\( \cot ^{-1}(\sqrt{3}) \)
\(\cot ^{-1}(\sqrt{3}) =\theta \)
\(\sqrt{3} =\cot \theta \)
\(\cot \frac{\pi}{6} =\cot \theta \)
\(\theta =\frac{\pi}{6} \)

It is known that cos^-1 x : [-1, 1]\(\rightarrow\)[0, \(\pi\)] is given by
cos⁻¹x = y if and only if x = cos y for -1\(\le x\le1\ and\ 0\le y \le\pi\) 
Thus, we have
cos^-1\((cos(-\frac{7\pi}{6}))\) = \(\frac{5\pi}{6}\), since cos\((\frac{7x}{6})=cos(\pi+\frac{\pi}{6})=-\frac{\sqrt3}{2}=cos(\frac{5\pi}{6})\in[0, \pi].\)

It is known that cos^-1 x : [-1, 1]\(\rightarrow\)[0, \(\pi\)] is given by
cos⁻¹x = y if and only if x = cos y for -1\(\le x\le1\ and\ 0\le y \le\pi\) 
Thus, we have
cos^-1\((cos(-\frac{\pi}{3}))\) = cos^-1\((cos(-\frac{\pi}{3}))\) = \(\frac{\pi}{3}\), since -\(\frac{\pi}{3}\)\(\notin \) [0, \(\pi\)], but \(\frac{\pi}{3}\in[0, \pi]\)

\({ sin }^{ -1 }\left( sin\left( \frac { 2\pi }{ 3 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi -\frac { \pi }{ 3 } \right) \right) \)   \(\left[ \because \frac { 2\pi }{ 3 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)which is the principal domain of sine function
= \({ sin }^{ -1 }\left( sin\left( \frac { \pi }{ 3 } \right) \right) \) = \(\frac { \pi }{ 3 }\)   \(\left[ \because sin\left( \pi -\theta \right) =sin\theta \right] \)

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 \(\frac{2\pi}{3}\)

Consider cos\(\left[ \frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1 }{ 8 } \right) \right]\).
Let \( { cos }^{ -1 }\left( \frac { 1 }{ 8 } \right) =\theta .\)
Then \( cos\theta =\frac { 1 }{ 8 } and\quad \theta \in \left[ 0,\pi \right] \)
Now, cos\(\theta=\frac{1}{8}\) implies 2 cos² \(\frac{\theta}{2}-1=\frac{1}{8}\).
Thus, cos\((\frac{\theta}{2})\) is positive
Thus, \(cos\left[ \frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1 }{ 8 } \right) \right] =cos\left( \frac { \theta }{ 2 } \right) =\frac { 3 }{ 4 } \)

We know that sin^-1: [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus
\({ sin }^{ -1 }\left( sin\left( \frac { 5\pi }{ 6 } \right) \right) \)= \({ sin }^{ -1 }\left( sin\left( \frac { \pi }{ 6 } \right) \right) \) = \({ sin }^{ -1 }\left( sin\frac { \pi }{ 6 } \right) \), since \(\frac{\pi}{6}\)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)

We know that sin^-1:  [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus,
\({ sin }^{ -1 }\left( sin\left( -\frac { \pi }{ 3 } \right) \right) \) = -\(\frac{\pi}{3}\), since -\(\frac{\pi}{3}\) \(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)

 sin^-1 [sin5]
We know that \({ sin }^{ -1 }\left( sin5 \right) =0\quad if\quad \theta \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ e } \right] \)
Consiideerning the approximation \(\frac { \pi }{ 2 } =\frac { 11 }{ 7 } \)
\(5=5\times \frac { 11 }{ 7 } \times \frac { 7 }{ 11 } =5\times \frac { \pi }{ 2 } \times \frac { 7 }{ 11 } =\frac { 35\pi }{ 22 } \notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
But \(5-2\pi =5-2\times \frac { 22 }{ 7 } =5-\frac { 44 }{ 7 } =\frac { 35z44 }{ 7 } \)
= \(\frac { -11 }{ 7 } \varepsilon \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(5-2\pi \notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
\({ sin }^{ -1 }\left( sin5 \right) ={ sin }^{ -1 }\left( sin(5-2\pi ) \right) \)
= \(5-2\pi \quad \) 

