Mathematics M.C.Q (1 Marks)

$\frac{1}{x} \in(-\infty,-2] \cup[0, \infty)$

$x \in\left[-\frac{1}{2}, 0\right) \cup(0, \infty)$

$x \in\left[-\frac{1}{2}, \infty\right)-\{0\}$

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

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

$\Rightarrow 2 \cos ^{-1} \mathrm{x}=\frac{\pi}{2}-\frac{2 \mathrm{a}}{\pi}$

$\Rightarrow \cos ^{-1}\left(2 \mathrm{x}^{2}-1\right)=\frac{\pi}{2}-\frac{2 \mathrm{a}}{\pi}$

$\Rightarrow 2 \mathrm{x}^{2}-1=\cos \left(\frac{\pi}{2}-\frac{2 \mathrm{a}}{\pi}\right)$

Taking tangent both sides :-

$\frac{(x+1)+(x-1)}{1-\left(x^{2}-1\right)}=\frac{8}{31}$

$\Rightarrow \frac{2 x}{2-x^{2}}=\frac{8}{31}$

$\Rightarrow 4 x^{2}+31 x-8=0$

$\Rightarrow x =-8, \frac{1}{4}$

But, if $x=\frac{1}{4}$

$\tan ^{-1}(x+1) \in\left(0, \frac{\pi}{2}\right)$

$\& \cot ^{-1}\left(\frac{1}{x-1}\right) \in\left(\frac{\pi}{2}, \pi\right)$

$\Rightarrow LHS >\frac{\pi}{2} \& RHS <\frac{\pi}{2}$

(Not possible)

Hence, $x=-8$

$\sin ^{-1}\left(\frac{3 x}{5} \sqrt{1-\frac{16 x^{2}}{25}}+\frac{4 x}{5} \sqrt{1-\frac{9 x^{2}}{25}}\right)=\sin ^{-1} x$

$\frac{3 x}{5} \sqrt{1-\frac{16 x^{2}}{25}}+\frac{4 x}{5} \sqrt{1-\frac{9 x^{2}}{25}}=x$

$x=0,3 \sqrt{25-16 x^{2}}+4 \sqrt{25-9 x^{2}}=25$

$4 \sqrt{25-9 x^{2}}=25-3 \sqrt{25-16 x^{2}}$ squaring we get

$16\left(25-9 x^{2}\right)=625+9\left(25-16 x^{2}\right)-150 \sqrt{25-16 x^{2}}$

$400=625+225-150 \sqrt{25-16 x^{2}}$

$\sqrt{25-16 x^{2}}=3 \Rightarrow 25-16 x^{2}=9$

$\Rightarrow x^{2}=1$

Put $x=0,1,-1$ in the original equation We see that all values satisfy the original equation.

Number of solution $=3$

$\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}$

Now, $\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]$ is defined if

$-1 \leq x^{2}+\frac{1}{3}<2 \Rightarrow \frac{-4}{3} \leq x^{2}<\frac{5}{3}$

$\Rightarrow 0 \leq x^{2}<\frac{5}{3} ......(1)$

and $\cos ^{-1}\left[ x ^{2}-\frac{2}{3}\right]$ is defined if

$-1 \leq x^{2}-\frac{2}{3}<2 \Rightarrow \frac{-1}{3} \leq x^{2}<\frac{8}{3}$

$\Rightarrow 0 \leq x^{2}<\frac{8}{3}......(2)$

So, form $(1)$ and $(2)$ we can conclude

$0 \leq x^{2}<\frac{5}{3}$

Case $- I$ if $0 \leq x ^{2}<\frac{2}{3}$

$\sin ^{-1}(0)+\cos ^{-1}(-1)=x^{2}$

$\Rightarrow x +\pi= x ^{2}$

$\Rightarrow x^{2}=\pi$

but $\pi \notin\left[0, \frac{2}{3}\right)$

$\Rightarrow$ No value of $'x'$

Case $- II$ if $\frac{2}{3} \leq x^{2}<\frac{5}{3}$

$\sin ^{-1}(1)+\cos ^{-1}(0)= x ^{2}$

$\Rightarrow \frac{\pi}{2}+\frac{\pi}{2}= x ^{2}$

$\Rightarrow x^{2}=\pi$

but $\pi \notin\left[\frac{2}{3}, \frac{5}{3}\right)$

$\Rightarrow$ No value of $'x'$

So, number of solutions of the equation is $zero.$

So, $a+b=\frac{\pi}{2 r}$

$\cos \left(\frac{\pi c }{ a + b }\right)=\cos \left(\frac{\pi \tan ^{-1} y }{\frac{\pi}{2 r } r }\right)$

