Mathematics M.C.Q (1 Marks)

Hence ${\sin ^{ - 1}}(x\sqrt {1 - x} - \sqrt x \,\sqrt {1 - {x^2}} )$

$ = {\sin ^{ - 1}}(\sin \theta \sqrt {1 - {{\sin }^2}\phi } - \sin \phi \sqrt {1 - {{\sin }^2}\theta } )$

$ = {\sin ^{ - 1}}(\sin \theta \cos \phi - \sin \phi \cos \theta ) = {\sin ^{ - 1}}\sin \,(\theta - \phi )$

$ = \theta - \phi = {\sin ^{ - 1}}(x) - {\sin ^{ - 1}}(\sqrt x )$.

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left[ {\frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right] = \frac{1}{2}\theta $    (Putting $x = \tan \theta )$

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left[ {\frac{{\tan \frac{\pi }{4} - \tan \theta }}{{1 + \tan \frac{\pi }{4}\tan \theta }}} \right] = \frac{\theta }{2}$

$ \Rightarrow \,\,{\tan ^{ - 1}}\tan \,\left( {\frac{\pi }{4} - \theta } \right) = \frac{\theta }{2}\,\, \Rightarrow \,\,\frac{\pi }{4} - \theta = \frac{\theta }{2}$

$ \Rightarrow \,\,\theta = \frac{\pi }{6} = {\tan ^{ - 1}}x\,\, $

$\Rightarrow \,\,x = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}$.

$ \Rightarrow \,\,\frac{{ - \pi }}{2} \le x - \pi \le \frac{\pi }{2}\,\, \Rightarrow \,\,\frac{{ - \pi }}{2} \le \pi - x \le \frac{\pi }{2}$

$ \Rightarrow \,\,{\sin ^{ - 1}}\{ \sin \,(\pi - x)\} = \pi - x$.

$ \Rightarrow \,\, - \pi \ge - x \ge - 2\pi \,\,\, \Rightarrow \,\,\pi \ge 2\pi - x \ge 0$

$ \Rightarrow \,\,{\cos ^{ - 1}}\,\left\{ {\cos \,(2\pi - x)} \right\} = 2\pi - x$.

$\Rightarrow \,0 < 10 - 3\pi < \frac{\pi }{2}$

$ \Rightarrow \,\,\frac{{ - \pi }}{2} < 3\pi - 10 < 0$

$ \Rightarrow \,\,{\sin ^{ - 1}}\left\{ {\sin \,(3\pi - 10)} \right\} = 3\pi - 10$.

Since $ - 1 \le x \le 0,$   therefore $\frac{{ - \pi }}{2} \le {\sin ^{ - 1}}x \le 0$ 

and so $\frac{{ - \pi }}{2} \le y \le 0$

We have $\cos y = \sqrt {1 - {{\sin }^2}y} $

$ \Rightarrow \,\,\cos y = \sqrt {1 - {x^2}} $, for $0 \le y \le \pi $   …..$(i)$

Now $ - \frac{\pi }{2} \le y \le 0\,\, \Rightarrow \,\,\frac{\pi }{2} \ge - y \ge 0$

$ \Rightarrow \,\,\cos \,\left( { - y} \right) = \sqrt {1 - {x^2}} $     {from $(i)$}

$ \Rightarrow \,\, - y = {\cos ^{ - 1}}\sqrt {1 - {x^2}} \,\, $

$\Rightarrow \,\,y = - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $.

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left[ {\frac{{x + y + z - xyz}}{{1 - xy - yz - xz}}} \right] = \frac{\pi }{2}$

$ \Rightarrow \,\,\left[ {\frac{{x + y + z - xyz}}{{1 - xy - yz - zx}}} \right] = \tan \frac{\pi }{2} = \frac{1}{0}$

Hence $xy + yz + zx - 1 = 0$.