 \({ tan }^{ -1 }\left( sin\left( -\frac { 5\pi }{ 2 } \right) \right) \)
\({ tan }^{ -1 }\left( sin\left( \frac { -5\pi }{ 2 } \right) \right) =x\)
\(\Rightarrow { tan }^{ -1 }\left( -1 \right) =x\)
\(\Rightarrow -1=tanx\)
\(\Rightarrow tan\ x=-1=-tan{ \frac { \pi }{ 4 } }\)
\(\Rightarrow tan\ x=tan\left( \frac { -\pi }{ 4 } \right) \)
\(\Rightarrow x=\frac { -\pi }{ 4 } \)\(\left[ \because \frac { -\pi }{ 4 } \epsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \) 
\(\therefore \ tan^-1\left( sin\left( \frac { -5\pi }{ 2 } \right) \right) =-\frac { \pi }{ 4 } \)

cosec^-1\((-\sqrt{2})\)
\(\Rightarrow -\sqrt { 2 } =cosex\theta \)
\(\Rightarrow sin\theta =\frac { -1 }{ \sqrt { 2 } } \)
\(\Rightarrow sin\theta =-sin\frac { \pi }{ 4 } \)
\(\Rightarrow sin\theta =sin\left( \frac { -\pi }{ 4 } \right) \)
\(\Rightarrow sin\theta =sin\left( \frac { -\pi }{ 4 } \right) \)
\(\Rightarrow \theta =-\frac { \pi }{ 4 } \)
\(\therefore\) \(\theta cosec^{ -1 }\left( -\sqrt { 2 } \right) =-\frac { \pi }{ 4 } \)

cot^-1 \((\sqrt{3})\)
Let \({ cot }^{ -1 }\left( \sqrt { 3 } \right) \)
\(\Rightarrow \sqrt { 3 } =cot\theta \)
\(\Rightarrow tan\theta =\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow tan\theta =tan\frac { \pi }{ 6 } \)
\(\Rightarrow \theta =\frac { \pi }{ 6 } \)

tan(tan^-1(0.2021))
= -0.2021[\(\because \) tan (tan^-1x) = x for any real number]

tan(tan^-1(1947))
= 1947

\(\tan^{-1}(tan(-\frac{\pi}{6}))\)
= \(-\frac { \pi }{ 6 } \)
Since \(-\frac { \pi }{ 6 } \epsilon \left( \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)

Let \(g(x)=\frac { 1 }{ 2 } { tan }^{ -1 }\left( 1-{ x }^{ 2 } \right) -\frac { \pi }{ 4 } \)
By the definition of tan^-1 x, it is a function with
the entire real line \(\left( -\infty ,\infty \right) \) as its domain.
\(\therefore \) Domain of \(g(x)=\frac { 1 }{ 2 } { tan }^{ -1 }\left( 1-{ x }^{ 2 } \right) -\frac { \pi }{ 4 } \)is R
\(\therefore \) Domain of g(x) is R.

\({ sin }^{ -1 }\left( sin\left( \frac { 5\pi }{ 4 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi +\frac { \pi }{ 4 } \right) \right) \) \(\because \frac { 5\pi }{ 4 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
= \({ sin }^{ -1 }\left( sin\left( -\frac { \pi }{ 4 } \right) \right) \)
= \( \frac {- \pi }{ 4 } \epsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) 

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 \((-2x)is2x=2\pi \Rightarrow =\frac { 2\pi }{ 2 } =\pi \)

 \(y=-sin\left( { \frac { 1 }{ 3 } x } \right) \)
 The amplitude sin x is 1
\(\Rightarrow \) amplitude of \(-sin\left( \frac { 1 }{ 3 } x \right) \) is also 1.
The period of \(-sin\left( \frac { 1 }{ 3 } x \right) \) is \(\frac { 1 }{ 3 } x=2\pi \Rightarrow x=6\pi \)

\({ sin }^{ -1 }\left( cos\pi \right) =x\)
Let \(\Rightarrow { sin }^{ -1 }\left( -1 \right) \left( cos\pi \right) =x\)
\(\Rightarrow -1=sinx\)
\(\Rightarrow sinx=-sin\frac { \pi }{ 2 } \)
\(\Rightarrow sinx=sin\left( \frac { -\pi }{ 2 } \right) \)
\(\Rightarrow x=\frac { -\pi }{ 2 } \)
\(\therefore { sin }^{ -1 }(cos\pi )=-\frac { \pi }{ 2 } \)