$=\cos \left(2 \tan ^{-1} y \right),$ let $\tan ^{-1} y =\theta$

$=\cos (2 \theta)$

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

$\Rightarrow \frac{a+b}{1-a b}=1$

$a+b=1-a b$

$(a+1)(b+1)=2$

Now $\left[ a -\frac{ a ^{2}}{2}+\frac{ a ^{3}}{3}+\ldots\right]+\left[ b -\frac{ b ^{2}}{2}+\frac{ b ^{3}}{3}+\ldots\right]$

$=\log _{e}(1+a)+\log _{e}(1+b)$

$\left(\because\right.$ expansion of $\left.\log _{ e }(1+ x )\right)$

$=\log _{e}[(1+a)(1+b)]$

$=\log _{e} 2$

$\sum_{r=1}^{50} \tan ^{-1}(2 r+1)-\tan ^{-1}(2 r-1)$

$\tan ^{-1}(101)-\tan ^{-1} 1 \Rightarrow \tan ^{-1} \frac{50}{51}$

$g(x) \in[1, \sqrt{2}]$ for $x \in[0, \pi / 2]$

$f(x)=\tan ^{-1}(\sin x+\cos x) \in\left[\frac{\pi}{4}, \tan ^{-1} \sqrt{2}\right]$

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

$-1 \leq \frac{x-1}{x+1} \leq 1 \Rightarrow 0 \leq x<\infty...(1)$

$-1 \leq \frac{3 x^{2}+x-1}{(x-1)^{2}} \leq 1 \Rightarrow x \in\left[\frac{-1}{4}, \frac{1}{2}\right] \cup\{0\}...(2)$

$(1) \,\&\,(2)$

$\Rightarrow \text { Domain }=\left[\frac{1}{4}, \frac{1}{2}\right] \cup\{0\}$

For equation to be defined,

$x^{3}+x \geq 0$

$\Rightarrow x^{2}+x+1 \geq 1$

$\therefore$ Only possibility that the equation is defined

$x^{2}+x=0 \Rightarrow x=0 ; x=-1$

None of these values satisfy

$\therefore$ No of roots $=0$

$=\sum_{n=1}^{100} \tan ^{-1}\left(\frac{2}{4 n^{2}}\right)$

$=\sum_{n=1}^{100} \tan ^{-1}\left(\frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)}\right)$

$=\sum_{n=1}^{100} \tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1)$

$=\tan ^{-1} 201-\tan ^{-1} 1$

$=\tan ^{-1}\left(\frac{200}{202}\right)$

$\therefore \cot ^{-1}(\alpha)=\cot ^{-1}\left(\frac{202}{200}\right)$

$\alpha=1.01$

$=2 \pi-\left(\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{16}{63}\right)\right)$

$=2 \pi-\left(\tan ^{-1}\left(\frac{63}{16}\right)+\tan ^{-1}\left(\frac{16}{63}\right)\right)$

$=2 \pi-\frac{\pi}{2}=\frac{3 \pi}{2}$

$S=\tan ^{-1}\left(\frac{2-1}{1+1.2}\right)+\tan ^{-1}\left(\frac{3-2}{1+2 \times 3}\right)+\tan ^{-1}$

$\left(\frac{4-3}{1+3 \times 4}\right)+\ldots+\tan ^{-1}\left(\frac{11-10}{1+10 \times 11}\right)$

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

$\left(\tan ^{-1} 4-\tan ^{-1} 3\right)+\ldots . .+\left(\tan ^{-1}(11)-\tan ^{-1}(10)\right)$

$\tan ( S )=\frac{11-1}{1+11 \times 1}=\frac{10}{12}=\frac{5}{6}$

For domain :

