos^2\theta} = 1 = cos 2 θ 1 − sin 2 θ = cos 2 θ cos 2 θ = 1
अंतिम उत्तर: सिद्ध हुआ।
PYQ 17 (5 अंक): सिद्ध कीजिए cos θ 1 − tan θ + sin 2 θ sin θ − cos θ = cos θ + sin θ \dfrac{\cos\theta}{1 - \tan\theta} + \dfrac{\sin^2\theta}{\sin\theta - \cos\theta} = \cos\theta + \sin\theta 1 − tan θ cos θ + sin θ − cos θ sin 2 θ = cos θ + sin θ
हल:
cos θ 1 − tan θ = cos 2 θ cos θ − sin θ \dfrac{\cos\theta}{1 - \tan\theta} = \dfrac{\cos^2\theta}{\cos\theta - \sin\theta} 1 − tan θ cos θ = cos θ − sin θ cos 2 θ sin 2 θ sin θ − cos θ = − sin 2 θ cos θ − sin θ \dfrac{\sin^2\theta}{\sin\theta - \cos\theta} = -\dfrac{\sin^2\theta}{\cos\theta - \sin\theta} sin θ − cos θ sin 2 θ = − cos θ − sin θ sin 2 θ योग = cos 2 θ − sin 2 θ cos θ − sin θ = ( cos θ − sin θ ) ( cos θ + sin θ ) cos θ − sin θ = cos θ + sin θ = \dfrac{\cos^2\theta - \sin^2\theta}{\cos\theta - \sin\theta} = \dfrac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}{\cos\theta - \sin\theta} = \cos\theta + \sin\theta = cos θ − sin θ cos 2 θ − sin 2 θ = cos θ − sin θ ( cos θ − sin θ ) ( cos θ + sin θ ) = cos θ + sin θ
अंतिम उत्तर: सिद्ध हुआ।
PYQ 18 (5 अंक): सिद्ध कीजिए tan θ + sec θ − 1 tan θ − sec θ + 1 = 1 + sin θ cos θ \dfrac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} = \dfrac{1 + \sin\theta}{\cos\theta} tan θ − sec θ + 1 tan θ + sec θ − 1 = cos θ 1 + sin θ
हल:
अंश में − 1 = − ( sec 2 θ − tan 2 θ ) -1 = -(\sec^2\theta - \tan^2\theta) − 1 = − ( sec 2 θ − tan 2 θ ) tan θ + sec θ − ( sec 2 θ − tan 2 θ ) \tan\theta + \sec\theta - (\sec^2\theta - \tan^2\theta) tan θ + sec θ − ( sec 2 θ − tan 2 θ )
= tan θ + sec θ − ( sec θ − tan θ ) ( sec θ + tan θ ) = ( tan θ + sec θ ) [ 1 − ( sec θ − tan θ ) ] = \tan\theta + \sec\theta - (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = (\tan\theta + \sec\theta)[1 - (\sec\theta - \tan\theta)] = tan θ + sec θ − ( sec θ − tan θ ) ( sec θ + tan θ ) = ( tan θ + sec θ ) [ 1 − ( sec θ − tan θ )] अंश = ( tan θ + sec θ ) ( 1 − sec θ + tan θ ) = (\tan\theta + \sec\theta)(1 - \sec\theta + \tan\theta) = ( tan θ + sec θ ) ( 1 − sec θ + tan θ ) ( tan θ − sec θ + 1 ) (\tan\theta - \sec\theta + 1) ( tan θ − sec θ + 1 )
अतः भिन्न = tan θ + sec θ = sin θ cos θ + 1 cos θ = 1 + sin θ cos θ = \tan\theta + \sec\theta = \dfrac{\sin\theta}{\cos\theta} + \dfrac{1}{\cos\theta} = \dfrac{1 + \sin\theta}{\cos\theta} = tan θ + sec θ = cos θ sin θ + cos θ 1 = cos θ 1 + sin θ
अंतिम उत्तर: सिद्ध हुआ।
PYQ 19 (5 अंक): 4 cot 2 30 ° + 1 sin 2 30 ° − 2 cos 2 45 ° − sin 2 0 ° \dfrac{4}{\cot^2 30°} + \dfrac{1}{\sin^2 30°} - 2\cos^2 45° - \sin^2 0° cot 2 30° 4 + sin 2 30° 1 − 2 cos 2 45° − sin 2 0°
हल:
