$$\tan(\theta) = \dfrac{y}{x}$$ Suppose initially the angle $\theta$ is $0$. As $\theta$ increases to $\pi/2$, the denominator $x$ decreases and the numerator $y$ increases, so $\tan(\theta)$ becomes arbitrarily large.     Exactly at $\theta=\pi/2$ the function becomes +infinity. Now if we increase $\theta$ even slightly, the denominator becomes $-ve$ and the function value suddenly changes to -infinity.     I'm finding it hard to wrap my head around this. Is there anything in real life that behaves like this? that is, changes its value from high positive to high negative abruptly..
113661
The way I have understood, $0/0$ is undefined or indeterminate because, if $c=0/0$ then $c\cdot 0=0$, where $c$ can be any finite number including $0$ itself.  If we also observe a fraction $F=a/b$ where $a,b$ are positive real numbers,  the value of $F$ increases with the decrements of $b$. Being $0$ the least non-negative integer, if $b$ tends $0$ then $F$ tends to $\infty$ which is greater than all finite numbers.    I have also heard that no number is equal to infinity, a variable can tend to infinity. Any ratio is also a variable or is resolved to a number.  Now,   $$\tan(\pi/2) = \frac{\sin(\pi/2)}{\cos(\pi/2)}=\frac{1}{0},$$    is it undefined or infinity?  Also we know   $$\tan2x= \frac{2\tan x}{1-\tan^2x}$$  Is this formula valid for $x=\pi/2$?  Any rectification is more than welcome.  Some references:
32691
$$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$ I tried using $\lim\limits_{x\to0} \frac{\sin x}x=1$. But it doesn't work :/
28612
I know $\int \tan^{1/3}\theta  d \theta$ is a non integrable but wolfram alpha does it easily. And can anyone explain how a fuction is non integrable? My argument is that there has to be a function which represents the area under a graph. It's just that we do not know it. Please shed a bit light on it. Also, how can we solve $\int \tan^{1/3}\theta  d \theta$?
16032
I am asked to find the following limit:  $$ \lim_{x \to 0} \frac{\tan 3x}{\tan 5x}$$  My problem is in simplifying the function. I followed two different approaches to solve  the problem. But both seems incorrect.  Apprach 1) Since $\tan \theta = \frac{sin \theta}{cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$, we have:  $$\frac{\tan 3x}{\tan 5x} = \frac{\sin 3x \times \cos 5x}{\sin 5x \times \cos 3x}$$  This approach does not work well, because I cannot simplify more.   Approach 2) Since $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \times \tan y)}$, we have:  $$\frac{\tan 3x}{\tan 5x} = \frac{\tan 3x}{\tan 3x + \tan 2x} = \frac{\tan 3x}{\frac{\tan 2 + \tan3}{1 - \tan 2 \times \tan 3}}$$  What am I doing wrong?
34440
I'm having a trouble with this integral expression:  $$\int_0^{2\pi} \frac{d\theta}{A+B \cos\theta}$$  I've done this substitution: $t= \tan(\theta/2)$  and get: $\displaystyle \cos\theta= \frac{1-t^2}{1+t^2}$ and $\displaystyle d\theta=\frac{2}{1+t^2}dt$ where $\displaystyle \cos^2\theta/2=\frac{1}{1+t^2}$  then the integral becomes:  $$\int\frac{2 \, dt}{(A-B)t^2+(A+B)}= \sqrt\frac{A+B}{A-B} \arctan \left(\left(\sqrt\frac{A+B}{A-B} \right) t\right)$$   However, I'm not sure about the new limits since $\tan$ has period $\pi$ so what I have to do at this point to decide the new limits? And of course if you find some mistake in what I've done before, please let me know!
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$$\tan(\theta_1+\theta_2+\cdots+\theta_n)=\frac{S_1-S_3+S_5+\cdots}{1-S_2+S_4+\cdots}$$  where  $S_i$ denotes the sum of product of tangent of angles taken $i$ at a time.     For example,$$\tan(\theta_1+\theta_2)=\frac{S_1}{1-S_2}=\frac{\tan\theta_1+\tan\theta_2} {1-\tan\theta_1\tan\theta_2}$$   (This formula is given  in my textbook with no derivation or background)  How to derive this?
118928
Prove that $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$$   I tried to rationalize the denominator but I always end up with a large fraction that won't cancel out. Is there something I'm missing?  Thanks in advance
26799
Verify the identity: $\tan^{-1} x + \tan^{-1} (1/x) = \frac\pi 2, x &gt; 0$  $$\alpha= \tan^{-1} x$$  $$\beta = \tan^{-1} (1/x)$$  $$\tan \alpha = x$$  $$\tan \beta = 1/x$$  $$\tan^{-1}[\tan(\alpha + \beta)]$$  $$\tan^{-1}\left [{\tan\alpha + \tan\beta\over 1 - \tan\alpha \tan\beta} \right]$$  $$\tan^{-1}\left[ {x + 1/x\over 1- x/x }\right]$$  $$\tan^{-1}\left[{x + (1/x)\over 0} \right]$$  I can't find out what I'm doing wrong..