\(sin\left[ \frac { \pi }{ 3 } -si{ n }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \right] =sin\left[ \frac { \pi }{ 3 } -\left( -\frac { \pi }{ 6 } \right) \right] =sin\left( \frac { \pi }{ 2 } \right) =1\)

\({ tan }^{ -1 }\left( \sqrt { 3 } \right) -{ sec }^{ -1 }(-2)\)
Let \({ tan }^{ 1 }\left( \sqrt { 3 } \right) =x\Rightarrow \sqrt { 3 } =tanx\)
\(\Rightarrow tanx=tan\frac { \pi }{ 3 } \)
\(\Rightarrow x=\frac { \pi }{ 3 } \)
Let sec^-1(-2) = y
\(\Rightarrow -2=sec\ y\Rightarrow cos\ y=\frac { -1 }{ 2 } \Rightarrow cos\ y=-cos\left( \frac { \pi }{ 3 } \right) \)
\(\Rightarrow cos\ y=cos\left( \pi -\frac { \pi }{ 3 } \right) \)
\(\Rightarrow cos\ y=cos\left( \frac { 2\pi }{ 3 } \right) \Rightarrow y=3\frac { 2\pi }{ 3 } \)
\(\therefore { tan }^{ -1 }\left( \sqrt { 3 } \right) -{ sec }^{ -1 }\left( -2 \right) \)
= \(\frac { \pi }{ 3 } -\frac { 2\pi }{ 3 } \)
= \(\frac { \pi -2\pi }{ 3 } =-\frac { \pi }{ 3 } \)
\(\therefore\) \({ tan }^{ -1 }\left( \sqrt { 3 } \right) -{ sec }^{ -1 }(-2)\)= \(-\frac { \pi }{ 3 } \)

sec^-1\((\frac{2}{\sqrt3})\)
Let \({ sec }^{ -1 }\left( { \frac { 2 }{ \sqrt { 3 } } } \right) \)
\(\Rightarrow \frac { 2 }{ \sqrt { 3 } } =sec\theta \Rightarrow cos\theta =\frac { \sqrt { 3 } }{ 2 } \)
\(\Rightarrow cos\theta =cos\left( \frac { \pi }{ 6 } \right) \)
\(\Rightarrow \theta =\frac { \pi }{ 6 } \)
\(\therefore { sec }^{ -1 }\left( \frac { 2 }{ \sqrt { 3 } } \right) =\frac { \pi }{ 6 } \)

Let sec^-1\(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\theta \).
Then, sec\(\theta\) = \(-\frac{2}{\sqrt3}\) where \(\theta\in[0,\pi]\)\{\(\frac{\pi}{2}\)}.
Thus, cos \(\theta =-\frac{\sqrt{3}}{2}\).
Now, \(cos\frac { 5\pi }{ 6 } =cos\left( \pi -\frac { \pi }{ 6 } \right) =-cos\left( \frac { \pi }{ 6 } \right) =-\frac { \sqrt { 3 } }{ 2 } .\) 
Hence, Sec-¹ \(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\frac { 5\pi }{ 6 } \)

Let cosec⁻¹(−1) = y. Then, cosec y = −1
Since the range of principal value branch of y = cosec^-1x is\(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)\{0} and
cosec \((-\frac{\pi}{2})=-1,\) 
We have y = \(-\frac{\pi}{2}\).
Note that \(-\frac{\pi}{2}\in\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)\{0}.
Thus, the principal value of cosec⁻¹(−1) is −\(\frac{\pi}{2}\)

\(tan(tan^{-1}(\frac{7\pi}{4}))\)
= \(tan(tan^{-1}(\frac{7\pi}{4}))\)
= \(\frac { 7\pi }{ 4 } \)[\(\therefore\) [tan (tan^-1(x)- x for any real number]

\(tan^{-1}(tan\frac{5\pi}{4})\)
= \({ tan }^{ -1 }\left( tan\left( \pi +\frac { \pi }{ 4 } \right) \right) \)
= \({ tan }^{ -1 }\left( tan\frac { \pi }{ 4 } \right) \) \(\left[ \because tan\left( \pi +\theta =tan\theta \right) \right] \)
= \(\frac { \pi }{ 4 } \varepsilon \left( \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)

