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3 years ago

(cosxtanx-tanx+2cosx-2)/tanx+2=cosx-1 can someone show me how to prove this?

Mathematics

1 answer:

uysha [10]3 years ago

7 0

Answer:

See detail below.

Step-by-step explanation:

A word of caution before getting to the actual problem: I believe there is an important set of brackets missing in the original post. The expression on the left hand side should be:

(cosxtanx-tanx+2cosx-2)/(tanx+2)

Without the brackets, it is left unclear whether the denominator is just tanx or tanx+2. I recommend to use brackets wherever any doubt could arise.

Now to the actual problem: \we can make the following transformations on the left hand side:

$\frac{\cos x \tan x + 2\cos x -2}{\tan x +2}=\frac{\cos x \frac{\sin x}{\cos x} + 2\cos x -2}{\frac{\sin x +2\sin x}{\cos x} }=\\=\frac{\cos x \sin x - \sin x + 2\cos^2 x - 2\cos x}{\sin x + 2\cos x}=\\=\frac{\cos x \sin x +2\cos^2 x }{\sin x + 2\cos x}-1=\\=\cos x -1$

which is shown to be the same as the right hand side, which was to be shown.

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Ples help me find slant assemtotes

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A polynomial asymptote is a function $p(x)$ such that

$\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0$

$(y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}$

Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form $p(x)=ax+b$ .

Ignore the negative root (we don't need it). If $y=2x-1+2\sqrt{x^2-x}$ , then we want to find constants $a,b$ such that

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

We have

$\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}$

since $x\to\infty$ forces us to have $x>0$ . And as $x\to\infty$ , the $\dfrac1x$ term is "negligible", so really $\sqrt{x^2-x}\approx x$ . We can then treat the limand like

$2x-1+2x-ax-b=(4-a)x-(b+1)$

which tells us that we would choose $a=4$ . You might be tempted to think $b=-1$ , but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of $b$ , we have to solve for it in the following limit.

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0$

$\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2$

We write

$(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}$

Now as $x\to\infty$ , we see this expression approaching $-\dfrac12$ , so that

$-\dfrac12=\dfrac{b+1}2\implies b=-2$

So one asymptote of the hyperbola is the line $y=4x-2$ .

The other asymptote is obtained similarly by examining the limit as $x\to-\infty$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

$\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0$

Reduce the "negligible" term to get

$\displaystyle\lim_{x\to-\infty}(-ax-(b+1))=0$

Now we take $a=0$ , and again we're careful to not pick $b=-1$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0$

$\displaystyle\lim_{x\to-\infty}(x+\sqrt{x^2-x})=\frac{b+1}2$

$(x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac
 x{x-(-x)\sqrt{1-\frac1x}}=\dfrac1{1+\sqrt{1-\frac1x}}$

This time the limit is $\dfrac12$ , so

$\dfrac12=\dfrac{b+1}2\implies b=0$

which means the other asymptote is the line $y=0$ .

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Step-by-step explanation:

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