$$\eqalign{ & {\sin ^{ - 1}}\left( {\frac{x}{5}} \right) + {\text{cose}}{{\text{c}}^{ - 1}}\left( {\frac{5}{4}} \right) = \frac{\pi }{2} \cr & \Rightarrow \,\,{\sin ^{ - 1}}\left( {\frac{x}{5}} \right) = \frac{\pi }{2} - {\text{cose}}{{\text{c}}^{ - 1}}\left( {\frac{5}{4}} \right) \cr & \Rightarrow \,\,{\sin ^{ - 1}}\left( {\frac{x}{5}} \right) = \frac{\pi }{2} - {\sin ^{ - 1}}\left( {\frac{4}{5}} \right) \cr & \left[ {\because \,\,{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x = \frac{\pi }{2}} \right] \cr & \Rightarrow \,\,{\sin ^{ - 1}}\left( {\frac{x}{5}} \right) = {\text{co}}{{\text{s}}^{ - 1}}\left( {\frac{4}{5}} \right) \cr & \Rightarrow \,\,{\sin ^{ - 1}}\frac{x}{5} = {\sin ^{ - 1}}\frac{3}{5} \cr & \Rightarrow \,\,\frac{x}{5} = \frac{3}{5} \cr & \Rightarrow \,\,x = 3 \cr} $$

$$\eqalign{ & \sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\cos \left( {{{\cot }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right\}}^2} - 1} \right]^{\frac{1}{2}}} \cr & = \sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\cos \left( {{{\cos }^{ - 1}}\left( {\frac{x}{{\sqrt {1 + {x^2}} }}} \right)} \right) + \sin \left( {{{\sin }^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right)} \right)} \right\}}^2} - 1} \right]^{\frac{1}{2}}} \cr & = \sqrt {1 + {x^2}} \left[ {{{\left\{ {x.\frac{x}{{\sqrt {1 + {x^2}} }} + \frac{1}{{\sqrt {1 + {x^2}} }}} \right\}}^2} - 1} \right] \cr & = \sqrt {1 + {x^2}} \left[ {{{\left( {\sqrt {1 + {x^2}} } \right)}^2} - 1} \right] \cr & = x\sqrt {1 + {x^2}} \cr} $$

If $$x = - 1,$$  L.H.S $$ = \frac{\pi }{2},$$  R.H.S. $$ = 2 \times \left( { - \frac{\pi }{2}} \right).$$   So, the formula does not hold.
If $$x < - 1,$$  the angle on the L.H.S. is in the second quadrant while the angle on the R.H.S. is 2 $$ \times $$ (angle in the fourth quadrant), which cannot be equal.
If $$x > 1,$$  the angle on the L.H.S. is in the second quadrant while the angle on the R.H.S. is 2 $$ \times $$ (angle in the first quadrant) and these two may be equal.
If $$- 1 < x < 0,$$   the angle on the L.H.S. is positive and that on the R.H.S. is negative and the two cannot be equal.

The given trigonometric expression is :
$$\eqalign{ & \cos \left[ {{{\tan }^{ - 1}}\left\{ {\tan \left( {\frac{{15\pi }}{4}} \right)} \right\}} \right] \cr & = \cos \left[ {{{\tan }^{ - 1}}\left\{ {\tan \left( {4\pi - \frac{\pi }{4}} \right)} \right\}} \right] \cr & = \cos \left[ {{{\tan }^{ - 1}}\left\{ { - \tan \frac{\pi }{4}} \right\}} \right] = \cos \left[ {{{\tan }^{ - 1}}\tan \left( {\frac{{ - \pi }}{4}} \right)} \right] \cr} $$
Since, $${\tan ^{ - 1}}\theta $$  is defined for $$\frac{{ - \pi }}{2} < \theta < \frac{{ - \pi }}{2}$$
$$\eqalign{ & = \cos \left( {\frac{{ - \pi }}{4}} \right) \cr & \cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}\,\,\left[ {{\text{since, }}\cos \left( { - \theta } \right) = \cos \theta } \right] \cr} $$

$$\eqalign{ & {\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2} \cr & \Rightarrow \,\,{\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2} - {\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) \cr & \Rightarrow \,\,{\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = {\cos ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) \cr & \Rightarrow \,\,{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - ..... = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - ..... \cr} $$
On both sides we have G.P. of infinite terms.
$$\eqalign{ & \therefore \,\,\frac{{{x^2}}}{{1 - \left( {\frac{{ - {x^2}}}{2}} \right)}} = \frac{x}{{1 - \left( { - \frac{x}{2}} \right)}} \cr & \Rightarrow \,\,\frac{{2{x^2}}}{{2 + {x^2}}} = \frac{{2x}}{{2 + x}} \cr & \Rightarrow \,\,2x + {x^3} = 2{x^2} + {x^3} \cr & \Rightarrow \,\,x\left( {x - 1} \right) = 0 \cr & \Rightarrow \,\,x = 0,1\,\,{\text{but }}0 < \left| x \right| < \sqrt 2 \cr & \Rightarrow \,\,x = 1. \cr} $$