 Let \(f(x)={ tan }^{ -1 }\sqrt { 9-{ x }^{ 2 } } \)
\(\sqrt { 9-{ x }^{ 2 } } \varepsilon R\) but \(\sqrt { 9-{ x }^{ 2 } } \varepsilon R\)
\(\therefore \ 9-{ x }^{ 2 }\ge 0\)
\(\Rightarrow { x }^{ 2 }-9\ge 0\)
\(\Rightarrow (x+3)(x-3)\le 0\)
\(\Rightarrow \) Domain is [-3, 3]

Let tan⁻¹(\(\sqrt3\)) = y
Then, tan y =\(\sqrt3\)
Thus, y = \(\frac{\pi}{3}\)
Since \(\in(-\frac{\pi}{2},\frac{\pi}{2})\)
Thus the principal value of tan⁻¹(\(\sqrt3\)) = \(\frac{\pi}{3}\)

Let \(x={ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(\Rightarrow cosx=\frac { 1 }{ 2 } \)
\(\Rightarrow cosx=\frac { 1 }{ 2 } \)    \(\left[ \because \frac { \pi }{ 3 } \varepsilon \left[ 0,\pi \right] \right] \)
\(\Rightarrow x=\frac { \pi }{ 3 } \)
\(\therefore \) The principal value of \(cos^{ -1 }\left( \frac { 1 }{ 2 } \right) \) is \(\frac { \pi }{ 3 } \)

cos^-1(-x) = \(\pi\)-cos⁻¹(x)
Let cos^-1(-x) = \(\theta \)   ..(1)
\(\Rightarrow -x=cos\theta \)
\(\Rightarrow x=-cos\theta =cos\theta =cos\left( \pi -\theta \right) \)
\(\Rightarrow \pi -\theta ={ cos }^{ -1 }\left( x \right) \)
\(\Rightarrow \theta =\pi -{ cos }^{ -1 }x\) ...(2)
From (1) & (2) \({ cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }\left( x \right) \)
\({ cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }\left( x \right) \) is true.

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 \(\theta\)]

It is known that cos^-1 x : [-1, 1]\(\rightarrow\)[0, \(\pi\)] is given by
cos⁻¹x = y if and only if x = cos y for -1\(\le x\le1 and 0\le y \le\pi\) 
Thus, we have
cos^-1 \((-\frac{1}{\sqrt2})\) = \(\frac{3\pi}{4}\), since \(\frac{3\pi}{4}\)\(\in[0,\pi]\)cos\(\frac{3\pi}{4}\) = cos\((\pi=\frac{\pi}{4})=-cos \frac{\pi}{4}=-\frac{1}{\sqrt2}\)

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).

The amplitude of sin x is 1 [Max of sin x curve is 1]
\(\Rightarrow \) amplitude of sin 7x is also 1
If p is the period of the function,
then f(x+p) = f(x)
Since the period of sine function is \(2\pi \)
The period of sin is \(\frac { 2\pi }{ 7 } \) 
amplitude = 1

 We know that sin^-1: [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus,
 \({ Sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \)= \(\frac{\pi}{4}\), Since \(\frac{\pi}{4}\)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)and sin \(\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)

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

Let sin^-1 \(\left( -\frac { 1 }{ 2 } \right) \) = y. Then sin y = -\(\frac{1}{2}\)
The range of the principal value of sin^-1x is \(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) and hence, Let us find y \(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) Such that sin y = -\(\frac{1}{2}\). Clearly, y = -\(\frac{\pi}{6}\)
Thus, the principal value of sin^-1\(\left( -\frac { 1 }{ 2 } \right) \) is -\(\frac{\pi}{6}\). This corresponds to -30^o.

We know that 2tan^-1 x = tan^-1\((\frac{2x}{1-x^{2}}, -1
So, \(2{ tan }^{ -1 }\frac { 1 }{ 2 } ={ tan }^{ -1 }\frac { 2\left( \frac { 1 }{ 2 } \right) }{ 1-\left( \frac { 1 }{ 2 } \right) ^{ 2 } } ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
Hence, \(2{ tan }^{ -1 }\frac { 1 }{ 2 } +{ tan }^{ -1 }\frac { 1 }{ 7 } ={ tan }^{ -1 }\frac { 4 }{ 3 } +{ tan }^{ -1 }\frac { 1 }{ 7 } ={ tan }^{ -1 }\left( \frac { \frac { 4 }{ 3 } +\frac { 1 }{ 7 } }{ 1-\left( \frac { 4 }{ 3 } \right) \left( \frac { 1 }{ 7 } \right) } \right) ={ tan }^{ -1 }\left( \frac { 31 }{ 17 } \right) \)