Trick : $x = y = z = \frac{1}{{\sqrt 3 }},$ so that

${\tan ^{ - 1}}\frac{1}{{\sqrt 3 }} + {\tan ^{ - 1}}\frac{1}{{\sqrt 3 }} + {\tan ^{ - 1}}\frac{1}{{\sqrt 3 }} = \frac{\pi }{2}$

Obviously $ (d)$ holds for these values of  $x, y, z.$

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left[ {\frac{{\frac{{x - 1}}{{x + 2}} + \frac{{x + 1}}{{x + 2}}}}{{1 - \left( {\frac{{x - 1}}{{x + 2}}} \right)\,\left( {\frac{{x + 1}}{{x + 2}}} \right)}}} \right] = \frac{\pi }{4}$

$ \Rightarrow \,\,\left[ {\frac{{2x\,(x + 2)}}{{{x^2} + 4 + 4x - {x^2} + 1}}} \right] = \tan \frac{\pi }{4}$

==> $\,\,\frac{{2x\,(x + 2)}}{{4x + 5}} = \tan \frac{\pi }{4} = 1$

$ \Rightarrow \,\,2{x^2} + 4x = 4x + 5$

$ \Rightarrow \,\,x = \pm \sqrt {\frac{5}{2}} $.

$ \Rightarrow \,{\tan ^{ - 1\,}}\left( {\frac{{\frac{{a + x}}{a} + \frac{{a - x}}{a}}}{{1 - \frac{{a + x}}{a}.\,\frac{{a - x}}{a}}}} \right) = \frac{\pi }{6}$

$ \Rightarrow \,\,\frac{{2{a^2}}}{{{x^2}}} = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}\,\, $

$\Rightarrow \,\,{x^2} = 2\sqrt 3 {a^2}$.

$ \Rightarrow {\tan ^{ - 1}}(x - 1) + {\tan ^{ - 1}}(x) = {\tan ^{ - 1}}3x - {\tan ^{ - 1}}(x + 1)$

$ \Rightarrow {\tan ^{ - 1}}\left[ {\frac{{(x - 1) + x}}{{1 - (x - 1)(x)}}} \right] = {\tan ^{ - 1}}\left[ {\frac{{3x - (x + 1)}}{{1 + 3x(x + 1)}}} \right]$

$ \Rightarrow \frac{{2x - 1}}{{1 - {x^2} + x}} = \frac{{2x - 1}}{{1 + 3{x^2} + 3x}}$

$ \Rightarrow (1 - {x^2} + x)(2x - 1) = (1 + 3{x^2} + 3x)(2x - 1)$

On simplification $x = 0,\, \pm \frac{1}{2}.$

$ = 2\, {\tan ^{ - 1}}\left[ {\frac{{\frac{2}{5}}}{{1 - \frac{1}{{25}}}}} \right] - {\tan ^{ - 1}}\frac{1}{{70}} + {\tan ^{ - 1}}\frac{1}{{99}}$

$ = 2\, {\tan ^{ - 1}}\left( {\frac{5}{{12}}} \right) - {\tan ^{ - 1}}\frac{1}{{70}} + {\tan ^{ - 1}}\frac{1}{{99}}$

$ = {\tan ^{ - 1}}\left[ {\frac{{\frac{5}{6}}}{{1 - \frac{{25}}{{144}}}}} \right] - {\tan ^{ - 1}}\frac{1}{{70}} + {\tan ^{ - 1}}\frac{1}{{99}}$

$ = {\tan ^{ - 1}}\left( {\frac{{120}}{{119}}} \right) - {\tan ^{ - 1}}\frac{1}{{70}} + {\tan ^{ - 1}}\frac{1}{{99}}$

$ = {\tan ^{ - 1}}\left( {\frac{{120}}{{119}}} \right) + {\tan ^{ - 1}}\left[ {\frac{{\frac{1}{{99}} - \frac{1}{{70}}}}{{1 + \frac{1}{{99}}.\frac{1}{{70}}}}} \right]$

$ = {\tan ^{ - 1}}\left( {\frac{{120}}{{119}}} \right) + {\tan ^{ - 1}}\left( {\frac{{ - 29}}{{6931}}} \right)$

$ = {\tan ^{ - 1}}\frac{{120}}{{119}} - {\tan ^{ - 1}}\frac{{29}}{{6931}} = {\tan ^{ - 1}}\frac{{120}}{{119}} - {\tan ^{ - 1}}\frac{1}{{239}}$

$ = {\tan ^{ - 1}}\left[ {\frac{{\frac{{120}}{{119}} - \frac{1}{{239}}}}{{1 + \frac{{120}}{{119}} \times \frac{1}{{239}}}}} \right] $

$= {\tan ^{ - 1}}(1) = \frac{\pi }{4}$.

