Let y = sec^-1 (-2). Then, sec y = -2
By the definition, the range of the principal value branch of y = sec⁻¹x is [0, \(\pi\)]\{\({{\frac{\pi}{2}}}\)}
Let us find y in [0, \(\pi\)] - {\({{\frac{\pi}{2}}}\)} such that sec y = -2
But, sec y = −2 \(\Rightarrow\) cos y = -\(\frac{1}{2}\)
Now, cos y = -\(\frac { 1 }{ 2 } =-cos\frac { \pi }{ 3 } =cos\left( \pi -\frac { \pi }{ 3 } \right) =cos\frac { 2\pi }{ 3 } \). Therefore, y = \(\frac{2\pi}{3}\)
since \(\frac{2\pi}{3}\in[0,\pi]\)\{\({{\frac{\pi}{2}}}\)}, the principal value of sec^-1(-2) is \(\frac{2\pi}{3}\)

tan(tan^-1(1947))
= 1947

y = 4 sin (-2x)
The amplitude of sin x is 1
\(\Rightarrow\) amplitude of sin (-2x) is 1
\(\therefore\) Amplitude 4 sin(-2x) is 4 \(\times\) 1 = 4.
The period of sin \((-2x)is2x=2\pi \Rightarrow =\frac { 2\pi }{ 2 } =\pi \)

cos^-1\([cos(-\frac{\pi}{6})]\neq \frac{\pi}{6}.\)
Since \(\frac { -\pi }{ 6 } \notin \left[ 0,\pi \right] \) which is the principal domain of cosine function. [\(\therefore \) cos -\(\theta\) = cos \(\theta\)]

Let cos⁻¹\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \) = y. Then, cos y = \( \frac { \sqrt { 3 } }{ 2 } \)
The range of the principal values of y = cos⁻¹x is [0, \(\pi\)].
So, let us find y in [0, \(\pi\)] such that cos y =\( \frac { \sqrt { 3 } }{ 2 } \)
But, cos\(\frac{\pi}{6}=\frac{\sqrt3}{2} and \frac{\pi}{6}\in[0,\pi]\). Therefore, y = \(\frac{\pi}{6}\)
Thus, the principal value of cos^-1 \(\left( \frac { \sqrt { 3 } }{ 2 } \right) is\frac { \pi }{ 6 } \)

Let y = sin^-1x
When y = 0, 0 = sin^-1Ix
\(\Rightarrow\) sin(0) = sin (sin^-1)(x))
\(\Rightarrow\)sin 0 = x
\(\Rightarrow\)x = 0
Hence, solution to (1) is x = 0. Also, graph of sin x and sin^-1x intersect at origin (0, 0).

Since the domain of y = sin^-1 is −[11], and 2\(\notin \)[-1, 1], sin⁻¹(2) does not exist.

sin^-1[sin10]
We know that sin^-1(sin \(\theta\)) = \(\theta\) is \(\theta \in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)Considering the approximation \(\frac{\pi}{2}=\frac{11}{7}\)
we conclude that 10\(\notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \), but (10-3\(\pi\)) \(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \).
Now, sin10 = sin(3\(\pi\)+(10−3\(\pi\))) = sin(\(\pi\)+(10−3\(\pi\)) = −sin(10−3\(\pi\)) = sin(3\(\pi\)-10)
Hence, sin^-1[sin10] = sin^-1[sin(3\(\pi\)-10)] = 3\(\pi\)-10, since(3\(\pi\)-10)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \).

\({ cos }^{ -1 }\left( cos\left( \frac { 13\pi }{ 3 } \right) \right) \).
The range of principal values of cos^-1x is [0, \(\pi\)].
Since \(\frac{13\pi}{3}\not \in[0,\pi]\), we write \(\frac{13\pi}{3} as \frac{13\pi}{3}=4\pi+\frac{\pi}{3}, whre \frac{\pi}{3}\in[0,\pi]\)
Now, cos\((\frac{13\pi}{3}=cos (4\pi+\frac{\pi}{3})=cos\frac{\pi}{3}\)
Thus,\({ cos }^{ -1 }\left( cos\left( \frac { 13\pi }{ 3 } \right) \right) ={ cos }^{ -1 }\left( cos\left( \frac { \pi }{ 3 } \right) \right) ,\ since\frac { \pi }{ 3 } \in [0,\pi ]\).