59734
$$ \int \sin \theta\cos \theta~d \theta= \int \frac {1} 2  \sin 2\theta~ d \theta=-\frac {1} 4  \cos 2\theta$$ But, if I let  $$ u=\sin \theta , \text{ then }du=\cos \theta~d\theta $$ Then $$ \int \sin \theta\cos \theta~d \theta= \int u ~ du =\frac { u^2 } 2 =\frac {1} 2 \sin^2 \theta $$ Since $$ \sin^2 \theta =\frac {1} 2 - \frac {1} 2 \cos 2\theta$$ The above can be written as $$ \int \sin \theta\cos \theta~d \theta= \frac {1} 2 \sin^2 \theta =\frac {1} 2 \left( \frac {1} 2 - \frac {1} 2 \cos 2\theta \right)=\frac {1} 4-\frac {1} 4 \cos 2\theta   $$ Why are the two results differ by the constant $1/4$? Thank you.
74654
I'm stuck in this problem. I can not integrate $\int \sec\theta\tan^2\theta \,d\theta$. Can anyone help me.
32720
$${\dfrac{1}{\pi^2}}\int_{0}^{\infty}\dfrac{{(\ln{x}})^2}{\sqrt{x}{(1-x)^2}} \mathrm d x$$   I tried substituting $1/x$ for $x$ but the the only change in the integral is that the $\sqrt{x}$ moves in the numerator from the denominator. I don't understand what to substitute. $\tan{x}$ doesn't seem to work.
117306
So I am looking at the integral $$ \int_0^{2\pi} \frac{1}{a + \cos\theta} \mathrm{d} \theta$$ for $a &gt; 1$. Evaluating the integral with complex analysis gives us $\frac{2 \pi}{\sqrt{a^2 - 1}} $ but I have failed to reproduce the result using standard substitution.  Here’s what I did: I found the antiderivative first, which is $$\frac{2}{\sqrt{a^2 - 1}}  \arctan \left(\sqrt{\frac{a - 1}{a + 1}} \tan\left(\frac{\theta}{2}\right) \right)$$ but substituting in the bounds would give us 0.   I used the standard substitution $t = \tan\left(\frac{\theta}{2}\right) $ to obtain the following result.  EDIT: Plotting the graph shows us that the area is definitely positive, so there’s no way the answer is $0$.  EDIT2: Is it because $2\pi$ corresponds to the second “cycle” so we should $\arctan 0$ to be $\pi$ for our upper bound instead?
116013
I have to prove the following result :  $$\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta$$  I tried converting $\tan\theta$ &amp; $\cot\theta$ into $\cos\theta$ and $\sin\theta$. That led to a huge expression which I wasn't able to simplify.  Please help!!!
4227
I'm trying to show that the derivative of $\sin\theta$ is equal to $\pi/180 \cos\theta$ if $\theta$ is measured in degrees. The main idea is that we need to convert $\theta$ to radians to be able to apply the identity $d/dx \sin x = \cos x $. So we need to express $ \sin \theta$  as  $$ \sin_{deg} \theta = \sin(\pi \theta /180), $$ where $\sin_{deg}$ is the $\sin$ function that takes degrees as input. Then applying the chain rule yields  $$ d/d\theta [ \sin(\pi\theta/180)] = \cos(\pi \theta/180) \pi/180 = \frac{\pi}{180}\cos_{deg}\theta. $$ Is this derivation formally correct?
67960
I proved as follows:  Let $f(x) = \tan x -  x - \frac{x^{3}}{3}.$ Then $f '(x) = \sec^{2}x - 1 - x^{2}$ and $f''(x) = 2\sec^{2}x\tan x-x &gt; \tan x - x,$ since $\\tan x &gt; x,$ $f''(x) &gt; 0$ so $f'(x)$ is increasing. Since $f '(0) = 0,$ $f '(x) &gt; 0$ for all $x \in (0, \frac{\pi}{2}).$ So $f(x)$ is increasing. since $f(0) = 0,$ $f(x) &gt; 0$ for all $x \in (0, \frac{\pi}{2}).$  But my proof seems complex. Is there any simpler way to prove this?
79662
I use tangent half-angle substitution to calculate this indefinite integral: $$ \int \frac{1}{2+\sin x}\,dx = \frac{2}{\sqrt{3}}\tan^{-1}\frac{2\tan \frac{x}{2}+1}{\sqrt{3}}+\text{constant}. $$  Wolfram Alpha also give the same answer. However, $\frac{2}{\sqrt{3}}\tan^{-1}\frac{2\tan \frac{x}{2}+1}{\sqrt{3}}$ is discontinuous on $(n+1)\pi$ where $n$ is any integer. Why is an anti-derivative of a continuous function discontinuous?
2761
$\displaystyle \int\nolimits_0^8 \sqrt{x^4+4x^2}\,dx$.   Alright, so I thought I had this figured out. Here's what I did:   I factor out an $x^2$ to get $\sqrt{x^2(x^2+4)}$. I let $x = 2\tan(\theta)$, therefore the integrand is $\sqrt{4\tan^2(\theta)  (4\tan^2(\theta) + 4)}$. Factor out a 4 and it becomes $\sqrt{(16\tan^2(\theta)  (\tan^2(\theta) + 1))}$ Which equals $\sqrt{16\tan^2(\theta) \sec^2(\theta)}$ This is easy to take the sqrt of. The integrand becomes $4\tan(\theta)\sec(\theta)$. Now, the integral of this is $4\sec(\theta)$ And it's evaluated from $0$ to $\arctan(4)$ right? Because as $x$ goes to $0$, so does $\theta$, and as $x$ goes to $8$, $\theta$ goes to $\arctan(4)$... But the end result $(4 (\sec(\arctan(4)) - 1) )$ isn't the correct answer   I put it into WolframAlpha and I get $(8/3) (17\sqrt{17} - 1)$, which is the right answer. How did they get that? (there's no "show steps" option)  Any help is greatly appreciated!  PS, what's the syntax for doing sqrts and exponentials?