$ = {\tan ^{ - 1}}\left[ {\frac{{\frac{{xy}}{{zr}} + \frac{{yz}}{{xr}} + \frac{{xz}}{{yr}} - \frac{{xyz}}{{{r^3}}}}}{{1 - \left( {\frac{{{x^2} + {y^2} + {z^2}}}{{{r^2}}}} \right)}}} \right] = {\tan ^{ - 1}}\infty = \frac{\pi }{2}$.

$\Rightarrow y = {\tan ^{ - 1}}33 - {\tan ^{ - 1}}3$

$ = {\tan ^{ - 1}}\frac{{33 - 3}}{{1 + 99}} = {\tan ^{ - 1}}\frac{{30}}{{100}}$

==> $y = {\tan ^{ - 1}}(0.3)$.

(b) ${\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(y) + {\tan ^{ - 1}}(z) = \pi $

==> ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi - {\tan ^{ - 1}}z$

==> $\frac{{x + y}}{{1 - xy}} = - z \Rightarrow x + y = - z + xyz$

==> $x + y + z = xyz$

Dividing by $ xyz,$ we get

$\frac{1}{{yz}} + \frac{1}{{xz}} + \frac{1}{{xy}} = 1$.

Note: Students should remember this question as a formula.

Dividing by $xyz$, we get
$\frac{1}{{yz}} + \frac{1}{{xz}} + \frac{1}{{xy}} = 1$.
Note: Students should remember this question as a formula.

==> ${\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{1 - {{\cos }^2}x}}} \right)$$= {\tan ^{ - 1}}(2\,{\rm{cosec }}x)$

$\frac{{2\cos x}}{{{{\sin }^2}x}} = 2{\rm{cosec}}\,x \Rightarrow 2\cos x = 2\sin x$

or $\sin x = \cos x$

$ \Rightarrow x = \frac{\pi }{4}$.

{Since $2{\cos ^{ - 1}}x = {\cos ^{ - 1}}(2{x^2} - 1)$}

$ = \tan {\cos ^{ - 1}}\left( { - \frac{7}{{25}}} \right) = \tan {\tan ^{ - 1}}\left[ {\sqrt {\frac{{1 - \frac{{49}}{{625}}}}{{ - \frac{7}{{25}}}}} } \right] = - \frac{{24}}{7}$

Therefore $\tan \left( {2{{\cos }^{ - 1}}\frac{3}{5}} \right) = - \frac{{24}}{7}$.

Now, $\frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right)$

$ = \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = \frac{1}{2}{\cos ^{ - 1}}\cos 2\theta = \frac{{2\theta }}{2} = \theta = {\tan ^{ - 1}}\sqrt x $.

$\therefore$ $4{\tan ^{ - 1}}\frac{1}{5} = 2\,\left[ {2{{\tan }^{ - 1}}\frac{1}{5}} \right] = 2{\tan ^{ - 1}}\frac{{\frac{2}{5}}}{{1 - \frac{1}{{25}}}}$

$ = 2{\tan ^{ - 1}}\frac{{10}}{{24}} = {\tan ^{ - 1}}\frac{{\frac{{20}}{{24}}}}{{1 - \frac{{100}}{{576}}}} = {\tan ^{ - 1}}\frac{{120}}{{119}}$

So, $4{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}} = {\tan ^{ - 1}}\frac{{120}}{{119}} - {\tan ^{ - 1}}\frac{1}{{239}}$

$ = {\tan ^{ - 1}}\frac{{\frac{{120}}{{119}} - \frac{1}{{239}}}}{{1 + \frac{{120}}{{119}}.\frac{1}{{239}}}} = {\tan ^{ - 1}}\frac{{(120 \times 239) - 119}}{{(119 \times 239) + 120}}$

==> ${\tan ^{ - 1}}1 = \frac{\pi }{4}$.