Given g(x) = sin^-1 x + cos^-1x
From the definition of sin^-1x.
\(-1\le x\le 1\) ...(1)
Also from the definition of cos^-1x
\(-1\le x\le 1\)   .........(2)
\(\therefore \) From (1) & (2),
Domain ofg(x) = [-1, 1] U [-1, 1]
= [-1, 1]
Hence the domain of g(x) is [-1, 1].

\({ cos }^{ -1 }\left( cos\frac { \pi }{ 7 } cos\frac { \pi }{ 17 } -sin\frac { \pi }{ 17 } sin\frac { \pi }{ 17 } \right) .\)
\({ cos }^{ -1 }\left( cos\left( \frac { \pi }{ 7 } +\frac { \pi }{ 17 } \right) \right) \)
\(\left[ \therefore cosA\ cosB-sinA\ sinB=cos(A+B) \right] \)
= \({ cos }^{ -1 }\left( cos\left( \frac { 24\pi }{ 119 } \right) \right) \)    \(\left[ \therefore \frac { 24\pi }{ 119 } \varepsilon \left[ 0,\pi \right] \right] \)
= \(\frac { 24\pi }{ 119 } \) 

\({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }(-1)\)
Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =x\) and -1 = sin y
\(\Rightarrow cosx=\frac { 1 }{ 2 } =cos\frac { \pi }{ 3 } \)
\(x=\frac { \pi }{ 3 } \)
\(siny=-1=-sin\left( \frac { -\pi }{ 2 } \right) \)
= \(-sin\left( \frac { -\pi }{ 2 } \right) \)
\(\left[ \because sin\left( -\theta \right) =-sin\ \theta\ and\ \frac { \pi }{ 2 } \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\(\Rightarrow y=\frac { -\pi }{ 2 } \)
\(\therefore { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( -1 \right) =x+y\)
= \(\frac { \pi }{ 3 } -\frac { \pi }{ 2 } =\frac { 2\pi -3\pi }{ 6 } =-\frac { \pi }{ 6 } \)
\(\therefore { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }(-1)=-\frac {\pi }{ 6 } \)

\(-5\pi \le x\le 5\pi \) and cos x = -1
cos x = -1
\(cosx=cos\pi \)  \([\because cos\theta =cos\alpha \Rightarrow 2n\pi +\alpha ,n\varepsilon Z]\)
\(x=2n\pi +\pi ,n\varepsilon Z\quad \)
= \((2n+1)\pi ,n\varepsilon Z\)
\(\Rightarrow But-5\pi \le x\le 5\pi \)
Putting,\(=0,\pm 1,\pm 2\) and -3 we get
\(\Rightarrow x=-5\pi ,-3\pi ,-\pi ,3\pi ,5\pi \)
\(x=(2n+1)\ {\pi},n=0,\pm 1,\pm 2 \) and -3

Now, tan^-1 2x + tan^-1 3x = \({ tan }^{ -1 }\left( \frac { 2x+3x }{ 1-6x^{ 2 } } \right) \), since 6x² < 1.
So, \(tan^{ -1 }\left( \frac { 5x }{ 1-6x^{ 2 } } \right) =\frac { \pi }{ 4 } \), Which implies \(\frac { 5x }{ 1-6x^{ 2 } } =tan\frac { \pi }{ 4 } \) = 1
Thus, 1-6x² = 5x , which gives 6x²+5x−1 = 0
Hence, x = \(\frac{1}{6},-1\). But x = -1 does not satisfy 6x² < 1.
Observe that x = −1 makes the left side of the equation negative whereas the right side is a positive number.
Thus, x = −1 is not a solution.
Hence, x = \(\frac{1}{6}\) is the only solution of the equation.