$-1 \leq \frac{|x|+5}{x^{2}+1} \leq 1$

since $|x|+5 \& x^{2}+1$ is always positive

So $\frac{|x|+5}{x^{2}+1} \geq 0 \forall x \in R$

So for domain :

$\frac{|x|+5}{x^{2}+1} \leq 1$

$\Rightarrow|x|+5 \leq x^{2}+1$

$\Rightarrow 0 \leq x^{2}-|x|-4$

$\Rightarrow 0 \leq\left(|x|-\frac{1+\sqrt{17}}{2}\right)\left(|x|-\frac{1-\sqrt{17}}{2}\right)$

$\Rightarrow|x| \geq \frac{1+\sqrt{17}}{2}$ or $|x| \leq \frac{1-\sqrt{17}}{2} \quad($ Rejected $)$

$\Rightarrow x \in\left(-\infty,-\frac{1+\sqrt{17}}{2}\right] \cup\left[\frac{1+\sqrt{17}}{2}, \infty\right)$

So, $a=\frac{1+\sqrt{17}}{2}$

$ = \cot \left[ {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + {n^2} + n} \right)} } \right]$

$ = \cot \left[ {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {\frac{1}{{1 + {n^2} + n}}} \right)} } \right]$

$ = \cot \left[ {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}1} } \right]$

$ = \cot \left[ {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right]$

$ = \cot \left( {{{\tan }^{ - 1}}\frac{{19}}{{21}}} \right)$

$ \Rightarrow \frac{{21}}{{19}}$

$ \Rightarrow {\cot ^{ - 1}}\left( x \right) \in \left( { - \infty ,2} \right) \cup \left( {5,\infty } \right)$

Put $0 < {\cot ^{ - 1}}\left( x \right) < \pi $

$ \Rightarrow {\cot ^{ - 1}}\left( x \right) \Rightarrow \left( {0,2} \right)$

$ \Rightarrow x \in \left( {\cot 2,\infty } \right)$

${\sin ^{ - 1}}\left( {x\sqrt {1 - {y^2}}  - y\sqrt {1 - {x^2}} } \right)$

$ = {\sin ^1}\left( {\frac{{33}}{{65}}} \right) = {\cos ^{ - 1}}\left( {\frac{{56}}{{65}}} \right) = \frac{\pi }{2} - {\sin ^{ - 1}}\left( {\frac{{56}}{{65}}} \right)$

${\cos ^{ - 1}}\left( {\frac{2}{{3x}} \times \frac{3}{{4x}} - \sqrt {1 - \frac{4}{{9{x^2}}}} \sqrt {1 - \frac{9}{{16{x^2}}}} } \right) = \frac{\pi }{2}$

$ \Rightarrow \frac{1}{{2{x^2}}} = \frac{{\sqrt {9{x^2} - 4} \sqrt {16{x^2} - 9} }}{{12{x^2}}}$

$ \Rightarrow 6 = \sqrt {9{x^2} - 4} \sqrt {16{x^2} - 9} $

Square both side

$36 = 144{x^4} - 81{x^2} - 64{x^2} + 36$

$ \Rightarrow 144{x^4} = 145{x^2}$

$ \Rightarrow {x^4} = \frac{{145{x^2}}}{{144}}\,\,\,\,\,\,\,\,\,\, \Rightarrow x =  \pm \frac{{\sqrt {145} }}{{12}},0$

$\therefore x > \frac{3}{4}$ hence $x = \frac{{\sqrt {145} }}{{12}}$

Taking tangent on both side, we get $\frac{{2x + 3x}}{{1 - 6{x^2}}} = 1$

$ \Rightarrow 6{x^2} + 5x - 1 = 0 \Rightarrow \left( {x + 1} \right)\left( {6x - 1} \right) = 0$

$ \Rightarrow x = \frac{1}{6}$  {$-1$is rejected as it does not satisfies the given equation}

Hence number of element in $S$ is one.

