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I will explain my approach, help me with the last step please! $$ \tan^{-1} {\left(\frac {x - \sqrt {1-x^2}}{x + \sqrt {1-x^2}}\right)}$$

substituting x = $\sin \theta$ (as learnt from book) and solving 1-$\sin^2 \theta$ = $\cos^2 \theta$ $$ \tan^{-1} {\left(\frac {\sin \theta - |\cos \theta|}{\sin \theta + |\cos \theta| }\right)}$$

For solving modulus, it was important to determine range of $\theta$ , therefore I defined it (as it is my variable,i can define it my way) for [-$\pi$/2 , $\pi$/2] so that sine covers all values from $-1$ to $1$ (as , $ -1 \le x \le 1 \,$ , from domain ) and $\cos \theta$ is positive , and hence $|\cos \theta| = \cos \theta$.

$$ \tan^{-1} {\left(\frac {\sin \theta - \cos \theta}{\sin \theta + \cos \theta }\right)}$$ = dividing by $\cos \theta$ $$ \tan^{-1} {\left(\frac {\tan \theta - 1}{\tan\theta + 1 }\right)}$$

= by formula of $\tan (\theta - \pi/4)$ $$ \tan^{-1}( \tan{\left(\theta - \pi/4\right)})$$

That's where I am stuck ,as according to the identity,$\quad$ $tan^{-1} ( \tan \alpha) = \alpha$ $\quad$ only when $\, -\pi/2 <\alpha < \pi/2$ . But here $$ -3\pi/4 \le \,(\theta-\pi/4) \, \le \pi/4 $$ Therefore, I am not going to get ($ \,\theta - \pi/4 $) out of the expression. What i get will be based on that graph of $\bf {\tan^{-1} (\tan x)}$ . $$ (\theta - \pi/4) +\pi \,$$ for $\,-3\pi/4 \le \, (\theta -\pi/4) \, < -\pi/2 \,\,$ and

$$\theta -\pi/4$$ for $\,-\pi/2 < \, (\theta -\pi/4) \, \le \pi/4 \,\,$

My teacher just cancelled arctan and tan and wrote $\theta - \pi/4$ and he didn't even include that modulus function over $\cos \theta$.

So what will be the exact answer because if everyone decide $\theta$ as per they like then there will not be a finite answer. Everyone will have their own answers and in each answer they have multiple cases as I just discussed above.

So please help me, very hopefully I signed up in stackexchange!

Found Solution :-

I was confused because I was thinking that there can be many solutions differing person to person, but even if you choose any value of $\theta$ , you are going to get two solutions which are in the asked question above. The problem resolves when we write $\theta$ in terms of $sin^{-1} x$ as then we would not simply write like $$ \theta = \sin^{-1} x $$ we would write an equation,$$ \sin^{-1} x = \sin^{-1} (\sin \theta)$$, now if $\theta$ is not in range of $-\pi/2$ and $\,\pi/2$ , then there would be some constant in $\pi$ (like , $\pi/4 , 2\pi$ etc. we would have to add or subtract according to the graph of 'sin inverse sin' and when we would put that value of $\theta$ , we would end with the solutions as answered by people. (I write the answer in this edit to help anyone who will reach here after searching web , thanks to everyone for answers)

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    $\begingroup$ Your answer is absolutely correct and not your teacher's. You should add one thing while substitution that $\theta\ne-π/4$. $\endgroup$ – SarGe Jul 17 '20 at 17:21
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The domain of the function is $x\in [-1,-\frac1{\sqrt2})\cup (-\frac1{\sqrt2},1]$. Then, with the substitution $\sin \theta =x$, we have $\theta \in[-\frac\pi2,-\frac\pi4)\cup (-\frac\pi4,\frac\pi2]$ and correspondingly

$$\tan^{-1} {\left(\frac {x - \sqrt {1-x^2}}{x + \sqrt {1-x^2}}\right)} =\tan^{-1}\left[\tan{\left(\theta - \frac\pi4\right)}\right]$$

$$= \begin{cases} \theta+\frac{3\pi}4 = \sin^{-1}x +\frac{3\pi}4 & x\in [-1,-\frac1{\sqrt2})\\ \theta -\frac\pi4 = \sin^{-1}x -\frac\pi4 & x\in (-\frac1{\sqrt2},1] \end{cases} $$

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  • $\begingroup$ As asked it in the question, what if everyone start defining $\theta$ on their own , say from $7\pi/2$ to $9\pi/2$ , then we will have different answers, how can be there infinite solutions to a single problem $\endgroup$ – Aryaman Jul 18 '20 at 4:28
  • $\begingroup$ @Aryaman - the choice of $\theta$ won’t change the final result which is expressed in $x$. You may use $7\pi/2$ to $9\pi/2 $ and need to revert back to $x$ accordingly $\endgroup$ – Quanto Jul 18 '20 at 12:32
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In the case of $$-\pi/2 \lt \theta-\pi/4 \le \pi/4 $$ There is no problem, and we get $\theta -\pi/4$.

