Given 
\({ tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) ={ sin }^{ -1 }\left( sin\frac { \pi }{ 6 } \right) \)
\(\Rightarrow { tan }^{ -1 }x=\frac { \pi }{ 6 } \)
\(\Rightarrow x=tan\frac { \pi }{ 6 } =\frac { 1 }{ \sqrt { 3 } } \)
\(\therefore x=\frac { 1 }{ \sqrt { 3 } } \)

Let \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\) where \(0\le y\le \pi \)
Then \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\Rightarrow cosy=\frac { -1 }{ 2 } \)
\(\Rightarrow cos\ y=-cos\frac { \pi }{ 3 } =cos\left( \pi -\frac { \pi }{ 3 } \right) =cos\left( \frac { 2\pi }{ 3 } \right) \)
\(\Rightarrow y=\frac { 2\pi }{ 3 } \left[ \because \frac { 2\pi }{ 3 } \in \left[ 0,\pi \right] \right] \)
\(\therefore \) The principal value of \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) \frac { 2\pi }{ 3 } \)

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 \(\frac{2\pi}{3}\)

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

Since tan(tan^-1 x) = x, x ∈ R,
We have tan(tan^-1(2019)) = 2019

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\((\pi=\frac{\pi}{4})=-cos \frac{\pi}{4}=-\frac{1}{\sqrt2}\)

 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 \(\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)

cos (tan^-1x) = \(sin\left( { cot }^{ -1 }\frac { 3 }{ 4 } \right) \)
\(\Rightarrow sin\left( { tan }^{ -1 }\frac { 4 }{ 3 } \right) =sin\left( { sin }^{ -1 }\frac { 4 }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } \right) \)
\(\left[ \because { tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right] \)
\(\Rightarrow sin\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)
\(\Rightarrow cos\left( { tan }^{ -1 }x \right) =cos\left(- { tan }^{ -1 }x \right) =\frac { 4 }{ 5 } \)
\(\left[ \because cosx=cos(-x) \right] \)
\(\Rightarrow { tan }^{ -1 }x={- tan }^{ -1 }x={ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \)
\(\Rightarrow \tan ^{-1} x=-\tan ^{-1} x=\tan ^{-1} \frac{3}{4}\)
\(\left[ \because { cos }^{ -1 }x={ tan }^{ -1 }\sqrt { \frac { 1-{ x }^{ 2 } }{ x } } ;{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) ={ tan }^{ -1 }\sqrt { \frac { 1-\frac { 16 }{ 25 } }{ \frac { 4 }{ 5 } } } \right] \)
\(\Rightarrow x=\frac { 3 }{ 4 } ,\frac { 3 }{ 4 } \) 

Given sin x = 0
\(\Rightarrow\) sin x = sin 0
\(\Rightarrow\) \(​​x=n\pi ,n\varepsilon z\)
Since \(-10\pi \le x\le 10\pi \) n can take the values only from -10 to +10.
\(\therefore\) \(x=n\pi ,\) When \(n=0,\pm ,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 7,\pm 8,\pm 9,\pm 10\)

cot(cot^-1(+600)) = -600

Given \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right) =a\)
\(\Rightarrow \frac { \left( \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) }{ \left( \sqrt { 1+{ x }^{ 3 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) } \)
= \(\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \frac { 2\sqrt { 1+{ x }^{ 2 } } }{ -2\sqrt { 1-{ x }^{ 2 } } } =\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { 1-tan\alpha }{ 1+tan\alpha } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \left( \frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \right) ^{ 2 }\)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } =\frac { 1-sin2\alpha }{ 1+sin2\alpha } \Rightarrow { x }^{ 2 }=sin2\alpha \)