$ \Rightarrow \tan \alpha  = \frac{4}{3}$

$ \Rightarrow \tan \left( {\alpha  - \beta } \right) = \frac{{\frac{4}{3} - \frac{1}{3}}}{{1 + \frac{4}{3},\frac{1}{3}}} = \frac{9}{{13}}$

$ \Rightarrow \sin \left( {\alpha  - \beta } \right) = \frac{9}{{5\sqrt {10} }}$

$ \Rightarrow \alpha  - \beta  = {\sin ^{ - 1}}\left( {\frac{9}{{5\sqrt {10} }}} \right)$

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

$ \Rightarrow x \times \frac{y}{2} + \sqrt {1 - {x^2}} \sqrt {1 - \frac{{{y^2}}}{4}}  = \cos \alpha $

$ \Rightarrow {\left( {\cos \alpha  - \frac{{xy}}{2}} \right)^2} = \left( {1 - {x^2}} \right)\left( {1 - \frac{{{y^2}}}{4}} \right)$

${x^2} + \frac{{{y^2}}}{4} - xy\cos \alpha  = 1 - {\cos ^2}\alpha  = {\sin ^2}\alpha $

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

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

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

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

$ = \frac{\pi }{4} + \frac{1}{2}{\cos ^{ - 1}}{x^2}$

$\cot \lambda  = 1 + x$

$\tan \beta  = x$

$\sin \lambda  = \cos \beta $

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

${x^2} + 2x + 2 = {x^2} + 1$

$\boxed{x =  - 1/2}$

$|x|<\frac{1}{\sqrt{3}}$

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

$\Rightarrow \quad \tan ^{-1} y=\tan ^{-1} x+2 \tan ^{-1} x$

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

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

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

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

$ \Rightarrow f\left( x \right) = \pi $

$ \Rightarrow f\left( 5 \right) = \pi $

${\tan ^{ - 1}}\left[ {\cot \frac{{43\pi }}{4}} \right]$

$ = {\tan ^{ - 1}}\left[ {\cot \left( {10\pi  + \frac{{43\pi }}{4}} \right)} \right]$

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

                          [$\because $ $\cot \left( {2n\pi  + \theta } \right) = \cot \theta $]

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

$ = \frac{\pi }{2} - \frac{{3\pi }}{4} = \frac{{2\pi  - 3\pi }}{4} = \frac{{ - \pi }}{4}$

$ \Rightarrow  - \frac{{3\pi }}{4} \le \left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right) \le \frac{\pi }{4}$

$0 \le {\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)^2} \le \frac{9}{{16}}{\pi ^2}\,\,\,\,\,\,\,.....\left( 1 \right)$

Statement $II$ is true 

${\left( {{{\sin }^{ - 1}}x} \right)^3} + {\left( {{{\cos }^{ - 1}}x} \right)^3} = a{\pi ^3}$

$\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right)\left[ {{{\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right)}^2} - 3{{\sin }^{ - 1}}x{{\cos }^{ - 1}}x} \right] = a{\pi ^3}$

$ \Rightarrow \frac{{{\pi ^2}}}{4} - 3{\sin ^{ - 1}}x{\cos ^{ - 1}}x = 2a{\pi ^2}$

$ \Rightarrow {\sin ^{ - 1}}x\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right) = \frac{{{\pi ^2}}}{{12}}\left( {1 - 8a} \right)$

$ \Rightarrow {\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)^2} = \frac{{{\pi ^2}}}{{12}}\left( {1 - 8a} \right) + \frac{{{\pi ^2}}}{{16}}$

$ \Rightarrow {\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)^2} = \frac{{{\pi ^2}}}{{48}}\left( {32a - 1} \right)$

Putting this value in equation $(1)$

$0 \le \frac{{{\pi ^2}}}{{48}}\left( {32a - 1} \right) \le \frac{9}{{16}}{\pi ^2}$

$ \Rightarrow 0 \le 32a - 1 \le 27$

$\frac{1}{{32}} \le a \le \frac{7}{8}$

Statement - $I$ is also true.

