The given expression simplifies to $$\sin^{-1}(\sin 2)-\cos^{-1}(\cos 2)+\tan^{-1}(\tan 4)-\cot^{-1}(\cot 4)+\sec^{-1}(\sec 6)-\csc^{-1} (\csc 6)$$

$$=(\pi-2)-2+(4-\pi)-(2\pi-4)+(2\pi-6)-(2\pi-6)$$ $$=-2\pi+4$$

But the given answer is $5\pi-16$. I rechecked all the the principle branches, and they all seem to be right. Where did use the wrong value?

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    $\begingroup$ Are the angle values in degrees or radians? $\endgroup$ – Teresa Lisbon Jul 7 '20 at 6:02
  • $\begingroup$ math.stackexchange.com/a/3747357/675453 $\endgroup$ – UmbQbify Jul 7 '20 at 6:26
  • $\begingroup$ @астон вілла тереса лисбон, rad. $\endgroup$ – UmbQbify Jul 7 '20 at 6:26
  • $\begingroup$ @Aditya, verify it on worlfram alpha or desmos. $\endgroup$ – UmbQbify Jul 7 '20 at 6:27
  • $\begingroup$ @астонвіллатересалисбон yes $\endgroup$ – Aditya Jul 7 '20 at 6:33

These type of questions are very easy to handle by graphs. Just remember their graphs and for finding the value at each point, see the location of that point and write the equation of the line by seeing it's slope and the point where it cuts the $x$ axis and you will get the value. For example, you have to find $\csc^{-1}(\csc 6)$, in it's graph note that $6$ lies in between $\frac{3\pi}{2}$ and $2\pi$ and the line has a positive slope and the cutting point with $x$ axis is $x=2\pi$ and hence the equation of the line would be $y=x-2\pi$.

So your mistakes were $\cot^{-1}(\cot 4)=\pi-4$, not $2\pi-4$, and secondly $\csc^{-1}(\csc 6)=6-2\pi$ and not $2\pi-6$. Hence the correct answer will come out to be $5\pi-16$.

  • $\begingroup$ Could you please elaborate on your method? It seems interesting but I don’t know what you mean by finding the slope on a non linear graph without derivatives. I also don’t know how you ended up with $y=x-2\pi-4$ $\endgroup$ – Aditya Jul 8 '20 at 11:57
  • $\begingroup$ I wrote the equation $y=x-2\pi$, not $y=x-2\pi-4$. See, what I am saying is , take for example you want to find the value of $\sin^{-1}(\sin 5)$, now here the first thing you are required to do is to draw (or look at) the graph of $y=\sin^{-1}(\sin x)$. You will see that the graph has triangular lines continuing from negative infinity to positive infinity. Now you are interested at $x=5$, right? So, you see that $5$ lies in that part of the graph where $x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right]$, it is a straight line. $\endgroup$ – Harry Potter Jul 8 '20 at 13:02
  • $\begingroup$ Now, for finding it's equation, see that it has a positive slope of ($+1$) with the cutting point of $x=2\pi$ on the $x$ axis, which is just a root. So now you can easily write the equation of the line as $y=x-2\pi$ because it becomes $0$ on substituting back $2\pi$ in the equation. Now just put $x=5$ in it to get $\sin^{-1}(\sin 5)=5-2\pi$. $\endgroup$ – Harry Potter Jul 8 '20 at 13:04
  • $\begingroup$ You can also refer to this $\endgroup$ – Harry Potter Jul 8 '20 at 13:08

I don't get your simplification. How did $\sin^{-1}(\cos 2) - \cos^{-1}(\sin 2)$ become $\sin^{-1}(\sin 2) - \cos^{-1}(\cos 2)$? I think the simplification part is off.

We must identify which quadrant each of these lie in. For example, $2$ lies in the second quadrant, because $\pi > 2 > \frac \pi 2$. $4$ lies in the third quadrant, and $6$ lies in the fourth quadrant.

With this in mind, $\sin^{-1}(\cos 2) = \frac{\pi}{2} - 2$. Similarly, $\cos^{-1}(\sin 2) = 2 - \frac \pi 2$, because the inverse cosine function is restricted to $(-\frac{\pi}{2},\frac{\pi}{2}]$.

  • $\tan^{-1}(\cot 4) = \frac{3 \pi}{2} - 4$.

  • $\cot^{-1} (\tan 4) = \frac{3 \pi }{2} - 4$.

  • $\sec^{-1}(\csc 6) = \frac{5 \pi}{2} - 6$.

  • $\csc^{-1}(\sec 6) = 6 - \frac{3 \pi}{2}$.

(You can convince yourself of these and finish the problem).

With the simplified expression :

  • $\sin^{-1}(\sin 2) = \pi- 2$

  • $\cos^{-1}(\cos 2) = 2$

  • $\tan^{-1}(\tan 4) = 4-\pi$.

  • $\color{green}{\cot^{-1}(\cot 4) = 4-\pi}$. Explanation : the principal region for $\cot^{-1}$ is $[-\frac \pi 2, \frac \pi 2]$, and $\cot 4$ is positive because $4$ is in the third quadrant. Thus, $\cot^{-1}(\cot 4)$ lies in the first quadrant, so the angle is positive, hence $4-\pi$ (and not $\pi -4$, which is a negative angle).

  • $\sec^{-1}(\sec 6) = 2\pi - 6$.

  • $\color{green}{\csc^{-1}(\csc 6) = 6-2\pi}$. Explanation : the principal region of $\csc^{-1}$ is $[-\frac {\pi}{2} , \frac{\pi}{2}]$ , and $6$ belongs in the fourth quadrant, so is already in the prescribed region, as long as we subtract $2\pi$, which gives $6 - 2\pi$ (and since the cosecants of both these angles are negative, this makes sense).

Now evaluating gives the right answer.

  • $\begingroup$ The simplification correct. I used $\cos^{-1}x + \sin^{-1}x=\frac{\pi}{2}$ $\endgroup$ – Aditya Jul 7 '20 at 10:28
  • $\begingroup$ Ok, then I will provide the corrected version for the expression you have written. $\endgroup$ – Teresa Lisbon Jul 7 '20 at 17:10

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