Find the value of (,tan1°tan4°tan6°........tan88°).

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Solution

tan 1° tan 4° tan 6° ...tan84° tan 86° tan 89°
=tan 1° tan 4° tan 6° ...tan(90−6)° tan (90−4)° tan (90−1)°
=tan 1° tan 4° tan 6° ...cot6° cot4° cot1°
=tan 1°×cot1°× tan 4° ×cot4°×tan 6°×cot6°× ...........×tan44°×cot44°×tan45°
=1

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Similar questions

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}

t a n 1^{\circ} . t a n 89^{\circ} . t a n 2^{\circ} . t a n 88^{\circ}

$t a n 1^{\circ} . t a n 89^{\circ} = t a n 2^{\circ} . t a n 88^{\circ}$

S t a t e t r u e o r f a l s e .

t a n 1^{\circ} . t a n 89^{\circ} = t a n 2^{\circ} . t a n 88^{\circ}

P-1:-tan1.tan2.tan3......tan87.tan88.tan89=1

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