The value of 1-tan 1-1-tan 2-1-tan 3- - - -1-tan 45- L MA- L - M -25B- L - M -44C- M - L 4-6D- M - L 4-7

The correct options are A L + M =25 D M L^4 =7If A + B = 45^∘ then (1 + tan A) (1+ tanB) = 2 Use above result ...

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Algebra of Limits

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Question

The value of $(1 + t a n 1^{\circ}) (1 + t a n 2^{\circ}) (1 + t a n 3^{\circ}) . . . . (1 + t a n 45^{\circ}) = L^{M}$

L + M =25

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L + M =44

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M - L^{4} = 6

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M - L^{4} = 7

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(1 + tan 1^{\circ}) \cdot (1 + tan 2^{\circ}) \cdot (1 + tan 3^{\circ}) \dots \cdot (1 + tan 45^{\circ}) = 2^{n}

^{'} n^{'}

A = (1 + tan 1^{\circ}) (1 + tan 2^{\circ}) (1 + tan 3^{\circ}) \dots (1 + tan 45^{\circ}) is

L = (4 tan 1^{\circ}) (4 tan 2^{\circ}) (4 tan 3^{\circ}) \dots (4 tan 10^{\circ}),

M = (tan 2^{\circ}) (tan 4^{\circ}) (tan 6^{\circ}) \dots (tan 20^{\circ}),

N = (1 + tan 1^{\circ}) (1 - tan 1^{\circ}) (1 + tan 2^{\circ}) (1 - tan 2^{\circ}) \dots (1 + tan 10^{\circ}) (1 - tan 10^{\circ})

\frac{L}{128 M N}

(1 + tan 1^{\circ}) (1 + tan 2^{\circ}) \dots (1 + tan 45^{\circ})

L = lim n \to \infty (2 \times 3^{2} \times 2^{3} \times 3^{4} . . . \times 2^{n - 1} \times 3^{n})^{\frac{1}{(n^{2} + 1)}}

L^{4}

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