中三數學... [ trigonometric Relations ] 急!!!

1. 1 + 1 / tan^2Θ

= 1 + (sin^2Θ + cos^2Θ) / tan^2Θ

= (sin^2Θ + cos^2Θ) x (cosΘ / sinΘ) + 1------第二部出了問題,留心看看是tan^2 Θ,不是tanΘ,你漏咗平方la,

以下步驟是在你的基礎上加上去的,即係幫你加翻d平方上去,請看:

= [ cos^2 Θ (sin^2Θ + cos^2Θ) / sin^2 Θ ] + (sin^2Θ + cos^2Θ) / 1

= [ cos^2 Θ (sin^2Θ + cos^2Θ) + sin^2 Θ (sin^2Θ + cos^2Θ)] / sin^2 Θ

= (cos^2 Θ + sin^2 Θ) / sin^2 Θ

=1/sin^2 Θ

=csc^2 Θ


這個方法雖然也能計出答案,不過,略嫌有點麻煩,其實一開始時無需將1化成sin^2 Θ+cos^2 Θ,因為這對題目沒甚麼幫助,以下兩個的方法,較為簡單,看看能否幫上你的忙......

方法一:

1 + 1 / tan^2Θ

=1+cot^2 Θ

=csc^2 Θ



方法二:

1 + 1 / tan^2Θ

=1 + cos^2 Θ /sin^2 Θ

=(sin^2 Θ +cos^2 Θ )/sin^2 Θ

=1/sin^2 Θ

=csc^2 Θ

1 + 1 / tan^2Θ
=1 + cot^2Θ
=csc^2Θ

2007-04-09 21:40:41 補充：
同埋你第2步已經有錯..= 1 + (sin^2Θ + cos^2Θ) / tan^2Θ = (sin^2Θ + cos^2Θ) x 【(cosΘ / sinΘ)】 + 1 ＜＝＝＝＝＝果個係tan^2Θ黎 少左個^2