please answer these 3 questions.
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Harshsingh111:
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(3 sinθ+5cosθ)²= 5²
Squaring on both sides.
(3sinθ)²+(5cosθ)²+2× 3sinθ 5cosθ= 25
[a+b= a²+b²+2ab]
9sin²θ+ 25cos²θ+30sinθcosθ= 25
9 (1-cos²θ) + 25(1-sin²θ)+30sinθcosθ=25
[sin²θ + cos²θ =1]
9-9cos²θ + 25-25sin²θ +30sinθcosθ=25
9+25 -(9cos²θ +25sin²θ -30sinθcosθ) =25
34 - (9cos²θ +25sin²θ -30sinθcosθ) =25
- (25sin²θ +9cos²θ-30sinθcosθ) =25-34
(25sin²θ+9cos²θ -30sinθcosθ) =9
(5sinθ - 3cosθ)²= 9
(5sinθ - 3cosθ)= √9
(5sinθ - 3cosθ)= ±3
L.H.S = R.H.S
==================================================================
Hope this will help you...
Squaring on both sides.
(3sinθ)²+(5cosθ)²+2× 3sinθ 5cosθ= 25
[a+b= a²+b²+2ab]
9sin²θ+ 25cos²θ+30sinθcosθ= 25
9 (1-cos²θ) + 25(1-sin²θ)+30sinθcosθ=25
[sin²θ + cos²θ =1]
9-9cos²θ + 25-25sin²θ +30sinθcosθ=25
9+25 -(9cos²θ +25sin²θ -30sinθcosθ) =25
34 - (9cos²θ +25sin²θ -30sinθcosθ) =25
- (25sin²θ +9cos²θ-30sinθcosθ) =25-34
(25sin²θ+9cos²θ -30sinθcosθ) =9
(5sinθ - 3cosθ)²= 9
(5sinθ - 3cosθ)= √9
(5sinθ - 3cosθ)= ±3
L.H.S = R.H.S
==================================================================
Hope this will help you...
but other 2?
plz send answers of other two questions also
thnx
ok wait
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I hope can help this answer thanks you
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schl ka namm kya h
rukmani devi jaipuria public school
its centre
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my school is A G D A V centenary public school
my school Indian modern school and centre meahroli mai sawaeshwati bal mandir m
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byw
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