F.3 maths
回答 (1)
2007-03-26 9:30 pm
✔ 最佳答案
LHS= tan²x + 1 /(tan²x)
= ( sin²x/cos²x ) + [1/(sin²x/cos²x) ]
= (sin²x/cos²x) + (cos²x/sin²x)
= (sin²x/cos²x)(sin²x/sin²x) + (cos²x/sin²x)(cos²x/cos²x)
= [(sin²x)²+ (cos²x)² ]/ sin²xcos²x
= [(sin²x)²+ (cos²x)² +2 sin²x cos²x - 2 sin²x cos²x]/ sin²x cos²x
= [(sin²x + cos²x)² - 2 sin²x cos²x ] / sin²x cos²x
= (1- 2 sin²x cos²x)/sin²x cos²x
= [1/(sin²x cos²x)] - 2
= RHS
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