Mathematics M.C.Q (1 Marks)

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

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

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

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

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

$\Rightarrow \frac{4-2 x^{2}}{3}=1$

$\Rightarrow 4-2 x^{2}=3$

$\Rightarrow 2 x^{2}=4-3=1$

$\Rightarrow x=\pm \frac{1}{\sqrt{2}}$

Hence, the value of $x$ is $\pm \frac{1}{\sqrt{2}}$

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

$\Rightarrow \frac{\pi}{4}=\frac{3}{2} \tan ^{-1} x$

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

$\Rightarrow x=\tan \frac{\pi}{6}$

$\therefore x=\frac{1}{\sqrt{3}}$

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

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

Let $\sin ^{-1} x=\theta \Rightarrow \sin \theta=x \Rightarrow \cos \theta=\sqrt{1-x^{2}}$

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

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

Therefore, from equation $( 1 )$, we have

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

Let, $x=\sin y .$ Then, we have:

$-2 \cos ^{-1}(\sqrt{1-\sin ^{2} y})=\cos ^{-1}(1-\sin y)$

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

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

$\Rightarrow 1-\sin y=\cos (-2 y)=\cos 2 y$

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

$\Rightarrow 2 \sin ^{2} y-\sin y=0$

$\Rightarrow \sin y(2 \sin y-1)=0$

$\Rightarrow \sin y=0$ or $\frac{1}{2}$

$\therefore x=0$ OR $x=\frac{1}{2}$

When $x=\frac{1}{2},$ it can be observed that

L.H.S. $=\sin ^{-1}\left(1-\frac{1}{2}\right)-\sin ^{-1} \frac{1}{2}$

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

$=-\sin ^{-1} \frac{1}{2}$

$=\frac{\pi}{6} \neq \frac{\pi}{2} \neq R . H . S$

$\therefore x=\frac{1}{2}$ is not a solution of the given equation. Thus, $x=0$

${\cos ^{ - 1}}\left( {\frac{{3 + 5\cos x}}{{5 + 3\cos x}}} \right) = {\cos ^{ - 1}}\left( {\frac{3}{5}} \right)$

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

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

we get $2{\tan ^{ - 1}}\left( {\frac{1}{2}\tan \frac{\pi }{4}} \right)$

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

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

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

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

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

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

$= {\cos ^{ - 1}}\left[ {\frac{{a\cos \theta + b}}{{a + b\cos \theta }}} \right]$.

$ \Rightarrow \tan \left(\tan ^{-1} \frac{x}{\pi}\right)<\tan \frac{\pi}{3}$

$\Rightarrow \quad \frac{\mathrm{x}}{\pi}<\sqrt{3}$

$ \Rightarrow \mathrm{x}<\sqrt{3} \pi=5.5$ (approxi)

The maximum value of $x$ is $5 .$

$ = {({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)^3}$$ - 3{\sin ^{ - 1}}x{\cos ^{ - 1}}x({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)$

$= \frac{{{\pi ^3}}}{8} - 3({\sin ^{ - 1}}x{\cos ^{ - 1}}x)\frac{\pi }{2}$

$= \frac{{{\pi ^3}}}{8} - \frac{{3\pi }}{2}{\sin ^{ - 1}}x\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right)$

$=  \frac{{{\pi ^3}}}{8} - \frac{{3{\pi ^2}}}{4}{\sin ^{ - 1}}x + \frac{{3\pi }}{2}{({\sin ^{ - 1}}x)^2}$

$=\frac{{{\pi ^3}}}{8} + \frac{{3\pi }}{2}\left[ {{{({{\sin }^{ - 1}}x)}^2} - \frac{\pi }{2}{{\sin }^{ - 1}}x} \right]$

$ = \frac{{{\pi ^3}}}{8} + \frac{{3\pi }}{2}\left[ {{{\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)}^2}} \right] - \frac{{3{\pi ^3}}}{{32}}$

$ = \frac{{{\pi ^3}}}{{32}} + \frac{{3\pi }}{2}{\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)^2}$

$\therefore$ The least value is $\frac{{{\pi ^3}}}{{32}}$

and since ${\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)^2} \le {\left( {\frac{{3\pi }}{4}} \right)^2}$

$\therefore$ The greatest value is $\frac{{{\pi ^3}}}{{32}} + \frac{{9{\pi ^2}}}{{16}} \times \frac{{3\pi }}{2} = \frac{{7{\pi ^3}}}{8}$.

