In a book I am reading I have:
$\displaystyle \frac{xcosx + 3sinx}{cos^2 x}$
which gets transformed to:
$\displaystyle \frac{x + 3sinx}{cos x}$
How is that possible? Can someone please explain how that works?
Angus
If you want to cancel a factor you have to factor denominator and numerator:
$\displaystyle \frac{\theta \cdot \cos(\theta) + 3 \sin(\theta)}{\cos(\theta) \cdot \cos(\theta)} = \frac{\cos(\theta) \cdot \left(\theta + 3 \cdot \frac{\sin(\theta)}{\cos(\theta)}\right)}{ \cos(\theta) \cdot \cos(\theta)}$
Hello, angypangy!
$\displaystyle \text{In a book I am reading I have: }\:\frac{x\cos x + 3\sin x}{\cos^2\!x}$
. . $\displaystyle \text{which gets transformed to: }\:\frac{x + 3\sin x}{\cos x}$
$\displaystyle \text{How is that possible?}$
It is not possible.
The choices are obvious:
. . you misread the statement,
. . the book has a typo,
. . or the author is a fulldecker.
Someone cancelled illegally: .$\displaystyle \frac{x\rlap{\:////}\cos x +3\sin x}{\cos^{\rlap{/}2}x} \;=\; \frac{x+3\sin x}{\cos x}\;\;??$
$\displaystyle \frac{x\cdot \cos{x}+\sin{x}}{\cos^2{x}}$
$\displaystyle \frac{x\cdot \cos{x}+\sin{x}}{\cos^2{x}}=$$\displaystyle \frac{x\cdot \cos{x} \cdot \sec{x} +\sec{x}\cdot\sin{x}}{\sec{x} \cdot\cos^2{x}}$Multiply the fraction by $\displaystyle {\sec{x} \over \sec{x}}$:
$\displaystyle = \frac{x \cdot \left( {\cos{x} \over \cos{x}}\right)+{\sin{x} \over \cos{x}}}{\left( {\cos^2{x} \over \cos{x} }\right)} = \frac{x +{\sin{x} \over \cos{x}}}{\left( {\cos{x}}\right)}$Put $\displaystyle \sec{x}={1 \over \cos{x}}$
$\displaystyle =\frac{x +3\tan{x}}{\cos{x}}$Put $\displaystyle {\sin{x} \over \cos{x}}=\tan{x}$