निम्नलिखित सर्वसमिका सिद्ध कीजिए, जहाँ वे कोण, जिनके लिए व्यंजक परिभाषित है, न्यूनकोण है :
(i) (cosec θ - cot θ)2 = 1-cos θ / 1+cos θ
(ii) cos A/1+sin A + 1+sin A/cos A =2sec A
(iii) _tan θ/ 1-cotθ + cot θ/ 1-tanθ = 1 + sec θ cosec θ
(iv)1+secA/ sec A = sin2 A/ 1-cosA
(v) सर्वसमिका cosec2 A= 1+ cot2 A को लागू करके
cos A - sin A +1/cos A + sin A - 1 = cosec A + cot A
(vi) √1+ sinA/1- sinA =secA+tanA
(vii) sin θ - 2 sin3 θ/2 cos3 θ - cos θ = tanθ
(viii) (sin A+ cosec A)2+ (cos A+ sec A)2 = 7+ tan2A+cot2A
(ix) (cosec A - sin A)(sec A - cos A) = 1/ tan A + cot A
(x) (1+tan2 A/1+cot2 A) = (1 - tan A/ 1 - cot A) = tan2A
Answers
Answer with Step-by-step explanation:
(i) दिया है :
(cosec θ - cot θ)² = (1-cos θ)/(1+cos θ)
L.H.S. = (cosec θ - cot θ)²
= (cosec²θ + cot²θ - 2cosec θ cot θ)
[(a + b)² = a² + b² + 2ab]
= (1/sin²θ + cos²θ/sin²θ - 2 × 1/sinθ × cos θ/sinθ)
= (1/sin²θ + cos²θ/sin²θ - 2 ×cos θ/sin²θ)
= (1 + cos²θ)/sin²θ - 2 × cos θ/sin²θ)
= [ (1 + cos²θ) - 2 × cos θ] /sin²θ
= (1 + cos²θ - 2cos θ)/(1 - cos²θ)
= (1- cos θ)²/(1 - cosθ)(1+ cos θ)
[a² + b² - 2ab = (a - b)² , a² - b² = (a + b) (a- b) ]
= (1 - cos θ)/(1 + cos θ)
= R.H.S.
(ii) दिया है:
cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
= [cos²A + (1+sin A)²]/(1+sin A)cos A
= [cos²A +(sin²A + 1 + 2sin A)]/(1+ sin A)cos A
[(a + b)² = a² + b² + 2ab]
= (1 + 1 + 2sin A)/(1+sin A)cos A
= (2+ 2sin A)/(1+sin A)cos A
= 2(1+sin A)/(1+sin A)cos A
= 2/cos A
= 2 sec A
= R.H.S.
(iii) दिया है :
tan θ/(1-cot θ) + cotθ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1 - cot θ) + cot θ/(1 - tan θ)
= [(sin θ/cos θ)/1 - (cos θ/sin θ)] + [(cos θ/sin θ)/1 - (sin θ/cos θ)]
= [(sin θ/cos θ)/(sin θ - cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ - sin θ)/cos θ] = sin²θ/[cos θ(sin θ - cos θ)] + cos²θ/[sin θ(cos θ - sin θ)]
= sin²θ/[cos θ(sin θ - cos θ)] - cos²θ/[sin θ(sin θ - cos θ)]
= 1/(sin θ-cos θ) [(sin²θ/cos θ) - (cos²θ/sin θ)]
= 1/(sin θ-cos θ) × [(sin³θ - cos³θ)/sin θ cos θ]
= [(sin θ - cos θ)(sin²θ + cos²θ + sin θ cos θ)]/[(sin θ - cos θ)sin θ cos θ]
[(a³ + b³) = (a + b) (a² + b² + ab]
= (1 + sin θ cos θ)/sin θ cos θ
= 1/sin θ cos θ + 1
= 1 + sec θ cosec θ
= R.H.S.
(iv) दिया है :
(1 + sec A)/sec A = sin²A/(1-cos A)
L.H.S. = (1 + sec A)/sec A
= (1 + 1/cos A)/1/cos A
= (cos A + 1)/cos A/1/cos A
= cos A + 1
R.H.S. = sin2A/(1-cos A)
= (1 - cos2A)/(1-cos A)
= (1-cos A)(1+cos A)/(1-cos A)
= cos A + 1
L.H.S. = R.H.S.
