Let cos⁻¹x = \(\alpha\) and cos-1 y = \(\beta\).
Then, x = cos\(\alpha\) and y cos =\(\beta\)
cos^-1x + x + cos^-1 y + cos^-1x = \(\pi\) gives \(\alpha\)+\(\beta\) = \(\pi\) -cos^-1z.
Now, cos(\(\alpha\)+\(\beta\)) = cos\(\alpha\)cos\(\beta\)-sin\(\alpha\)sin\(\beta\) = xy-\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
-cos(cos^-1 z) = xy\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
so, \(-z=xy-\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \), Which gives -xy - z = -\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
Squaring on both sides and simplifying, we get x² + y² + z² + 2xyz = 1.

g(x) = 2sin⁻¹(2x−1)−\(\frac{\pi}{4}\)
From the definition of sin^-1x,
\(-1\le 2x-1\le 1\)
\(\Rightarrow 1+1\le 2x\le 1+1\)
\(\Rightarrow 0\le 2x\le 2\)
\(\Rightarrow 0\le x\le 1\)
\(\therefore \) Domain = [0, 1]

sin(cos⁻¹(1-x))
we know that \({ cos }^{ -1 }x={ sin }^{ -1 }\left( \sqrt { 1-{ x }^{ 2 } } \right) \text {if}\ 0\le x\le 1\)
\(\therefore { cos }^{ -1 }\left( 1-x \right) ={ sin }^{ 1 }\sqrt { 1-\left( 1-x \right) ^{ 2 } } \left[ \because 0\le x\le 1 \right] \)
= \({ sin }^{ -1 }\left( \sqrt { 1-\left( 1+{ x }^{ 2 }-2x \right) } \right) \)
= \({ sin }^{ -1 }\left( \sqrt { 1-1-{ x }^{ 2 }+2x } \right) ={ sin }^{ -1 }\left( \sqrt { 2x-{ x }^{ 2 } } \right) \)
\(\therefore sin\left( { cos }^{ -1 }\left( 1-x \right) \right) =sin\left( { sin }^{ -1 }\left( \sqrt { 2x-{ x }^{ 2 } } \right) \right) \)
= \(\sqrt { 2x-{ x }^{ 2 } } \)

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.

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