$$\eqalign{ & {\text{Let}}\frac{1}{2}{\cos ^{ - 1}}\frac{a}{b} = \theta ;{\text{ then}} \cr & {\cos ^{ - 1}}\frac{a}{b} = 2\theta ; \cr & \Rightarrow \cos 2\theta = \frac{a}{b}{\text{ then expression}} \cr & = \tan \left( {\frac{\pi }{4} + \theta } \right) + \tan \left( {\frac{\pi }{4} - \theta } \right) \cr & = \frac{{1 + \tan \theta }}{{1 - \tan \theta }} + \frac{{1 - \tan \theta }}{{1 + \tan \theta }} \cr & = \frac{{{{\left( {1 + \tan \theta } \right)}^2} + {{\left( {1 - \tan \theta } \right)}^2}}}{{\left( {1 - \tan \theta } \right)\left( {1 + \tan \theta } \right)}} \cr & = \frac{{2 + 2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }} = \frac{{2\left( {1 + {{\tan }^2}\theta } \right)}}{{1 - {{\tan }^2}\theta }} \cr & = \frac{{2\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}}{{\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)}} \cr & = \frac{2}{{\cos 2\theta }} = \frac{2}{{\frac{a}{b}}} = \frac{{2b}}{a} \cr} $$

$$\eqalign{ & {\text{Let, }}\,{\sin ^{ - 1}}a = x \cr & \therefore a = \sin x \cr & {\sin ^{ - 1}}b = y \cr & \therefore b = \sin y;{\sin ^{ - 1}}c = z \cr & \therefore c = \sin z \cr & \therefore a\sqrt {1 - {a^2}} + b\sqrt {1 - {b^2}} + c\sqrt {1 - {c^2}} \cr & = \sin x\cos x + \sin y\cos y + \sin z\cos z \cr & = \left( {\frac{1}{2}} \right)\left( {\sin 2x + \sin 2y + \sin 2z} \right) = \left( {\frac{1}{2}} \right)\left( {4\sin x\sin y\sin z} \right) \cr & = 2\sin x\sin y\sin z = 2abc \cr} $$

$$\eqalign{ & {\tan ^{ - 1}}x + {\tan ^{ - 1}}\frac{1}{y} = {\tan ^{ - 1}}3\,\,{\text{or,}}\,{\tan ^{ - 1}}\frac{1}{y} = {\tan ^{ - 1}}3 - {\tan ^{ - 1}}x\,\,\,{\text{or, }}{\tan ^{ - 1}}\frac{1}{y} = {\tan ^{ - 1}}\frac{{3 - x}}{{1 + 3x}} \cr & \Rightarrow \,\,y = \frac{{1 + 3x}}{{3 - x}}. \cr} $$
As $$x, y$$  are positive integers, $$x = 1, 2$$  and corresponding $$y = 2, 7.$$
∴ solutions are $$\left( {x,y} \right) = \left( {1,2} \right),\left( {2,7} \right).$$

$$\eqalign{ & {\text{We have,}} \cr & 3\,{\tan ^{ - 1}}\frac{1}{2} + 2\,{\tan ^{ - 1}}\frac{1}{5} + {\sin ^{ - 1}}\frac{{142}}{{65\sqrt 5 }} \cr & = 2\left( {{{\tan }^{ - 1}}\frac{1}{2} + {{\tan }^{ - 1}}\frac{1}{5}} \right) + {\tan ^{ - 1}}\frac{1}{2} + {\tan ^{ - 1}}\frac{{142}}{{31}} \cr & = 2\,{\tan ^{ - 1}}\frac{7}{9} + \pi + {\tan ^{ - 1}}\frac{{\frac{1}{2} + \frac{{142}}{{31}}}}{{1 - \left( {\frac{1}{2}} \right)\left( {\frac{{142}}{{31}}} \right)}} \cr & = {\tan ^{ - 1}}\frac{{\frac{{14}}{9}}}{{1 - \frac{{49}}{{81}}}} + \pi - {\tan ^{ - 1}}\frac{{315}}{{80}} \cr & = \pi - {\tan ^{ - 1}}\frac{{63}}{{16}} + {\tan ^{ - 1}}\frac{{63}}{{16}} = \pi \cr} $$

$$\eqalign{ & {\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right) + {\cos ^{ - 1}}\left( {\frac{3}{{4x}}} \right) = \frac{\pi }{2};\left( {x > \frac{3}{4}} \right) \cr & \Rightarrow \,\,{\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right)\frac{\pi }{2} - {\cos ^{ - 1}}\left( {\frac{3}{{4x}}} \right) \cr & \Rightarrow \,\,{\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right) = {\sin ^{ - 1}}\left( {\frac{3}{{4x}}} \right) \cr & \left[ {\because \,\,{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x = \frac{\pi }{2}} \right] \cr & {\text{Put }}{\sin ^{ - 1}}\left( {\frac{3}{{4x}}} \right) = \theta \cr & \Rightarrow \,\,\sin \theta = \frac{3}{{4x}} \cr & \Rightarrow \,\,\cos \theta = \sqrt {1 - {{\sin }^2}\theta } \cr & = \sqrt {1 - \frac{9}{{16{x^2}}}} \cr & \Rightarrow \,\,\theta = {\cos ^{ - 1}}\left( {\frac{{\sqrt {16{x^2} - 9} }}{{4x}}} \right) \cr & \therefore \,\,{\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right) \cr & = {\cos ^{ - 1}}\left( {\frac{{\sqrt {16{x^2} - 9} }}{{4x}}} \right) \cr & \Rightarrow \,\,\frac{2}{{3x}} = \frac{{\sqrt {16{x^2} - 9} }}{{4x}} \cr & \Rightarrow \,\,{x^2} = \frac{{64 + 81}}{{9 \times 16}} \cr & \Rightarrow \,\,x = \pm \sqrt {\frac{{145}}{{144}}} \cr & \Rightarrow \,\,x = \frac{{\sqrt {145} }}{{12}}\,\,\left( {\because \,\,x > \frac{3}{4}} \right) \cr} $$