Let tan⁻¹(−1) = y. Then, tan y = -1 = -tan\(\frac{\pi}{4}=tan(-\frac{\pi}{4})\)
As -\(\frac{\pi}{4}\in (-\frac{\pi}{2},\frac{\pi}{2}), tan^-1(-1)=-\frac{\pi}{ 3}\)
Now, cos^-1\((\frac{1}{2})\) = y implies cos y = \(\frac{1}{2}\) = cos\(\frac{\pi}{3}\)
As \(\frac{\pi}{3}\)\(\in\)[0, \(\pi\)], cos^-1 \((\frac{1}{2})=\frac{\pi}{3}\)
Now, sin^-1\((-\frac{1}{2})\) = y implies sin y = -\(\frac{1}{2}\) = sin(-\(\frac{\pi}{3}\)).
As -\(\frac{\pi}{6}\in[-\frac{\pi}{2},\frac{\pi}{2}], sin^-1(-\frac{1}{2})=-\frac{\pi}{6}\)
Therefore, tan⁻¹(−1)+cos^-1\((\frac{1}{2})+sin^-1(-\frac{1}{2})=-\frac{\pi}{4}+\frac{\pi}{3}-\frac{\pi}{6}=-\frac{\pi}{12}\)

\(2{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =x\) and\({ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(\Rightarrow cosx=\frac { 1 }{ 2 } \)
\(\Rightarrow cosx=cos\frac { \pi }{ 3 } \) \(\left[ \therefore \frac { \pi }{ 3 } \varepsilon \left[ 0,\pi \right] \right] \)
\(\Rightarrow x=\frac { \pi }{ 3 } \) 
[\(\therefore\) Principal domain of sin is \(\therefore \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) and principal domain of cos is\(\left[ 0,\pi \right] \)
\(siny=\frac { 1 }{ 2 } \)
\(siny=sin\frac { \pi }{ 6 } \)  \(\left[ \because \frac { \pi }{ 6 } \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\(\Rightarrow x=\frac { \pi }{ 3 } \) and \(y=\frac { \pi }{ 6 } \)
\(\therefore \quad 2{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
= \(2\left( \frac { \pi }{ 3 } \right) +\frac { \pi }{ 6 } =\frac { 2\pi }{ 3 } +\frac { \pi }{ 6 } \)
= \(\frac { 4\pi +\pi }{ 6 } =\frac { 5\pi }{ 6 } \)

By definition, the domain of yx = cos^-1 x is -1. This leads to \(-1\le\frac{2+sinx}{3}\le1\) which is same as -3\(\le\)2+sinx\(\le\)3
so, -5\(\le sin\ x\le1 \) reduces to -1\(\le sin\ x\le1 \), which gives
-sin^-1(1)\(\le x\le sin^-1(1) or -\frac{\pi}{2}\le x\le \frac{\pi}{2}\)
Thus, the domain of cos-1\((\frac{2+sin\ x}{3}) is [-\frac{\pi}{2},\frac{\pi}{2}].\)

We know that the domain of sin⁻¹(x) is [-1, 1].
This leads to −1\(\le\)2 - 3x²\(\le\)1, Which implies -3\(\le\) -3x²\(\le\)-1
Now, -3\(\le\) -3x², gives x²\(\le\)1 and ............(1)
-3\(\le\)-3x²\(\le\)-1, gives x²\(\ge\)\(\frac{1}{2}\) .........(2)
Combining the equations (1) and (2), we get \(\frac{1}{3}\le x^2\le 1\). That is \(\frac{1}{\sqrt3}\le |x|\le1\), Which gives \(x\in[-1,-1\frac{1}{\sqrt3}]\cup[\frac{1}{\sqrt3},1]\)
since  a\(\le|x|\le b\) implies x \(\in[-b,-a]\cup[a,b]\).

Now, \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =tan^{ -1 }\left[ \frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \right] =\frac { \pi }{ 4 } \)
Thus, \(\frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \) = 1, which on simplification gives 2x²−4 = −3
Thus, x² = \(\frac{1}{2}\)gives x = \(\pm \frac { 1 }{ \sqrt { 2 } } \)