Here $a < \frac{1}{{32}}$. So, number of solution is zero.

$..... + {\tan ^{ - 1}}\frac{1}{{{c_n}}}$

$= {\tan ^{ - 1}}\left( {\frac{{\frac{x}{y} - \frac{1}{{{c_1}}}}}{{1 + \frac{x}{y}.\frac{1}{{{c_1}}}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{\frac{1}{{{c_1}}} - \frac{1}{{{c_2}}}}}{{1 + \frac{1}{{{c_1}{c_2}}}}}} \right)$

$ + {\tan ^{ - 1}}\left( {\frac{{\frac{1}{{{c_2}}} - \frac{1}{{{c_3}}}}}{{1 + \frac{1}{{{c_2}{c_3}}}}}} \right) + ....... + {\tan ^{ - 1}}\frac{1}{{{c_n}}}$

$ = {\tan ^{ - 1}}\frac{x}{y} - {\tan ^{ - 1}}\frac{1}{{{c_1}}} + {\tan ^{ - 1}}\frac{1}{{{c_1}}} - {\tan ^{ - 1}}\frac{1}{{{c_2}}} + {\tan ^{ - 1}}\frac{1}{{{c_2}}}$

$ - {\tan ^{ - 1}}\frac{1}{{{c_3}}} + ... + {\tan ^{ - 1}}\frac{1}{{{c_{n - 1}}}} - {\tan ^{ - 1}}\frac{1}{{{c_n}}} + {\tan ^{ - 1}}\frac{1}{{{c_n}}}$

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

$ = \sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left( {\frac{{2m}}{{1 + ({m^2} + m + 1)({m^2} - m + 1)}}} \right)} $

$ = \sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left( {\frac{{({m^2} + m + 1) - ({m^2} - m + 1)}}{{1 + ({m^2} + m + 1)({m^2} - m + 1)}}} \right)} $

$= \sum\limits_{m = 1}^n {[{{\tan }^{ - 1}}({m^2} + m + 1) - {{\tan }^{ - 1}}({m^2} - m + 1)]} $

$ = ({\tan ^{ - 1}}3 - {\tan ^{ - 1}}1) + ({\tan ^{ - 1}}7 - {\tan ^{ - 1}}3) + $
$({\tan ^{ - 1}}13 - {\tan ^{ - 1}}7) + ...... + [{\tan ^{ - 1}}({n^2} + n + 1)$
$ - {\tan ^{ - 1}}({n^2} - n + 1)]$

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

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

Let ${\cos ^{ - 1}}\sqrt {\frac{2}{3}}  = \theta  \Rightarrow \cos \theta  = \sqrt {\frac{2}{3}} $

$ \Rightarrow \sin \theta  = \sqrt {1 - {{\cos }^2}\theta }  = \sqrt {1 - \frac{2}{3}}  = \sqrt {\frac{1}{3}} $

$\therefore {\tan ^{ - 1}}\left[ {\sin \left( {{{\cos }^{ - 1}}\sqrt {\frac{2}{3}} } \right)} \right] = {\tan ^{ - 1}}\left[ {\sin \theta } \right]$

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

      $ = \frac{\pi }{6}$

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

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

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

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

$\Rightarrow \frac{x}{5}=\sin \left(\sin ^{-1} \frac{3}{5}\right)$

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

$\Rightarrow x=3$

Then $\frac{{{x^2}}}{{{a^2}}}\cos \theta + \frac{{{y^2}}}{{{b^2}}} = {\sin ^2}\theta $

Here ${\cos ^{ - 1}}\frac{x}{1} + {\cos ^{ - 1}}\frac{y}{2} = \alpha ;$

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

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

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

==>${\tan ^{ - 1}}\left[ {\frac{{\frac{1}{{\sqrt {\cos \alpha } }} - \sqrt {\cos \alpha } }}{{1 + \frac{{\sqrt {\cos \alpha } }}{{\sqrt {\cos \alpha } }}}}} \right] = x$

