Math, asked by ujjwalsinghg3, 3 months ago

cot 12 cot 38 cot 52 cot 60 cot 78 = 1/root 3

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Answered by AbhilabhChinchane

1

Answer:

Answer:cot12° cot38° cot52° cot60° cot78°

Answer:cot12° cot38° cot52° cot60° cot78°= (cot12°× cot78°)(cot38° cot52°) cot60°

Answer:cot12° cot38° cot52° cot60° cot78°= (cot12°× cot78°)(cot38° cot52°) cot60° =[cot12°×cot(90° - 12º)][cot38°×cot(90° - 38º) cot60°

Answer:cot12° cot38° cot52° cot60° cot78°= (cot12°× cot78°)(cot38° cot52°) cot60° =[cot12°×cot(90° - 12º)][cot38°×cot(90° - 38º) cot60° = cot12°× tan12°× cot38°× tan38°× cot60°

Answer:cot12° cot38° cot52° cot60° cot78°= (cot12°× cot78°)(cot38° cot52°) cot60° =[cot12°×cot(90° - 12º)][cot38°×cot(90° - 38º) cot60° = cot12°× tan12°× cot38°× tan38°× cot60° = 1 × 1× cot60°

Answer:cot12° cot38° cot52° cot60° cot78°= (cot12°× cot78°)(cot38° cot52°) cot60° =[cot12°×cot(90° - 12º)][cot38°×cot(90° - 38º) cot60° = cot12°× tan12°× cot38°× tan38°× cot60° = 1 × 1× cot60° = 1/2

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