a-maths

2007-07-08 11:27 pm
証明恆等式
(secA-cosA)(1+cotA+tanA)=sec²A/cosec²A+secA/cosec²A

回答 (1)

2007-07-09 12:18 am
✔ 最佳答案
The question should be :

(secA-cosA)(1+cotA+t anA)=sec²A/cosecA + secA/cosec²A

L.H.S. = (secA-cosA)(1+cotA+t anA)

= secA + sec A cot A + sec A tan A - cos A - cos A cot A - cos A tan A

= sec A + cosec A + sin A sec²A - cos A - cos²A cosec A - sin A

= 1 / cos A + 1 / sin A + sin A /cos²A - cos A - cos²A / sin A - sin A

= ( cos A + sin A - cos^3 A ) / cos²A + ( 1 - cos²A - sin²A) / sin A

= { cos A ( 1 - cos²A) + sin A } / cos²A

= ( cos A sin²A + sin A ) / cos²A

= sin²A sec A + sin A sec²A

R.H.S. = sec²A / cosec A + sec A / cosec²A

= ( sec²A cosec A + sec A ) / cosec²A

= ( 1 / cos²A sin A + 1 / cos A ) / ( 1 / sin²A )

= sinA / cos²A + sin²A / cos A

= sin²A sec A + sin A sec²A

= L.H.S.

參考: My Maths Knowledge


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