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

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

Let \({ tan }^{ -1 }\left( \frac { -1 }{ \sqrt { 3 } } \right) =y\) where \(\frac{-\pi}{2}<y<\frac{\pi}{2}\)
\(\Rightarrow tan\ y=\frac { -1 }{ 3 } =-tan\frac { \pi }{ 6 } =tan\left( \frac { -\pi }{ 6 } \right) \)
\(y=\frac { -\pi }{ 6 } \) \(\left[ \because \frac { -\pi }{ 6 } \epsilon \left( \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
The principal value of \({ tan }^{ -1 }\left( \frac { -1 }{ \sqrt { 3 } } \right) =6\)

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

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

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 \(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\frac { 5\pi }{ 6 } \)

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

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

\({ sec }^{ -1 }\left( sec\left( \frac { 5\pi }{ 3 } \right) \right) \)
Note that \(\frac{5\pi}{3}\) is not in [0, \(\pi\)]\{\(\frac{\pi}{2}\)}, the principal range of sec^-1 x.
we write \(\frac { 5\pi }{ 3 } =2\pi -\frac { \pi }{ 3 } \).
Now, sec\(\left( \frac { 5\pi }{ 3 } \right) =sec\left( 2\pi -\frac { \pi }{ 3 } \right) =sec\left( \frac { \pi }{ 3 } \right) and\frac { \pi }{ 3 } \in [0,\pi ]\)\{\(\frac{\pi}{2}\)}
Hence, sec^-1\(\left( sec\left( \frac { 5\pi }{ 3 } \right) \right) ={ sec }^{ -1 }\left( sec\left( \frac { \pi }{ 3 } \right) \right) =\frac { \pi }{ 3 } \)

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

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, \(\frac{\pi}{2}\le sin^-{1}x+2cos^{-1}x\le \frac{3\pi}{2}\)

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

We know that \({ tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \)
\(\therefore LHS={ tan }^{ -1 }(x)+{ tan }^{ -1 }(y)+{ tan }^{ -1 }(z)\)
= \({ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) +{ tan }^{ -1 }\left( z \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x+y }{ 1-xy } +z }{ 1-z\left( \frac { x+y }{ 1-xy } \right) } \right) \) by(1)
= \({ tan }^{ -1 }\left( \frac { \frac { x+y+z(1-xy) }{ 1-xy } }{ \frac { (1-xy)-z(x+y) }{ 1-xy } } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x+y+z-xyz }{ 1-xy } }{ \frac { 1-xy-zx-zy }{ 1-xy } } \right) \)

= \({ tan }^{ -1 }\left( \frac { x+y+z-xyz }{ 1-xy-yz-zx } \right) =RHS\)
Hence proved.