Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =\theta \Rightarrow cos\theta =\frac { 1 }{ 2 } \)
\(\Rightarrow sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 1 }{ 4 } } =\sqrt { \frac { 3 }{ 4 } } =\frac { \sqrt { 3 } }{ 2 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right) =\frac { \sqrt { 3 } }{ 2 } \)

LHS = \(2{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) +{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { 2 }{ 3 } +\cfrac { 2 }{ 3 } }{ 1-\left( \frac { 2 }{ 3 } \right) \left( \frac { 2 }{ 3 } \right) } \right) ={ tan }^{ -1 }\left( \frac { \frac { 4 }{ 3 } }{ \frac { 9-2 }{ 9 } } \right) \)
= \({ tan }^{ -1 }\left( \frac { 4 }{ 3 } \times \frac { 9 }{ 7 } \right) ={ tan }^{ -1 }\left( \frac { 12 }{ 7 } \right) \)
= RHS
Hence proved

L.H.S = \({ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 13 } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 1 }{ 7 } +\frac { 1 }{ 3 } }{ 1-\left( \frac { 1 }{ 7 } \right) \left( \frac { 1 }{ 13 } \right) } \right) ={ tan }^{ -1 }\left( \cfrac { \frac { 13+7 }{ 91 } }{ \frac { 91-1 }{ 91 } } \right) \)

\(=\tan ^{-1}\left(\frac{\frac{20}{91}}{\frac {90}{91}}\right)=\tan ^{-1}\left(\frac{20}{\not 91} \times \frac{\not 91}{90}\right) \)
\(=\tan ^{-1}\left(\frac{\not 20^{2}}{\not 90^{9}}\right)=\tan ^{-1}\left(\frac{2}{9}\right)=\mathrm{RHS} \)
Hence proved.

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

Given
\({ cot }^{ -1 }\left( \frac { 1 }{ 7 } \right) =0\Rightarrow \theta =\frac { 1 }{ 7 } \)
\(\Rightarrow tan\theta =7\)
\(\Rightarrow sec\theta =\sqrt { 1+{ tan }^{ 2 }\theta } =\sqrt { 1+{ 7 }^{ 2 } } =\sqrt { 50 } =5\sqrt { 2 } \)
\(\Rightarrow cos\theta ={ \frac { 1 }{ 5\sqrt { 2 } } }\)

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, \(cos\left[ \frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1 }{ 8 } \right) \right] =cos\left( \frac { \theta }{ 2 } \right) =\frac { 3 }{ 4 } \)

tan(tan^-1(1947))
= 1947

sin x = -1
\(\Rightarrow sinx=sin\left( \frac { -\pi }{ 2 } \right) \Rightarrow x=\frac { -\pi }{ 2 } ,\frac { 3\pi }{ 2 } ,\frac { 7\pi }{ 2 } ...\)
\(\Rightarrow x=(4n-1)\frac { \pi }{ 2 } ,n\varepsilon Z.\)
\(\Rightarrow x=(4n-1)\frac { \pi }{ 2 } \)
n takes the values \(0,\pm 1,\pm 2,\pm 3and\pm 4\)
since when n = -4, \(x=\frac { -17\pi }{ 2 } <-8\pi \)

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

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 \(x={ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(\Rightarrow cosx=\frac { 1 }{ 2 } \)
\(\Rightarrow cosx=\frac { 1 }{ 2 } \)    \(\left[ \because \frac { \pi }{ 3 } \varepsilon \left[ 0,\pi \right] \right] \)
\(\Rightarrow x=\frac { \pi }{ 3 } \)
\(\therefore \) 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 \(\theta\)]

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)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \).

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 \(\theta\) = \(\frac{1}{5\sqrt2}\).