${\tan ^{ - 1}}\left( {\frac{c}{{a + b}}} \right) + {\tan ^{ - 1}}\left( {\frac{b}{{a + c}}} \right)$

$ = {\tan ^{ - 1}}\left[ {\frac{{\frac{c}{{a + b}} + \frac{b}{{a + c}}}}{{1 - \left( {\frac{c}{{a + b}}} \right)\left( {\frac{b}{{a + c}}} \right)}}} \right]$

$ = {\tan ^{ - 1}}\left[ {\frac{{ca + {c^2} + ab + {b^2}}}{{{a^2} + ab + ca + bc - bc}}} \right]$

$ = {\tan ^{ - 1}}\left[ {\frac{{{a^2} + ab + ca}}{{{a^2} + ab + ca}}} \right]$

$ = {\tan ^{ - 1}}(1) = \frac{\pi }{4}$.

${\sin ^{ - 1}}2x = {\sin ^{ - 1}}\left( {x\sqrt {\left( {1 - \frac{3}{4}} \right)} - \frac{{\sqrt 3 }}{2}\sqrt {1 - {x^2}} } \right)$

$2x = \left( {\frac{x}{2} - \frac{{\sqrt 3 }}{2}\sqrt {1 - {x^2}} } \right)$

$\frac{{\sqrt 3 }}{2}\sqrt {1 - {x^2}} = \frac{x}{2} - 2x = \frac{{ - 3x}}{2}$

$\frac{{3(1 - {x^2})}}{4} = \frac{{9{x^2}}}{4}$

$ \Rightarrow 3 - 3{x^2} = 9{x^2}$

==> ${x^2} = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}$.

Let ${\cot ^{ - 1}}\frac{1}{2} = \phi \Rightarrow \frac{1}{2} = \cot \phi $

$ \Rightarrow \sin \phi = \frac{1}{{\sqrt {1 + {{\cot }^2}\phi } }} = \frac{2}{{\sqrt 5 }}$

Let ${\cos ^{ - 1}}x = \theta \Rightarrow \sec \theta = \frac{1}{x} $

$\Rightarrow \tan \theta = \sqrt {{{\sec }^2}\theta - 1} $

$ \Rightarrow \tan \theta = \sqrt {\frac{1}{{{x^2}}} - 1} $

$\Rightarrow \tan \theta = \frac{{\sqrt {1 - {x^2}} }}{x}$

So, $\tan \{ {\cos ^{ - 1}}(x)\} = \sin \left( {{{\cot }^{ - 1}}\frac{1}{2}} \right)$

$ \Rightarrow \tan \left( {{{\tan }^{ - 1}}\frac{{\sqrt {1 - {x^2}} }}{x}} \right) = \sin \left( {{{\sin }^{ - 1}}\frac{2}{{\sqrt 5 }}} \right)$

$ \Rightarrow \frac{{\sqrt {1 - {x^2}} }}{x} = \frac{2}{{\sqrt 5 }} \Rightarrow \sqrt {(1 - {x^2})5} = 2x$

Squaring both sides, we get $x = \pm \frac{{\sqrt 5 }}{3}$.

$ = {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + \cos 2\theta } + \sqrt {1 - \cos 2\theta } }}{{\sqrt {1 + \cos 2\theta } - \sqrt {1 - \cos 2\theta } }}} \right]$

(Putting ${x^2} = \cos 2\theta \Rightarrow \theta = \frac{1}{2}{\cos ^{ - 1}}{x^2})$

$= {\tan ^{ - 1}}\left[ {\frac{{\sqrt 2 \cos \theta + \sqrt 2 \sin \theta }}{{\sqrt 2 \cos \theta - \sqrt 2 \sin \theta }}} \right]$

$ = {\tan ^{ - 1}}\left[ {\frac{{1 + \tan \theta }}{{1 - \tan \theta }}} \right] = {\tan ^{ - 1}}\left[ {\frac{{\tan \frac{\pi }{4} + \tan \theta }}{{1 - \tan \frac{\pi }{4}\tan \theta }}} \right]$

$ = {\tan ^{ - 1}}\tan \left( {\frac{\pi }{4} + \theta } \right) = \frac{\pi }{4} + \theta = \frac{\pi }{4} + \frac{1}{2}{\cos ^{ - 1}}{x^2}$.