(v) दिया है : (cos A – sin A + 1)/ (cos A + sin A - 1) = cosec A + cot A
LHS:
(cos A – sin A + 1)/ (cos A + sin A - 1)
= {(cos A – sin A + 1)/sin A}/{(cos A + sin A - 1)/sin A}
[हर और अंश को sin A से भाग करने पर]
= {(cos A/sin A – sin A/sin A + 1/sin A)}/{(cos A/sin A + sin A/sin A - 1/sin A)}
= (cot A – 1 + cosec A)/ (cot A + 1 - cosec A)
= {cot A + cosec A - 1 }/ (cot A + 1 - cosec A)
= [cot A + cosec A - (cosec² A - cot²A)]/ (cot A + 1 - cosec A)
[1 = cosec² A - cot2 A]
= (cot A + cosec A) - [(cosec A - cot A) (cosec A + cot A)]/ (cot A + 1 - cosec A)
[a² - b² = (a +b) (a - b)]
= (cosec A + cot A) [(1 - cosec A + cot A)]/ (cot A + 1 - cosec A)
= cosec A + cot A
= RHS
(vi) दिया है : √[(1 + sin A)/(1 – sin A)] = sec A + tan A
LHS:
√[(1 + sin A)/(1 – sin A)]
= √[(1 + sin A) × (1 + sin A)] / [(1 - sin A)× (1 + sin A)]
[अंश और हर को (1 + sin A) से गुणा करने पर]
= √[(1 + sin A)² /(1 – sin² A)]
= √[(1 + sin A)²/cos² A]
[ (1 – sin² A) = cos² A]
= (1 + sin A)/cos A
= 1/cos A + sin A/cos A
= sec A + tan A
= RHS
(vii) दिया है :
(sin θ - 2sin³θ)/(2cos³θ-cos θ) = tan θ
L.H.S. = (sin θ - 2sin³θ)/(2cos³θ - cos θ)
= [sin θ(1 - 2sin²θ)]/[cos θ(2cos²θ - 1)]
= sin θ[1 - 2(1- cos²θ)]/[cos θ(2cos²θ - 1)]
[ (sin² A = 1 - cos² A]
= [sin θ(2cos²θ -1)]/[cos θ(2cos²θ -1)]
= sin θ/cos θ
= tan θ
= R.H.S.
(viii) दिया है :
(sin A + cosec A)²+ (cos A + sec A)² = 7+ tan²A + cot²A
L.H.S. = (sin A + cosec A)²+ (cos A + sec A)²
= (sin²A + cosec²A + 2 sin A cosec A) + (cos²A + sec²A + 2 cos A sec A)
[(a +b)² = a² + b² + 2ab]
= (sin²A + cos²A + 2 sin A cosec A) + (cosec²A + sec²A + 2 cos A sec A)
= (sin²A + cos²A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan²A + 1 + cot²A [sin²A + cos²A = 1 , cosecA = 1/sinA , secA = 1/cosA, cosec²A = 1 + cot²A ,sec²A = 1 + tan²A ]
= 1 + 2 + 2 + 2 + tan²A + cot²A
= 7 + tan²A + cot²A
= R.H.S.
(ix) दिया है :
(cosec A – sin A)(sec A – cos A) = 1/(tan A + cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
= (1/sin A - sin A)(1/cos A - cos A)
[ cosecA = 1/sinA , secA = 1/cosA]
= [(1 - sin²A)/sin A][(1 - cos²A)/cos A]
= (cos²A/sin A) × (sin²A/cos A)
[(1 – sin² A) = cos² A, (1 – cos² A) = sin² A]
= cos A sin A
R.H.S. = 1/(tan A + cotA)
= 1/(sin A/cos A + cos A/sin A)
= 1/[(sin²A + cos²A)/sin A cos A]
= 1/1/(sin A cos A)
[(sin²A + cos²A) = 1]
= 1 × (sin A cos A)
= cos A sin A
L.H.S.= R.H.S.
(x) दिया है :
(1 + tan²A/1 + cot²A) = (1 - tanA /1 - cotA)² = tan²A
L.H.S. = (1 + tan²A/1 + cot²A)
= (1 + tan²A /1 + 1/tan²A)
= 1 + tan²A/[(1 + tan²A)/tan²A]
= tan²A
[(1 - tan A)/(1 - cot A)]²
= [(1 - sin A/cos A)/(1 - cos A/sin A)]²
= [{(cos A - sin A)/cos A} /{(sin A - cos A)/sin A)}]²
= [{(cos A - sin A)/cos A} × {sin A/(sin A - cos A)/sin A}]²
= [-{(sin A - cos A)/cos A} × {sin A/(sin A - cos A)/sin A}]²
= (-sin A/cos A)²
= sin² A/cos² A
= tan² A
आशा है कि यह उत्तर आपकी अवश्य मदद करेगा।।।।
इस पाठ से संबंधित कुछ और प्रश्न :
मान निकालिए :
(i) sin2 63° +sin2 27° / cos 17° +cos2 73°
(ii) sin 25° cos 65° + cos 25° sin 65°
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