k = 0 ∏ 4 5 ( 1 + tan ( 1 8 0 k π ) ) = 2 m m = ?
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can you prove your statement
( 1 + tan x ) ( 1 + tan y ) = 2 , when x+y=45
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@Lakshya Sinha w h e n x + y = 4 5 º 1 − tan ( x ) × tan ( y ) tan ( x ) + tan ( y ) = 1 ∴ tan ( x ) + tan ( y ) = 1 − tan ( x ) × tan ( y ) tan ( x ) + tan ( y ) + tan ( x ) × tan ( y ) ( + 1 ) = 1 ( + 1 ) tan ( x ) + tan ( y ) + tan ( x ) × tan ( y ) + 1 = 2 = ( 1 + tan ( x ) ) ( 1 + tan ( y ) )
Did the same way !
( tan 45° +tan1°)( tan 45°+tan2°)......
((Sin45°/cos 45° )+(sin1°/cos1°))....
On simplifying. Using sin(A+B) formula
((Sin46°.sin47°....)/(cos45°.cos 1°.cos45°.cos2°......)
(Sin46°.sin47°....)/(((cos45°)^46).cos1.cos2......))
On simplifying further using sin(90-x)=cosx
(√2)^46
=> m=23
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P = ( 1 + tan ( 1 ∘ ) ) ( 1 + tan ( 2 ∘ ) ) ( 1 + tan ( 3 ∘ ) ) . . . ( 1 + tan ( 4 5 ∘ ) ) Using ( 1 + tan ( x ) ) ( 1 + tan ( y ) ) = 2 , when x + y = 4 5 ∘ ( 1 + tan ( 1 ∘ ) ) ( 1 + tan ( 4 4 ∘ ) ) ( 1 + tan ( 2 ∘ ) ( 1 + tan ( 4 3 ∘ ) ) . . . ( 1 + tan ( 4 5 ∘ ) ) P = 2 3 times 2 × 2 × ⋯ × 2 = 2 2 3 2 m = 2 2 3 m = 2 3