==> ${\cos ^{ - 1}}x + ({\cos ^{ - 1}}x + {\sin ^{ - 1}}x) = \frac{{11\pi }}{6}$

==> ${\cos ^{ - 1}}x + \frac{\pi }{2} = \frac{{11\pi }}{6}$

$ \Rightarrow {\cos ^{ - 1}}x = 4\pi /3$

which is not possible as ${\cos ^{ - 1}}x \in [0,\,\pi ]$.

==> ${\tan ^{ - 1}}\left[ {\frac{{x + y + z - xyz}}{{1 - xy - yz - zx}}} \right] = \pi $

==> $x + y + z - xyz = 0$

==> $x + y + z\,\, = xyz$.

$\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}})$

$=\tan ^{-1}(\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}})$

$=\tan ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)$

$=\tan ^{-1}\left(\tan \frac{x}{2}\right)$

$=\frac{x}{2}$

$=\tan ^{-1}\left(\frac{1-\left(\frac{\sin x}{\cos x}\right)}{1+\left(\frac{\sin x}{\cos x}\right)}\right)$

$=\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$

$=\tan ^{-1}(1)-\tan ^{-1}(\tan x)$ $\left[\because \frac{-y}{x y}=\tan ^{-1} x-\tan ^{-1} y\right]$

$=\frac{\pi}{4}-x$

Let, $x=a \sin \theta \Rightarrow \frac{x}{a}=\sin \theta \Rightarrow \sin ^{-1}\left(\frac{x}{a}\right)$

$\therefore \tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}$

$=\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right)$

$=\tan ^{-1}\left(\frac{a \sin \theta}{a \sqrt{1-\sin ^{2} \theta}}\right)$

$=\tan ^{-1}\left(\frac{a \sin \theta}{a \cos \theta}\right)$

$=\tan ^{-1}(\tan \theta)=\theta=\sin ^{-1} \frac{x}{a}$

Let $x=a \tan \theta \Rightarrow \frac{x}{a}=\tan \theta$ $\Rightarrow \theta=\tan ^{-1}\left(\frac{x}{a}\right)$

$\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)$

$=\tan ^{-1}\left(\frac{3 a^{2} \cdot a \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot a^{2} \tan ^{2} \theta}\right)$

$=\tan ^{-1}\left(\frac{3 a^{3} \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a^{3} \tan ^{2} \theta}\right)$

$=\tan ^{-1}(\tan 3 \theta)$

$=3 \theta$

$=3 \tan ^{-1} \frac{x}{a}$

Then, $\sin x=\frac{1}{2}=\sin \left(\frac{\pi}{6}\right)$

$\therefore \sin ^{-1} \frac{1}{2}=\frac{\pi}{6}$

$\therefore \tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]$

$=\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]$

$=\tan ^{-1}\left[2 \cos \frac{\pi}{3}\right]$

$=\tan ^{-1}\left[2 \times \frac{1}{2}\right]$

$=\tan ^{-1} 1=\frac{\pi}{4}$

Then, $\theta=\tan ^{-1} x$

$\therefore \sin ^{-1} \frac{2 x}{1+x^{2}}=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$

$=\sin ^{-1}(\sin 2 \theta)$

$=2 \theta$

$=2 \tan ^{-1} x$

Let $y=\tan \theta .$ Then, $\theta=\tan ^{-1} y$

$\therefore \cos ^{-1}\left(\frac{1-y^{2}}{1+y^{2}}\right)$

$=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$

$=\cos ^{-1}(\cos 2 \theta)$

$=2 \theta=2 \tan ^{-1} y$

$\therefore \tan \frac{1}{2}\left[\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-y^{2}}{1+y^{2}}\right)\right]$

$=\tan \frac{1}{2}\left[2 \tan ^{-1} x+2 \tan ^{-1} y\right]$

$=\tan \left[\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

$=\frac{x+y}{1-x y}$

$=\tan ^{-1} \frac{a}{b}-\tan ^{-1}(\tan x)=\tan ^{-1} \frac{a}{b}-x$

$=\tan ^{-1}\left[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\left(\frac{x}{y}\right)\left(\frac{x-y}{x+y}\right)}\right]$

$=\tan ^{-1}\left[\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}\right]$