cot 2 30 ° = 3 \cot^2 30° = 3 cot 2 30° = 3 4 3 \tfrac43 3 4 sin 2 30 ° = 1 4 \sin^2 30° = \tfrac14 sin 2 30° = 4 1 1 1 / 4 = 4 \tfrac{1}{1/4} = 4 1/4 1 = 4 2 cos 2 45 ° = 1 2\cos^2 45° = 1 2 cos 2 45° = 1 sin 2 0 ° = 0 \sin^2 0° = 0 sin 2 0° = 0 4 3 + 4 − 1 − 0 = 4 3 + 3 = 13 3 \tfrac43 + 4 - 1 - 0 = \tfrac43 + 3 = \tfrac{13}{3} 3 4 + 4 − 1 − 0 = 3 4 + 3 = 3 13
अंतिम उत्तर: 13 3 \tfrac{13}{3} 3 13
PYQ 20 (5 अंक): यदि sec θ + tan θ = m \sec\theta + \tan\theta = m sec θ + tan θ = m m 2 − 1 m 2 + 1 = sin θ \dfrac{m^2 - 1}{m^2 + 1} = \sin\theta m 2 + 1 m 2 − 1 = sin θ
हल:
m 2 = sec 2 θ + 2 sec θ tan θ + tan 2 θ m^2 = \sec^2\theta + 2\sec\theta\tan\theta + \tan^2\theta m 2 = sec 2 θ + 2 sec θ tan θ + tan 2 θ m 2 − 1 = tan 2 θ + 2 sec θ tan θ + tan 2 θ = 2 tan θ ( tan θ + sec θ ) = 2 tan θ m m^2 - 1 = \tan^2\theta + 2\sec\theta\tan\theta + \tan^2\theta = 2\tan\theta(\tan\theta + \sec\theta) = 2\tan\theta\, m m 2 − 1 = tan 2 θ + 2 sec θ tan θ + tan 2 θ = 2 tan θ ( tan θ + sec θ ) = 2 tan θ m sec 2 − 1 = tan 2 \sec^2 - 1 = \tan^2 sec 2 − 1 = tan 2 m 2 + 1 = sec 2 θ + 2 sec θ tan θ + sec 2 θ = 2 sec θ ( sec θ + tan θ ) = 2 sec θ m m^2 + 1 = \sec^2\theta + 2\sec\theta\tan\theta + \sec^2\theta = 2\sec\theta(\sec\theta + \tan\theta) = 2\sec\theta\, m m 2 + 1 = sec 2 θ + 2 sec θ tan θ + sec 2 θ = 2 sec θ ( sec θ + tan θ ) = 2 sec θ m m 2 − 1 m 2 + 1 = 2 tan θ m 2 sec θ m = tan θ sec θ = sin θ \dfrac{m^2 - 1}{m^2 + 1} = \dfrac{2\tan\theta\, m}{2\sec\theta\, m} = \dfrac{\tan\theta}{\sec\theta} = \sin\theta m 2 + 1 m 2 − 1 = 2 sec θ m 2 tan θ m = sec θ tan θ = sin θ
अंतिम उत्तर: सिद्ध हुआ।
PYQ 21 (3 अंक): यदि cos θ + sin θ = 2 cos θ \cos\theta + \sin\theta = \sqrt2\cos\theta cos θ + sin θ = 2 cos θ cos θ − sin θ = 2 sin θ \cos\theta - \sin\theta = \sqrt2\sin\theta cos θ − sin θ = 2 sin θ
हल:
दिए से, sin θ = ( 2 − 1 ) cos θ \sin\theta = (\sqrt2 - 1)\cos\theta sin θ = ( 2 − 1 ) cos θ
cos θ − sin θ = cos θ − ( 2 − 1 ) cos θ = ( 2 − 2 ) cos θ = 2 ( 2 − 1 ) cos θ \cos\theta - \sin\theta = \cos\theta - (\sqrt2 - 1)\cos\theta = (2 - \sqrt2)\cos\theta = \sqrt2(\sqrt2 - 1)\cos\theta cos θ − sin θ = cos θ − ( 2 − 1 ) cos θ = ( 2 − 2 ) cos θ = 2 ( 2 − 1 ) cos θ = 2 sin θ = \sqrt2\sin\theta = 2 sin θ sin θ = ( 2 − 1 ) cos θ \sin\theta = (\sqrt2-1)\cos\theta sin θ = ( 2 − 1 ) cos θ
अंतिम उत्तर: सिद्ध हुआ।
PYQ 22 (3 अंक): cos 48 ° csc 42 ° \cos 48° \csc 42° cos 48° csc 42°
हल:
csc 42 ° = sec 48 ° \csc 42° = \sec 48° csc 42° = sec 48° cos 48 ° csc 42 ° = cos 48 ° sec 48 ° = 1 \cos 48°\csc 42° = \cos 48°\sec 48° = 1 cos 48° csc 42° = cos 48° sec 48° = 1
अंतिम उत्तर: 1।
PYQ 23 (3 अंक): यदि tan 2 A = cot ( A − 18 ° ) \tan 2A = \cot(A - 18°) tan 2 A = cot ( A − 18° ) 2 A 2A 2 A A A A
हल:
cot ( A − 18 ° ) = tan ( 90 ° − ( A − 18 ° ) ) = tan ( 108 ° − A ) \cot(A - 18°) = \tan(90° - (A - 18°)) = \tan(108° - A) cot ( A − 18° ) = tan ( 90° − ( A − 18° )) = tan ( 108° − A ) 2 A = 108 ° − A ⇒ 3 A = 108 ° ⇒ A = 36 ° 2A = 108° - A \Rightarrow 3A = 108° \Rightarrow A = 36° 2 A = 108° − A ⇒ 3 A = 108° ⇒ A = 36°