For the case $$-3\pi/4 \le \theta-\pi/4 \lt -\pi/2 $$ You need to adjust by adding $\pi$, so that $$\pi/4 \le \theta +3\pi/4 \lt \pi/2 $$ The answer should really be $$\begin{cases} \theta-\pi/4 , & -\pi/4 \lt \theta \le \pi/2 \\ \theta+3\pi/4, & -\pi/2\le \theta \lt \pi/4 \end{cases}$$ Your teacher is wrong.

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  • $\begingroup$ Can you answer that last part of the question "So what will be ......above" $\endgroup$ – Aryaman Jul 18 '20 at 4:41
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    $\begingroup$ @Aryaman It doesn’t matter what range you define $\theta$ in, (as long as it covers every possible value of $\sin x$), because the range of $x$ corresponding to that will be the same. So eventually you do need to make cases according the the $x$ values, as in Quanto’s answer. $\endgroup$ – Tavish Jul 18 '20 at 8:15
  • $\begingroup$ So Tavish , we shouldn't be expecting a unique answer as one can define $\theta$ (for every possible value of sin x) and give different answers for the same range of $x$ $\endgroup$ – Aryaman Jul 18 '20 at 9:17
  • $\begingroup$ @Aryaman There is a unique answer. Notice, there is no mention of $\theta$ in the question, we have to express the answer in terms of $x$ only. (In my answer, I haven’t made the conversion, but you must) $\endgroup$ – Tavish Jul 18 '20 at 9:19
  • $\begingroup$ Tavish , see I made some edits , I got your point and it helped me reach the conclusion that I made in edit. $\endgroup$ – Aryaman Jul 18 '20 at 11:25
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I have another answer : let $x \in (-1,1)\setminus\left\{-\frac{1}{\sqrt{2}}\right\}$. Then the derivative of the formula with respect to $x$ is : \begin{align} \frac{d}{dx}\left(\arctan\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)\right) &= \frac{d}{dx}\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)\times \dfrac{1}{1+\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)^2} \end{align} We compute the first term : \begin{align} \frac{d}{dx}\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)&= \dfrac{\left(1 +\dfrac{x}{\sqrt{1-x^2}}\right)\left(x + \sqrt{1-x^2}\right) - \left(x - \sqrt{1-x^2}\right)\left(1-\dfrac{x}{\sqrt{1-x^2}}\right) }{\left(x+\sqrt{1-x^2}\right)^2} \\ &= \dfrac{\dfrac{\sqrt{1-x^2}+x}{\sqrt{1-x^2}}\left(x+\sqrt{1-x^2}\right)+\left(x-\sqrt{1-x^2}\right)\dfrac{x-\sqrt{1-x^2}}{\sqrt{1-x^2}}}{\left(x+\sqrt{1-x^2}\right)^2} \\ &= \dfrac{\left(x+\sqrt{1-x^2}\right)^2 + \left(x-\sqrt{1-x^2}\right)^2}{\sqrt{1-x^2}\left(x+\sqrt{1-x^2}\right)^2} \\ &= \dfrac{x^2 + 2x\sqrt{1-x^2} + 1-x^2 + x^2 -2x\sqrt{1-x^2} + 1-x^2}{\sqrt{1-x^2}\left(x+\sqrt{1-x^2}\right)^2} \\ &= \dfrac{2}{\sqrt{1-x^2}\left(x+\sqrt{1-x^2}\right)^2} \end{align} From there, we have : \begin{align} \frac{d}{dx}\left(\arctan\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)\right) &= \dfrac{2}{\sqrt{1-x^2}\left(x+\sqrt{1-x^2}\right)^2} \times \dfrac{1}{1+\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)^2} \\ &= \dfrac{2}{\sqrt{1-x^2}}\times \dfrac{1}{\left(x+\sqrt{1-x^2}\right)^2 + \left(x-\sqrt{1-x^2}\right)^2} \\ &= \dfrac{1}{\sqrt{1-x^2}} \end{align} Therefore, the solution is an antiderivative of $x\in (-1,1)\setminus\left\{-\frac{1}{\sqrt{2}}\right\}\mapsto \frac{1}{\sqrt{1-x^2}}$. It follows that there exists two constant $C$ and $C'$ such that: \begin{align} \forall x \in \left(-1,-\dfrac{1}{\sqrt{2}}\right) ,~ \arctan\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right) &= \arcsin x + C \\ \forall x \in \left(-\dfrac{1}{\sqrt{2}},1\right) ,~ \arctan\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right) &= \arcsin x + C' \end{align} To get the value of $C$ and $C'$, evaluate the limits at $-1$ and $1$ : \begin{align} C &=\arctan 1- \arcsin( -1)\\ &= \frac{\pi}{4}+\frac{\pi}{2} \\ &= \frac{3\pi}{4} \\ C' &=\arctan 1- \arcsin 1\\ &= \frac{\pi}{4}-\frac{\pi}{2} \\ &= -\frac{\pi}{4} \end{align} It seems like your teacher forgot to check what happens around $-\frac{1}{\sqrt2}$.

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  • $\begingroup$ Thanks Dldier, for providing a calculus view of the problem, I am new to calculus so it may take me time to appreciate your answer. $\endgroup$ – Aryaman Jul 18 '20 at 11:29

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