$ \Rightarrow \cos e{c^2}\left( {{{\cot }^{ - 1}}\left( {1 + x} \right)} \right) = {\sec ^2}\left( {{{\tan }^{ - 1}}x} \right)$

$ \Rightarrow 1 + {\left[ {\cot \left( {{{\cot }^{ - 1}}\left( {1 + x} \right)} \right)} \right]^2}$ $ = 1 + {\left[ {\sec \left( {{{\tan }^{ - 1}}x} \right)} \right]^2}$

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

$ \Rightarrow {\sin ^{ - 1}}x > \frac{\pi }{2} - {\sin ^{ - 1}}x$

$ \Rightarrow 2{\sin ^{ - 1}}x > \frac{\pi }{2} \Rightarrow {\sin ^{ - 1}}x > \frac{\pi }{4}$

$ \Rightarrow x > \sin \frac{\pi }{2} \Rightarrow x > \frac{1}{{\sqrt 2 }}$

MNaximum value of ${\sin ^{ - 1}}x$ is $\frac{\pi }{2}$

So, maximum value of $x$ is $1$. So,

$x \in \left( {\frac{1}{{\sqrt 2 }},1} \right)$.

${\tan ^{ - 1}}\frac{1}{{1 + 2}} + {\tan ^{ - 1}}\frac{1}{{1 + 2 \times 3}} + {\tan ^{ - 1}}\frac{1}{{1 + 3 \times 4}} + ... + $

    ${\tan ^{ - 1}}\frac{1}{{1 + \left( {n - 1} \right)n}} + {\tan ^{ - 1}}\frac{1}{{1 + n\left( {n + 1} \right)}} + .... + $

          ${\tan ^{ - 1}}\frac{1}{{1 + \left( {n + 19} \right)\left( {n + 20} \right)}} = {\tan ^{ - 1}}\frac{{n + 19}}{{n + 21}}$

$ \Rightarrow {\tan ^{ - 1}}\frac{{n + 1}}{{n + 1}} + {\tan ^{ - 1}}\frac{1}{{1 + n\left( {n + 1} \right)}} + {\tan ^{ - 1}}\frac{1}{{1 + n\left( {n + 1} \right)\left( {n + 2} \right)}}$

          $ + .... + {\tan ^{ - 1}}\frac{1}{{1 + \left( {n + 19} \right)\left( {n + 20} \right)}} = {\tan ^{ - 1}}\frac{{n + 19}}{{n + 21}}$

${\tan ^{ - 1}}\frac{1}{{1 + n\left( {n + 1} \right)}} + {\tan ^{ - 1}}\frac{1}{{1 + n\left( {n + 1} \right)\left( {n + 2} \right)}} + ... + $

${\tan ^{ - 1}}\frac{1}{{1 + \left( {n + 19} \right)\left( {n + 20} \right)}} = {\tan ^{ - 1}}\frac{{n + 19}}{{n + 21}} - {\tan ^{ - 1}}\frac{{n - 1}}{{n + 1}}$

         ${\tan ^{ - 1}}\left( {\frac{1}{{{n^2} + n + 1}}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{{{n^2} + 3n + 3}}} \right) + .... + $

                ${\tan ^{ - 1}}\frac{1}{{1 + \left( {n + 19} \right)\left( {n + 20} \right)}}$

$ = {\tan ^{ - 1}}\left( {\frac{{\frac{{n + 19}}{{n + 21}} - \frac{{n - 1}}{{n + 1}}}}{{1 + \frac{{n + 19}}{{n + 21}} \times \frac{{n - 1}}{{n + 1}}}}} \right)$

$ = {\tan ^{ - 1}}\frac{{20}}{{{n^2} + 20n + 1}} = S$

$\therefore {\tan ^{ - 1}}S = \frac{{20}}{{{n^2} + 20n + 1}}$

$\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\pi+\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3}$

$2 \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=-\frac{\pi}{3}$

$y^2-6 \sqrt{3} y-9=0 \Rightarrow y=3 \sqrt{3}-6(\because y \in(-3,0))$

$\text { Case-I : } y \in(0,3)$

$2 \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3} \Rightarrow \sqrt{3} y^2+6 y-9 \sqrt{3}=0$

$y=\sqrt{3} \text { or } y=-3 \sqrt{3} \text { (rejected) }$

$\text { sum }=\sqrt{3}+3 \sqrt{3}-6=4 \sqrt{3}-6$

$|\cos x|=\tan ^{-1} \tan x$

No. of solutions $=3$

$\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\pi+\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3}$

