85. ค่าของ \(\frac{(\sin\theta-\cos\theta)^{2}-1}{\tan\theta-\sin\theta\cos\theta}\) ตรงกับข้อใดต่อไปนี้
- \(\cot\theta\)
- \(2\cot\theta\)
- \(2\)
- \(\frac{1}{\tan^{2}\theta}\)
- \(2\cot^{2}\theta\)
วิธีทำ ข้อนี้สิ่งที่ต้องรู่คือ
\(\sin^{2}\theta +\cos^{2}\theta =1\) ดังนั้น \(sin^{2}\theta =1-\cos^{2}\theta\)
\(\cot\theta=\frac{\cos\theta}{\sin\theta}\)
\(\tan\theta=\frac{\sin\theta}{\cos\theta}\)
เอาละมาเริ่มทำกันเลย
\begin{array}{lcl}\frac{(\sin\theta + \cos\theta )^{2}-1}{\tan\theta -\sin\theta\cos\theta}&=&\frac{\sin^{2}+2\sin\theta\cos\theta +\cos^{2}\theta -1}{\tan\theta -\sin\theta\cos\theta}\\&=&\frac{2\sin\theta\cos\theta +\sin^{2}\theta +\cos^{2}\theta - 1}{\tan\theta -\sin\theta\cos\theta}\\&=&\frac{2\sin\theta\cos\theta +1-1}{\tan\theta -\sin\theta\cos\theta}\\&=&\frac{2\sin\theta\cos\theta}{\tan\theta -\sin\theta\cos\theta}\\&=&\color{red}{\frac{\cos\theta}{\cos\theta}}\times \frac{2\sin\theta\cos\theta}{tan\theta -\sin\theta\cos\theta}\\&=&\frac{2\sin\theta\cos^{2}\theta}{\sin\theta-\sin\theta\cos^{2}\theta}\\&=&\frac{\sin\theta(2\cos^{2}\theta)}{\sin\theta(1-\cos^{2}\theta)}\\&=&\frac{2\cos^{2}\theta}{1-\cos^{2}\theta}\\&=&\frac{2\cos^{2}\theta}{\sin^{2}\theta}\\&=&2\cot^{2}\theta\quad\underline{Ans}\end{array}