Let sin⁻¹= \(\theta\). Then, x = sin\(\theta\) and x \(\neq\) 0, we get \(\theta \in\left[\frac{-\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right]\)
Hence, \(\cos \theta \geq 0 \text { and } \)\(\cos \theta=\sqrt{1-\sin ^{2} \theta}=\)\( \sqrt { 1-x^{ 2 } } \)
Thus, \(cot(sin^{ -1 }x)=cot\theta =\frac { \sqrt { 1-x^{ 2 } } }{ x } ,|x|\le1\ and\ x\neq0\)

Given that sin^-1x > cos^-1x. Note that -1\(\le x\le\)
Adding both sides by sin^-1x, we get
sin^-1 x + sin^-1 x > x cos^-1 x + sin^-1x, Which resucess to 2 sin^-1 x > \(\frac{\pi}{2}\)
As sine function increases in the interval \(\left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] \), we have x > sin\(\frac{\pi}{4} or x> \frac{1}{\sqrt2}\)
Thus, the solution set is the interval \(\left[ \frac { 1 }{ \sqrt { 2 } } ,1 \right] \)

We know that tan^-1 x + tan^-1 y = tan^-1\(\frac{x+y}{1-xy}\), xy < 1
Thus, tan⁻¹\(\frac{1}{2}+tan^{-1}\frac{1}{3}={ tan }^{ -1 }\frac { \frac { 1 }{ 2 } +\frac { 1 }{ 3 } }{ 1-\left( \frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 3 } \right) } ={ tan }^{ -1 }(1)=\frac { \pi }{ 4 } \)

Let sin⁻¹ x = \(\theta\).
Then, x = sin \(\theta\) and-1\(\le x\le1\)
Now, tan(sin^-1 x) = tan \(\theta\) = \(\frac{sin\theta}{cos\theta}=\frac{sin\theta}{\sqrt{1-sin^{2}\theta}}=\frac{x}{\sqrt{1-x2}},|x|<1\)

sin^-1x + 2cos^-1x = sin^-1x + cos^-1x = \(\frac{\pi}{2}\) + cos^-1x
We know that 0\(\le\)cos^-1x\(\le\pi\)
Thus, \(\frac{\pi}{2}+0\le cos^{-1}x\)x+\(\frac{\pi}{2}\le\pi+\frac{\pi}{2}\)
Thus, \(\frac{\pi}{2}\le sin^-{1}x+2cos^{-1}x\le \frac{3\pi}{2}\)

By definition, cot^-1x\(\in(0,\pi)\)
Therefore, cot^-1\((\frac{1}{7})\) = \(\theta\) implies \(\theta \in(0,\pi)\)
But cot^-1\((\frac{1}{7})\) = \(\theta\) implies cot \(\theta\) = \(\frac{1}{7}\) and hence tan \(\theta\) = 7 and \(\theta\) is acute.
Using tan \(\theta\) = \(\frac{7}{1}\), We construct a right triangle as shown .
Then, we have, cos \(\theta\) = \(\frac{1}{5\sqrt2}\).

Given \(\frac { \pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\pi \)
\((cos2\frac { \pi }{ 2 } <3x-1\))
\(\Rightarrow 0<3x-1<-1\)
\(0+1<3x<-1+1\)
\(1<3x<0\)
This inequality is true, only when 0 < x < \(\frac { 1 }{ 3 }\)

\(2{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
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 \(y=\frac { \pi }{ 6 } \)
\(\therefore \quad 2{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
= \(2\left( \frac { \pi }{ 3 } \right) +\frac { \pi }{ 6 } =\frac { 2\pi }{ 3 } +\frac { \pi }{ 6 } \)
= \(\frac { 4\pi +\pi }{ 6 } =\frac { 5\pi }{ 6 } \)

cos x = 0
\(\Rightarrow x=\left( 2n+1 \right) \frac { \pi }{ 2 } ,n\varepsilon Z\)
But \(-6\pi \le x\le 6\pi \)
\(\therefore \) n can take values from  
\(x=(2n+1)\frac { \pi }{ 2 } ,n=0\pm 1,\pm 2,...\pm 5\), -6

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 = \(\pm \frac { 1 }{ \sqrt { 2 } } \)

tan^-1 \(\left( \frac { 1-x }{ 1+x } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }\)x gives tan^-1 1-tan^-1 x = \(\frac{1}{2}\)tan^-1x. 
Therefore, \(\frac{\pi}{4}=\frac{3}{2}tan^{-1}\)x, which in turn reduces to tan⁻¹ = \(\frac{\pi}{6}\)
Thus, x = tan\(\frac{\pi}{6}=\frac{1}{\sqrt3}\)