$2 \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=-\frac{\pi}{3}$

$y^2-6 \sqrt{3} y-9=0 \Rightarrow y=3 \sqrt{3}-6(\because y \in(-3,0))$

$\text { Case-I : } y \in(0,3)$

$2 \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3} \Rightarrow \sqrt{3} y^2+6 y-9 \sqrt{3}=0$

$y=\sqrt{3} \text { or } y=-3 \sqrt{3} \text { (rejected) }$

$\text { sum }=\sqrt{3}+3 \sqrt{3}-6=4 \sqrt{3}-6$

$\sin ^{-1}\left(\frac{2 \sqrt{2} \pi}{2+\pi^2}\right)=\sin ^{-1}\left(\frac{2 \times \frac{\pi}{\sqrt{2}}}{1+\left(\frac{\pi}{\sqrt{2}}\right)^2}\right)$

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

$\left(\text { As, } \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=\pi-2 \tan ^{-1} x, x \geq 1\right)$

$\text { and } \tan ^{-1} \frac{\sqrt{2}}{\pi}=\cot ^{-1}\left(\frac{\pi}{\sqrt{2}}\right)$

$\therefore \text { Expression }=\frac{3}{2}\left(\tan ^{-1} \frac{\pi}{\sqrt{2}}\right)+\frac{1}{4}\left(\pi-2 \tan ^{-1} \frac{\pi}{\sqrt{2}}\right)+\cot ^{-1}\left(\frac{\pi}{\sqrt{2}}\right)$

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

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

$=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3 \pi}{4}$

$=2.35 \text { or } 2.36$

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

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

$\text { (A) } S_{10}(x)=\tan ^{-1} \frac{10 x}{1+11 x^2}=\frac{\pi}{2}-\tan ^{-1}$$\left(\frac{1+11 x^2}{10 x}\right)(x>0)$

$(B)$ $\lim _{n \rightarrow \infty} \cot \left(S_n(x)\right)=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}+\left(1+\frac{1}{n}\right) x^2}{x}=x(x>0)$

$(C)$ $S_3(x)=\tan ^{-1} \frac{3 x}{1+4 x^2}=\frac{\pi}{4} \Rightarrow 4 x^2-3 x+1=0 \Rightarrow x \notin R$

$(D)$ $\tan \left( S _{ n }( x )\right)=\frac{ nx }{1+( n +1) x ^2} ; \forall n \geq 1 ; x >0$

We need to check the validity of $\frac{ nx }{1+( n +1) x ^2} \leq \frac{1}{2} \forall n \geq 1 ; x >0 ; n \in N$

$\Rightarrow 2 nx \leq( n +1) x ^2+1$

$\Rightarrow( n +1) x ^2-2 nx +1 \geq 0 \forall n \geq 1 ; x >0 ; n \in N$

Discriminant of $y=(n+1) x^2-2 n x+1$ is

$D=4 n^2-4(n+1) \text { and } n \in N$

$D >0$ for $n \geq 2 \Rightarrow \exists$ some $x >0$

for which $y <0$ as both roots of $y =0$ will be positive.

$y =( n +1) x ^2-2 nx +1, n \geq 2$

So, $y \geq 0 \forall n \geq 1 ; \forall x >0 ; n \in N$ is false.

$\Rightarrow \sin \beta<0 ; \cos \alpha < 0$

$\Rightarrow \cos (\alpha+\beta) > 0$

$f(x)=\frac{10-x}{10}=1-\frac{x}{10}$

$\cot \sum \cot ^{-1}(1+n(n+1)) $

$\cot \sum \tan ^{-1} \frac{(n+1)-n}{1+n(n+1)} $

$\cot \sum_{n=1}^{23}\left(\tan ^{-1}(n+1)-\tan ^{-1} n\right) $

$\cot \left(\tan ^{-1} 24-\tan ^{-1} 1\right) $

$\cot \left(\tan ^{-1} \frac{24-1}{1+24}\right) $

$\cot \left(\cot ^{-1} \frac{25}{23}\right)=\frac{25}{23}$

$=\left[\frac{1}{y^2}\left[\frac{\left(\frac{1}{\sqrt{1+y^2}}+\frac{y \cdot y}{\sqrt{1+y^2}}\right)}{\left(\frac{\sqrt{1-y^2}}{y}+\frac{y}{\sqrt{1-y^2}}\right)}\right]^2+y^4\right]^{1 / 2}$