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

$=\tan ^{-1}\left(\frac{x^{2}+y^{2}}{x^{2}+y^{2}}\right)=\tan ^{-1} 1=\frac{\pi}{4}$

$ \Rightarrow \,\,\tan \theta = \sqrt {{{\sec }^2}\theta - 1} = \sqrt {\frac{1}{{{x^2}}} - 1} = \frac{1}{x}\sqrt {1 - {x^2}} $

Now $\sin \,\,{\cot ^{ - 1}}\tan \theta = \sin \,\,{\cot ^{ - 1}}\,\left( {\frac{1}{x}\sqrt {1 - {x^2}} } \right)$

Again, putting $x = \sin \theta $

$\sin \,\,{\cot ^{ - 1}}\left( {\frac{1}{x}\sqrt {1 - {x^2}} } \right) = \sin \,\,{\cot ^{ - 1}}\left( {\frac{{\sqrt {1 - {{\sin }^2}\theta } }}{{\sin \theta }}} \right)$

$ = \sin \,\,{\cot ^{ - 1}}\,(\cot \theta ) = \sin \theta = x$.

${\tan ^{ - 1}}\,\left( {\frac{{1 - x}}{{1 + x}}} \right) = {\tan ^{ - 1}}1 - {\tan ^{ - 1}}x = \frac{\pi }{4} - {\tan ^{ - 1}}x$

Since $0 \le x \le 1\,\, \Rightarrow \,\,0 \le {\tan ^{ - 1}}x \le \frac{\pi }{4}$

$ \Rightarrow \,\,0 \ge - {\tan ^{ - 1}}x \ge \frac{{ - \pi }}{4} $

$\Rightarrow \,\,\frac{\pi }{4} \ge \frac{\pi }{4} - {\tan ^{ - 1}}x \ge 0$

$ \Rightarrow \,\,\frac{\pi }{4} \ge {\tan ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) \ge 0$.

==> ${\sin ^{ - 1}}x = {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}}$

$ \Rightarrow \frac{{2x}}{{1 + {x^2}}} = x$ ==> ${x^3} - x = 0$

==> $x(x + 1)(x - 1) = 0$ ==> $x = \left\{ { - 1,\,\,1,\,\,0} \right\}$.

Put ${\sin ^{ - 1}}x = \alpha $, ${\sin ^{ - 1}}y = \beta ,$ ${\sin ^{ - 1}}z = \gamma $

$\therefore $ $\alpha + \beta + \gamma = \frac{\pi }{2}$,   (Given)

or $\alpha + \beta = \frac{\pi }{2} - \gamma $ or $\cos (\alpha + \beta ) = \cos \left( {\frac{\pi }{2} - \gamma } \right)$

$\cos \alpha \cos \beta - \sin \alpha \sin \beta = \sin \gamma $ …..$(i)$

and, we have $\sin \alpha = x$ ==> $\cos \alpha = \sqrt {1 - {x^2}} $

Similarly, $\cos \beta = \sqrt {1 - {y^2}} $

$\therefore $ From equation $(i),$ we get $\sqrt {1 - {x^2}} .\sqrt {1 - {y^2}} = xy + z$

Squaring both sides, we have ${x^2} + {y^2} + {z^2} + 2xyz = 1$.

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

$ = \sin \,\left[ {{{\cot }^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}} \right] = \sin \,\left[ {{{\sin }^{ - 1}}\sqrt {\frac{{1 + {x^2}}}{{2 + {x^2}}}} } \right]$

$ = \sqrt {\frac{{1 + {x^2}}}{{2 + {x^2}}}} $.

Since $\frac{{ - \pi }}{2} \le {\tan ^{ - 1}}x \le \frac{\pi }{2} $

$\Rightarrow 0 \le \frac{\pi }{2} + {\tan ^{ - 1}}x \le \pi $

$\therefore$ $K = \pi ,k = 0$.

$\because 0 \le {\cos ^{ - 1}}x \le \pi $

$\therefore 0 \le {\cos ^{ - 1}}y \le \pi $ and $0 \le {\cos ^{ - 1}}z \le \pi $

Here ${\cos ^{ - 1}}x = {\cos ^{ - 1}}y = {\cos ^{ - 1}}z = \pi $

$ \Rightarrow x = y = z = \cos \pi = - 1$

$\therefore$ $xy + yz + zx$$ = ( - 1)( - 1) + ( - 1)( - 1) + ( - 1)( - 1)$

$ = 1 + 1 + 1 = 3$.

