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 \(\frac{2\pi}{3}\)

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

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 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 \(\left( \frac { \sqrt { 3 } }{ 2 } \right) is\frac { \pi }{ 6 } \)

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

\(cos\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) -{ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \right) \)
\(\Rightarrow \frac { 4 }{ 5 } =sinx\)

\(\therefore cosx=\frac { adj }{ hyp } =\frac { 3 }{ 5 } \)
Let \({ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) =y\)
\(\Rightarrow tany=\frac { 3 }{ 4 } \)

\(\Rightarrow siny=\frac { opp }{ hyp } =\frac { 3 }{ 5 } \ cosy=\frac { adj }{ hup } =\frac { 4 }{ 5 } \)
\(\therefore cos\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) -{ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \)
= cos (x - y) = cos x cos y + sin x sin y
= \(\frac { 3 }{ 5 } .\frac { 4 }{ 5 } +\frac { 4 }{ 5 } .\frac { 3 }{ 5 } \)
= \(\frac { 12 }{ 25 } +\frac { 12 }{ 25 } =\frac { 24 }{ 25 } \)
= \(\frac { 24 }{ 25 } \)

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 \(\left[ \frac { 1 }{ \sqrt { 2 } } ,1 \right] \)

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 < \(\frac { 1 }{ 3 }\)