$=\left(\frac{1}{y^2} \cdot y^2\left(1-y^4\right)+y^4\right)^{1 / 2}=1$

$(Q)$ $\quad $$ \cos x+\cos y=-\cos z $

$ \sin x+\sin y=-\sin z $

$ 2+2 \cos (x-y)=1 $

$\Rightarrow \quad  \cos (x-y)=-1 / 2$

$(R)\quad$ $\cos 2 x\left(\cos \left(\frac{\pi}{4}-x\right)-\cos \left(\frac{\pi}{4}+x\right)\right)+2 \sin ^2 x=2 \sin x \cos x $

$\cos 2 x(\sqrt{2} \sin x)+2 \sin ^2 x=2 \sin x \cos x $

$\sqrt{2} \sin x[\cos 2 x+\sqrt{2} \sin x-\sqrt{2} \cos x]=0 $

$\text { Either } \sin x=0 \text { OR } \cos ^2 x-\sin ^2 x=\sqrt{2}(\cos x-\sin x) $

$\sec x=1 \quad \text { OR } \cos x=\sin x $

$\Rightarrow \quad \sec x=\sqrt{2}$

$(S)$ $\quad \cot \left(\sin ^{-1} \sqrt{1-x^2}\right)=\sin \left(\tan ^{-1}(x \sqrt{6})\right)$

$Image$

$\frac{x}{\sqrt{1-x^2}}  =\frac{x \sqrt{6}}{\sqrt{1+6 x^2}} $

$ \Rightarrow \quad 1+6 x^2=6-6 x^2 $

$ \Rightarrow \quad 12 x^2=5 $

$ x=\sqrt{\frac{5}{12}}==\frac{1}{2} \sqrt{\frac{5}{3}}$

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

$ =\sqrt{1+x^2}\left[\left(\frac{x^2}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\right)^2-1\right]^{1 / 2} $

$ =\sqrt{1+x^2}\left(x^2+1-1\right)^{1 / 2}=x \sqrt{1+x^2}$

then $\sin ^{-1} x+\cos ^{-1} y=0$

$ \Rightarrow \sin ^{-1} x=-\cos ^{-1} y $

$ \Rightarrow x^2+y^2=1 .$

$(B) $If $a=1$ and $b=1$, then

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

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

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

(taking sine on both the sides)

$(C)$ If $\mathrm{a}=1, \mathrm{~b}=2$

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

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

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

$ \Rightarrow x^2+y^2=1 \text { (on squaring). }$

$(D)$ If $a=2$ and $b=2$ then

$ \sin ^{-1}(2 x)+\cos ^{-1}(y)+\cos ^{-1}(2 x y)=\frac{\pi}{2} $

$ \Rightarrow 2 x y+\sqrt{1-4 x^2} \sqrt{1-y^2}=2 x y $

$ \Rightarrow\left(4 x^2-1\right)\left(y^2-1\right)=0 .$

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

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

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

$ \Rightarrow {x^2} + 2x + 2 = 1 + {x^2}$

$ \Rightarrow $ $x = - \frac{1}{2}$.

==> $2x + {x^3} = 2{x^2} + {x^3}$ ==> $x = {x^2}$
$\therefore \,x - {x^2} = 0\, \Rightarrow x(1 - x) = 0\, \Rightarrow x = 0$ and $x = 1$,
but $x \ne 0.$ So, $x = 1$.

${\tan ^{ - 1}}\sqrt {x(x + 1)} $ is defined when

$x(x + 1) \ge 0$…..$(i)$

${\sin ^{ - 1}}\sqrt {{x^2} + x + 1} $ is defined when

$0 \le x(x + 1) + 1 \le 1$ or $0 \le x(x + 1) \le 0$…..$(ii)$

From $(i)$ and $(ii),$ $x(x + 1) = 0$

or $x = 0$ and   $-1. $

Hence number of solution is  $2.$

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

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

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