$ = \sin \left[ {2{{\tan }^{ - 1}}\frac{3}{4}} \right] = \sin {\sin ^{ - 1}}\left( {\frac{{2 \times (3/4)}}{{1 + (9/16)}}} \right)$

$ = \frac{3}{2} \times \frac{{16}}{{25}} = \frac{{24}}{{25}}$ 

$\left( {\because 2{{\tan }^{ - 1}}x = {{\sin }^{ - 1}}\frac{{2x}}{{1 + {x^2}}}} \right)$

Putting $x = \tan \theta $

$3{\sin ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) - 4{\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ + 2{\tan ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right) = \frac{\pi }{3}$

==> $3{\sin ^{ - 1}}(\sin 2\theta ) - 4{\cos ^{ - 1}}(\cos 2\theta )$

$ + 2{\tan ^{ - 1}}(\tan 2\theta ) = \frac{\pi }{3}$

==> $3(2\theta ) - 4(2\theta ) + 2(2\theta ) = \frac{\pi }{3} \Rightarrow 6\theta - 8\theta + 4\theta = \frac{\pi }{3}$

==> $\theta = \frac{\pi }{6} \Rightarrow {\tan ^{ - 1}}x = \frac{\pi }{6} \Rightarrow x = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}$

$= \sin \left[ {{{\tan }^{ - 1}}\frac{{2/3}}{{1 - 1/9}}} \right] + \cos \,[{\tan ^{ - 1}}(2\sqrt 2 )]$

$ = \sin [{\tan ^{ - 1}}3/4] + \cos \,[{\tan ^{ - 1}}2\sqrt 2 ]$

$ = \frac{3}{5} + \frac{1}{3} = \frac{{14}}{{15}}$.

==> ${\cos ^{ - 1}}2x = - \pi - {\cos ^{ - 1}}x$

$ \Rightarrow 2x = \cos (\pi + {\cos ^{ - 1}}x)$

==> $2x = \cos \pi (\cos {\cos ^{ - 1}}x) - \sin \pi \sin ({\cos ^{ - 1}}x)$

$2x = - x \Rightarrow x = 0$

But $x = 0$ does not satisfy the given equation.

No solution will exist.

But ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$

$\therefore$ ${\sin ^{ - 1}}x = \frac{\pi }{3}$ and ${\cos ^{ - 1}}x = \frac{\pi }{6}$

==> $x = \frac{{\sqrt 3 }}{2}$ is the unique solution.

==> ${\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{1 - {{\cos }^2}x}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{{{{\sin }^2}x}}} \right)$

$ \Rightarrow \frac{{2\cos x}}{{{{\sin }^2}x}} = \frac{1}{{{{\sin }^2}x}}$

==> $2\cos x = 1$

$ \Rightarrow x = \frac{\pi }{3}$.

$=\sin \left[\sin ^{-1}\left(\frac{4}{5} \sqrt{1-\frac{4}{5}}+\frac{2}{\sqrt{5}}\right)\right]$

$=\sin \left[\sin ^{-1}\left(\frac{10}{5 \sqrt{5}}\right)\right]$

$=\frac{2}{\sqrt{5}}$

$=\tan ^{-1}(1)+\tan ^{-1}(2)+\tan ^{-1}(3)$

$=\frac{\pi}{4}+\left[\frac{\pi}{2}-\cot ^{-1} 2\right]+\frac{\pi}{2}-\cot ^{-1} 3$

$=\frac{\pi}{4}+\frac{\pi}{2}+\frac{\pi}{2}-\left[\cot ^{-1}(2)+\cot ^{-1}(3)\right]$

$=\frac{5 \pi}{4}-\left[\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)\right]$

$=\frac{5 \pi}{4}-\left[\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\right)\right]$

$=\frac{5 \pi}{4}-\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)$

$=\frac{5 \pi}{4}-\tan ^{-1}(1)$

$=\frac{5 \pi}{4}-\frac{\pi}{4}$

$=\pi$