अंतिम उत्तर: A = 36 ° A = 36° A = 36°
PYQ 24 (5 अंक): सिद्ध कीजिए sin θ cot θ + csc θ = 2 + sin θ cot θ − csc θ \dfrac{\sin\theta}{\cot\theta + \csc\theta} = 2 + \dfrac{\sin\theta}{\cot\theta - \csc\theta} cot θ + csc θ sin θ = 2 + cot θ − csc θ sin θ
हल:
RHS दूसरा पद: sin θ cot θ − csc θ = sin θ cos θ − 1 sin θ = sin 2 θ cos θ − 1 \dfrac{\sin\theta}{\cot\theta - \csc\theta} = \dfrac{\sin\theta}{\frac{\cos\theta - 1}{\sin\theta}} = \dfrac{\sin^2\theta}{\cos\theta - 1} cot θ − csc θ sin θ = s i n θ c o s θ − 1 sin θ = cos θ − 1 sin 2 θ
= 1 − cos 2 θ cos θ − 1 = − ( cos 2 θ − 1 ) cos θ − 1 = − ( cos θ + 1 ) = \dfrac{1 - \cos^2\theta}{\cos\theta - 1} = \dfrac{-(\cos^2\theta - 1)}{\cos\theta - 1} = -(\cos\theta + 1) = cos θ − 1 1 − cos 2 θ = cos θ − 1 − ( cos 2 θ − 1 ) = − ( cos θ + 1 ) LHS: sin θ cot θ + csc θ = sin 2 θ cos θ + 1 = 1 − cos 2 θ cos θ + 1 = 1 − cos θ \dfrac{\sin\theta}{\cot\theta + \csc\theta} = \dfrac{\sin^2\theta}{\cos\theta + 1} = \dfrac{1 - \cos^2\theta}{\cos\theta + 1} = 1 - \cos\theta cot θ + csc θ sin θ = cos θ + 1 sin 2 θ = cos θ + 1 1 − cos 2 θ = 1 − cos θ
RHS = 2 + [ − ( cos θ + 1 ) ] = 2 − cos θ − 1 = 1 − cos θ = = 2 + [-(\cos\theta + 1)] = 2 - \cos\theta - 1 = 1 - \cos\theta = = 2 + [ − ( cos θ + 1 )] = 2 − cos θ − 1 = 1 − cos θ =
अंतिम उत्तर: सिद्ध हुआ।
PYQ 25 (3 अंक): sin 2 5 ° + sin 2 10 ° + sin 2 80 ° + sin 2 85 ° \sin^2 5° + \sin^2 10° + \sin^2 80° + \sin^2 85° sin 2 5° + sin 2 10° + sin 2 80° + sin 2 85°
हल:
sin 80 ° = cos 10 ° \sin 80° = \cos 10° sin 80° = cos 10° sin 85 ° = cos 5 ° \sin 85° = \cos 5° sin 85° = cos 5° sin 2 5 ° + cos 2 5 ° + sin 2 10 ° + cos 2 10 ° = 1 + 1 = 2 \sin^2 5° + \cos^2 5° + \sin^2 10° + \cos^2 10° = 1 + 1 = 2 sin 2 5° + cos 2 5° + sin 2 10° + cos 2 10° = 1 + 1 = 2
अंतिम उत्तर: 2।
PYQ 26 (5 अंक): यदि x = a sin θ + b cos θ x = a\sin\theta + b\cos\theta x = a sin θ + b cos θ y = a cos θ − b sin θ y = a\cos\theta - b\sin\theta y = a cos θ − b sin θ x 2 + y 2 = a 2 + b 2 x^2 + y^2 = a^2 + b^2 x 2 + y 2 = a 2 + b 2
हल:
x 2 = a 2 sin 2 θ + 2 a b sin θ cos θ + b 2 cos 2 θ x^2 = a^2\sin^2\theta + 2ab\sin\theta\cos\theta + b^2\cos^2\theta x 2 = a 2 sin 2 θ + 2 ab sin θ cos θ + b 2 cos 2 θ y 2 = a 2 cos 2 θ − 2 a b sin θ cos θ + b 2 sin 2 θ y^2 = a^2\cos^2\theta - 2ab\sin\theta\cos\theta + b^2\sin^2\theta y 2 = a 2 cos 2 θ − 2 ab sin θ cos θ + b 2 sin 2 θ जोड़ें: x 2 + y 2 = a 2 ( sin 2 θ + cos 2 θ ) + b 2 ( cos 2 θ + sin 2 θ ) = a 2 + b 2 x^2 + y^2 = a^2(\sin^2\theta + \cos^2\theta) + b^2(\cos^2\theta + \sin^2\theta) = a^2 + b^2 x 2 + y 2 = a 2 ( sin 2 θ + cos 2 θ ) + b 2 ( cos 2 θ + sin 2 θ ) = a 2 + b 2
अंतिम उत्तर: सिद्ध हुआ।
