# Thread: How does this algebraic work?

1. ## How does this algebraic work?

In a book I am reading I have:

$\frac{xcosx + 3sinx}{cos^2 x}$

which gets transformed to:

$\frac{x + 3sinx}{cos x}$

How is that possible? Can someone please explain how that works?

Angus

2. ## Re: How does this algebraic work?

Let $\theta = x$

$\frac{\theta \cdot cos\theta + 3sin\theta}{cos\theta \cdot cos\theta}$

Top and bottom Cos theta's cancel out leaving you with the answer

3. ## Re: How does this algebraic work?

Originally Posted by BobBali
Let $\theta = x$

$\frac{\theta \cdot cos\theta + 3sin\theta}{cos\theta \cdot cos\theta}$

Top and bottom Cos theta's cancel out <--- No! you can't cancel summands!
If you want to cancel a factor you have to factor denominator and numerator:

$\frac{\theta \cdot \cos(\theta) + 3 \sin(\theta)}{\cos(\theta) \cdot \cos(\theta)} = \frac{\cos(\theta) \cdot \left(\theta + 3 \cdot \frac{\sin(\theta)}{\cos(\theta)}\right)}{ \cos(\theta) \cdot \cos(\theta)}$

4. ## Re: How does this algebraic work?

Hello, angypangy!

$\text{In a book I am reading I have: }\:\frac{x\cos x + 3\sin x}{\cos^2\!x}$

. . $\text{which gets transformed to: }\:\frac{x + 3\sin x}{\cos x}$

$\text{How is that possible?}$

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: . $\frac{x\rlap{\:////}\cos x +3\sin x}{\cos^{\rlap{/}2}x} \;=\; \frac{x+3\sin x}{\cos x}\;\;??$

5. ## Re: How does this algebraic work?

Yeah, maybe there's a typo.

I am sure the author meant to say:

$\frac{x\cdot \cos{x}+3\sin{x}}{\cos^2{x}}= \frac{x +3\tan{x}}{\cos{x}}$

6. ## Re: How does this algebraic work?

Originally Posted by sbhatnagar
Yeah, maybe there's a typo.

I am sure the author meant to say:

$\frac{x\cdot \cos{x}+3\sin{x}}{\cos^2{x}}= \frac{x +3\tan{x}}{\cos{x}}$

Sorry all, sbhatnager is correct, yes I meant to say
$\frac{x +3\tan{x}}{\cos{x}}$

Angus

7. ## Re: How does this algebraic work?

Originally Posted by BobBali
Let $\theta = x$

$\frac{\theta \cdot cos\theta + 3sin\theta}{cos\theta \cdot cos\theta}$

Top and bottom Cos theta's cancel out leaving you with the answer
what is the point in your assumption $x = \theta$ ??

8. ## Re: How does this algebraic work?

Originally Posted by angypangy
Sorry all, sbhatnager is correct, yes I meant to say
$\frac{x +3\tan{x}}{\cos{x}}$

Angus
$\frac{x\cdot \cos{x}+\sin{x}}{\cos^2{x}}$

Multiply the fraction by ${\sec{x} \over \sec{x}}$:
$\frac{x\cdot \cos{x}+\sin{x}}{\cos^2{x}}=$ $\frac{x\cdot \cos{x} \cdot \sec{x} +\sec{x}\cdot\sin{x}}{\sec{x} \cdot\cos^2{x}}$

Put $\sec{x}={1 \over \cos{x}}$
$= \frac{x \cdot \left( {\cos{x} \over \cos{x}}\right)+{\sin{x} \over \cos{x}}}{\left( {\cos^2{x} \over \cos{x} }\right)} = \frac{x +{\sin{x} \over \cos{x}}}{\left( {\cos{x}}\right)}$

Put ${\sin{x} \over \cos{x}}=\tan{x}$
$=\frac{x +3\tan{x}}{\cos{x}}$