Question

☆☆☆☆Need Help☆☆☆☆

Answer the given question.
*With expalnation*.

Answer 1

Hey !!!

Here is ur solution :-
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cos²¢ + 2sin²¢ + 3cos2¢²¢ + 4sin²¢ ....+200th term = 10025

means 100 term of sin²¢ 's identity and 100term of cos²¢ identities .

=> (cos²¢ + 3cos²¢ ......100term )+(2sin²¢ +4cos²¢ ....+100th term ) = 10025

Now by using AP
in cos²¢ 's identity
a = 1
d = 3 -1 =2
n = 100

S100 = n/2{2a+(n-1)d }
= 100/2{(2*1 +(100-1)2
= 50×{2+198)
=50×200
= 10000

now in the term of sin²¢
a = 2
d = 4-2 =2
n = 100

s100= 100/2+(2×2 + (100-1)2
=> 50(4 + 198)
=> 50×202
=> 10100 it can also written as 100² + 101

now from our question .

=> 10000cos²¢ + (10100 sin²¢ ) = 10025

=> 100²cos²¢ + {100²+ 100(sin²¢)} = 10025

=> 100²cos²¢ + 100²sin²¢+ 100sin²¢=10025

=> 100² (cos²¢+sin²¢) + 100sin²¢=10025

=> 10000×1 + 100sin²¢ = 10025

=> 100sin²¢ = 10025

=> sin²¢ = 10025-10000

=> 100sin²¢ = 25

=> sin²¢ = √25/100

=> sin¢ = 5/10 = b/2 =p/h -----1)

now by using Pythagoras theorem
b = √h² -p²
b= √2² - 1²
b= √4-1 = √3

cos¢ =b/h = √3/2-------2)

so, sin¢ - cos¢
=> 1/2 - √3/2
=> 1 -√3/2 Answer

Hence option a ) 1 - √3/2 is ur answer

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Hope it helps you !!!

@Rajukumar111

Answer 2

HELLO DEAR,


we know that:-

IN an AP

$s_{n} = \frac{n}{2} (2a + (n - 1)d)$


cos²∅ + 2sin²∅ + 3cos²∅ + 4sin²∅ + ........(200term) = 10025


=> [cos²∅ + 3cos²∅ + 5cos²∅ +......(100term)] +
[ 2sin²∅ + 4sin²∅ + 6sin²∅ +......(100term) = 10025


=> [ 100/2×( 2cos²∅ + (100-1)2cos²∅ ] + [100/2
(2×(2sin²∅) + (100-1)2sin²∅)] = 10025


=> [ 50×{2cos²∅ ( 1 + 99)} ] + [ 50×{2sin²∅(2 +
99)}] = 10025


=> [ 100cos²∅×100 ] + [ 100sin²∅ ×101] 10025


=> 100 [ 100cos²∅ + 101sin²∅ ] = 10025


=> 4 [ 100cos²∅ + (100sin²∅ + sin²∅) ] = 401


=> 4 [ 100(cos²∅ + sin²∅) + sin²∅ ] = 401


=> 400×(1) + 4sin²∅ = 401


=> 4sin²∅ = 401 - 400 = 1


=> sin²∅ = 1/4

=> sin∅ = 1/2

=> sin∅ = sin30°

[ sin30° = 1/2 ]

=> ∅ = 30°

AND,

cos∅ = cos30° = √3/2 ,sin∅ = sin30° = 1/2

=> sin∅ - cos∅ = 1/2 - √3/2

=> (1 - √3)/2


hence, option (A) is correct



I HOPE ITS HELP YOU DEAR,
THANSK