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Question : 21 of 160
Marks:
+1 ,
-0
Solution:
1 sin ‌ 45 ∘ ⋅ sin ‌ 4 ‌ 6 ∘
+ 1 sin ‌ 4 ‌ 7 ∘ ⋅ sin ‌ 48 ∘
+ . . . + 1 sin ‌ 133 ∘ ⋅ sin ‌ 134 ∘
= [ sin ( 46 ∘ − 45 ∘ ) sin ‌ 45 ∘ ⋅ sin ‌ 46 ∘
+ sin ( 48 ∘ − 47 ∘ ) sin ‌ 47 ∘ ⋅ sin ‌ 48 ∘
+ . . . + . . sin ( 134 ∘ − 133 ∘ ) sin ‌ 133 ∘ ⋅ sin ‌ 134 ∘
] = [ ( cot ‌ 45 ∘ − cot ‌ 46 ∘ ) + ( cot ‌ 47 ∘ − cot ‌ 48 ∘ ) . + . . . + ( cot ‌ 133 ∘ − cot ‌ 134 ∘ ) ] = [ ( cot ‌ 45 ∘ − cot ‌ 46 ∘ ) + ( cot ‌ 47 ∘ − cot ‌ 48 ∘ ) . + . . . + ( − cot ‌ 47 ∘ + cot ‌ 46 ∘ ) ] [ ∵ cot ‌ 133 ∘ = cot ( 180 ∘ − 47 ∘ ) = − cot ‌ 47 ∘ and
cot ‌ 134 ∘ = cot ( 180 ∘ − 46 ∘ ) = − cot ‌ 46 ∘ and ‌ so ‌ on ] = × cot ‌ 45 ∘ = According to question,
= ∴ n = 1
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