Given,

$${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right)$$

$$ \Rightarrow {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{x + {{2x} \over {1 - {x^2}}}} \over {1 - x\left( {{{2x} \over {1 - {x^2}}}} \right)}}} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\tan ^{ - 1}}\left( {{{x - {x^3} + 2x} \over {1 - {x^2} - 2{x^2}}}} \right)$$

$$\therefore$$ $$\,\,\,$$ $${\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{3x - {x^2}} \over {1 - 3{x^2}}}} \right)$$

$$ \Rightarrow y = {{3x - {x^3}} \over {1 - 3{x^2}}}$$

Given that, $$x,y,z\,\,$$ are in $$AP$$

So, $$\,\,\,$$ $$2y = x + y$$

Also given that,

$${\tan ^{ - 1}}x,{\tan ^{ - 1}}y\,\,$$ and $$\,\,\,{\tan ^{ - 1}}z\,\,$$ are in $$AP$$

So, $$2{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}z$$

$$ \Rightarrow {\tan ^{ - 1}}\left( {{{2y} \over {1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {{{x + 7} \over {1 - xz}}} \right)$$

$$ \Rightarrow {{2y} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}$$

$$ \Rightarrow {{x + z} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}$$

[ as $$\,\,\,\,$$ $$2y = x + z$$]

$$ \Rightarrow 1 - {y^2} = 1 - xz$$

$$ \Rightarrow $$ $${y^2} = xz$$

As we get $${y^2} = xz,$$ so, $$x,y,z$$ are in $$GP.$$

According to the question $$x,y,z$$ are $$AP.$$

$$x,y,z$$ both can be $$AP$$ as well as $$GP$$

when $$x=y=z.$$

Given,

$$Cot\left( {so{{\sec }^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$

$$ = \cot \left( {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$

$$ = cot\left( {{{\tan }^{ - 1}}{{{3 \over 4} + {2 \over 3}} \over {1 - {3 \over 4} - {2 \over 3}}}} \right)$$

$$ = \cot \left( {{{\tan }^{ - 1}}\left( {{{17} \over 6}} \right)} \right)$$

$$ = \cot \left( {{{\cot }^{ - 1}}{6 \over {17}}} \right)$$

$$ = {6 \over {17}}$$

Given sin^-1$$\left( {{x \over 5}} \right)$$ + cosec^-1$$\left( {{5 \over 4}} \right)$$ = $${\pi \over 2}$$

$$ \Rightarrow $$ sin^-1$$\left( {{x \over 5}} \right)$$ + sin^-1$$\left( {{4 \over 5}} \right)$$ = $${\pi \over 2}$$

$$ \Rightarrow $$ sin^-1$$\left( {{x \over 5}} \right)$$ = $${\pi \over 2}$$ - sin^-1$$\left( {{4 \over 5}} \right)$$

$$ \Rightarrow $$ sin^-1$$\left( {{x \over 5}} \right)$$ = cos^-1$$\left( {{4 \over 5}} \right)$$

$$ \Rightarrow $$ $${x \over 5}$$ = sin(cos^-1$$ {{4 \over 5}}$$)

$$ \Rightarrow $$ $${x \over 5}$$ = sin(sin^-1$$ {{3 \over 5}}$$)

$$ \Rightarrow $$ $${x \over 5}$$ = $${3 \over 5}$$

$$ \Rightarrow $$ x = 3