Math, asked by arts3095, 1 year ago

निम्नलिखित के मान निकालिए:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60°
(iii) cos 45°/ sec 30°+ cosec30°
(iv) sin 30° + tan45° - cosec 60° / sec 30° + cos 60° + cot 45°
(v) 5cos2 60° + 4sec2 30° - tan2 45° / sin2 30° + cos2 30°

Answers

Answered by Amisha2005
4

PLZZ MARK AS BRAINLIEST ANSWER...

Attachments:
Answered by nikitasingh79
5

Answer with Step-by-step explanation:

(i) sin 60° cos 30° + sin 30° cos 60°

= √3/2 × √3/2 + ½ + ½

= (√3/2)² + (½)²

= ¾ + ¼

= (3 + 1)/4

= 4/4

= 1

sin 60° cos 30° + sin 30° cos 60° = 1

(ii) 2 tan² 45° + cos² 30° - sin² 60°

= 2 (1)² + (√3/2)² - (√3/2)²

= 2 × 1 + ¾ - ¾

= 2 + 0

= 2

2 tan² 45° + cos² 30° - sin² 60° = 2

(iii) cos 45°/ sec 30°+ cosec30°

= (1/√2) /(2/√3 + 2 )

= 1/√2/ (2 + 2√3)/√3

= 1/√2 × √3/(2 + 2√3)

= √3/√2(2 + 2√3)

= √3/ 2√2 + 2√2√3)

= √3/[2√2( 1 + √3)]

= √3 × (√3 - 1) /[2√2 (1+ √3)] × (√3 - 1)

= (3 - √3) /[2√2 × (√3² - 1²)]

= (3 - √3) / [2√2 × (3 - 1)]

= (3 - √3) / [2√2 × 2]

= (3 - √3) / (4√2)

= √2 (3 - √3) / (4√2) × √2

= (3√2 - √6)/(4 × 2)

= (3√2 - √6)/8

cos 45°/ sec 30°+ cosec30° =  (3√2 - √6)/8

 

(iv) sin 30° + tan45° - cosec 60° / sec 30° + cos 60° + cot 45°

= [(1/2 + 1 - 2/√3)]/[( 2/√3+ 1/2 + 1)]

= [(3/2 - 2/√3)] /[( 2/√3 + 3/2)

= (3√3 - 4)/(4 + 3√3)]

= [(3√3 - 4) × (3√3 - 4)] /(3√3 + 4) × (3√3 - 4)]

= [(3√3)² + 4² - 2×3√3 × 4]/[(3√3)² - 4²]

= [27 + 16 - 24√3]/ [27 - 16]

= (43 - 24√3)/11

sin 30° + tan45° - cosec 60° / sec 30° + cos 60° + cot 45° = (43 - 24√3)/11

 

(v) 5cos² 60° + 4sec² 30° - tan² 45° / sin² 30° + cos² 30°

= [5(½)² + 4(2/√3)² - 1²] /[(½)² + (√3/2)²]

= [(5 ×1/4 + 4 × 4/3 - 1)]/[( 1/4 + 3/4)]

= [5/4 + 16/3 - 1]/ [(1 +3)/4

= [(15 + 64 - 12)/12 /(4/4)

= 67/12

5cos² 60° + 4sec² 30° - tan² 45° / sin² 30° + cos² 30° = 67/12  

आशा है कि यह उत्तर आपकी अवश्य मदद करेगा।।।।

इस पाठ से संबंधित कुछ और प्रश्न :  

सही विकल्प चुनिए और अपने विकल्प का औचित्य दीजिए :

(i) 2 tan 30° / 1+tan2 30°

(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°

(ii) 1-tan245° / 1+tan2 45°

(A)tan90° (B)1 (C)sin45° (D)0

(iii) sin 2A = 2 sin A तब सत्य होता है, जबकि A बराबर है :

(A) 0° (B) 30° (C) 45° (D) 60°

(iv) 2 tan 30° / 1-tan2 30° बराबर है :

(A)cos60° (B)sin60° (C)tan60° (D)sin 30

https://brainly.in/question/12659816

यदि tan(A+B)= √3 और tan (A - B) 1 /√3 है; 0°<A+B ≤ 90; A > B तो A ओर B मान ज्ञात कीजिए।

https://brainly.in/question/12659814

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