Question

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​

Answer 1

1.

➥$\sf\pink{दिया \:हुआ\: है}$

• (cosec θ - cot θ)² = (1-cos θ)/(1+cos θ)  

अतः,

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

LHS = RHS

➥$\sf\purple{इति\: सिद्धम}$

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2.

➥$\sf\green{दिया \:हुआ\: है}$

• cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A  

अतः,

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.

LHS = RHS

➥$\sf\orange{इति \:सिद्धम}$

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3.

➥$\sf\blue{दिया \:हुआ\: है}$

• tan θ/(1-cot θ) + cotθ/(1-tan θ)  = 1 + sec θ cosec θ

अतः,

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.

LHS = RHS

➥$\sf\red{इति\:सिद्धम}$

Answer 2

Step-by-step explanation:

Answer of (ii)

• take lcm of (1+sinA) and (cosA)
• solve it using trigonometric identity