==> $\tan x = \frac{{1 - \cos \alpha }}{{2\sqrt {\cos \alpha } }}$

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

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

$x=1,-5(\text { reject })$

$\sin ^{-1} 2 \mathrm{x}=\sin ^{-1} \sqrt{1-\mathrm{x}^{2}}$

$\Rightarrow \quad 2 \mathrm{x}=\sqrt{1-\mathrm{x}^{2}}$

$\Rightarrow \quad 4 \mathrm{x}^{2}=1-\mathrm{x}^{2}$

$\Rightarrow \quad \mathrm{x}=\pm \frac{1}{\sqrt{5}}$

$\mathrm{x}=\frac{1}{\sqrt{5}}\left(\because \quad \mathrm{x} \neq-\frac{1}{\sqrt{5}}\right)$

$\tan ^{-1}(\tan 100)=100-32 \pi$

$\cos ^{-1}(\cos 100)=32 \pi-100$

$\cot ^{-1}(\cot 100)=100-31 \pi$

$\therefore 32 \pi-\frac{\pi}{2}<100<32 \pi$

than $1$

$\Rightarrow x^{2}=1$

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

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

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

${=-x \Rightarrow x=0}$

$\mathrm{x}<0$

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

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

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

$2-2 x^{2}=2-x$

$\mathrm{x}=\frac{1}{2}$

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

$2=\tan \theta$

$x=\sin 2 \theta$

$x=2 \sin \theta \cos \theta$

$x=2 \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}=\frac{4}{5}$

$\because \quad 1-x=1 / 5$

and $ \quad \because y^{2}=\frac{1}{5}$

$\tan ^{-1} 4 / 3=\alpha$

$4 / 3=\tan \alpha$

$y=\sin \frac{\alpha}{2}$

$y=\sqrt{\frac{1-\cos \alpha}{2}}=\sqrt{\frac{1-3 / 5}{2}}$

$y=\frac{1}{\sqrt{5}} \quad \therefore \quad y^{2}=1-x$

put $x = \cos \theta $          $\because $ $\theta  \in \left( {0,\frac{\pi }{2}} \right)\quad \therefore \quad 0 < x < 1$

$\Rightarrow \quad \cot ^{-1}\left(\frac{\cos 2 \theta}{2 \cos \theta|\sin \theta|}\right)$

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

$\because 2 \theta \in(0, \pi)$

$\cos ^{-1}\left[\cot \left\{\sin ^{-1}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)+\frac{\pi}{6}+\frac{\pi}{4}\right\}\right]$

$\cos ^{-1}\left[\cot \left\{\frac{\pi}{12}+\frac{\pi}{6}+\frac{\pi}{4}\right\}\right]$

$\cos ^{-1}\left[\cot 90^{\circ}\right]=\cos ^{-1} 0=\frac{\pi}{2}$

$=\cos ^{-1}\left\{\cot \left(\sin ^{-1} \frac{\sqrt{3}-1}{2 \sqrt{2}}+\cos ^{-1} \frac{\sqrt{3}}{2}+\sec ^{-1} \sqrt{2}\right)\right\}$

$=\cos ^{-1}\left\{\cot \left(\frac{\pi}{12}+\frac{\pi}{6}+\frac{\pi}{4}\right)\right\}$

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

$=\cos ^{-1} 0=\frac{\pi}{2}$

$\sec ^{-1}(1)<\sec ^{-1}(2)<\sec ^{-1}(-2)<\sec ^{-1}(-1)$

$\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}+\mathrm{T}_{4}+\mathrm{T}_{5}=\tan ^{-1} 6-\tan ^{-1} 1$

$=\tan ^{-1}\left(\frac{5}{7}\right)$

$p+q=5+7=12$

and $\sin ^{-1} u-\sin ^{-1} v \in[-\pi, \pi]$

$\therefore $ maximum value of expression is $[\pi]-[-\pi]=7$

$\Rightarrow 0 \leq x \leq 2,2 \leq x \leq 4, x \neq \pm \sqrt{2}$

$\therefore x=2$

$\therefore \mathrm{RHS}=\sin ^{-1}(1)+\cos ^{-1}(-1)+\tan ^{-1}(-1)$

$=\frac{\pi}{2}+\pi-\frac{\pi}{4}=\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{5 \pi}{4}$

$\tan ^{-1}(\tan 2)=2-\pi.$

$ \Rightarrow \quad 1+\cos ^{2} x=\cos ^{2} x  \Rightarrow \text { No solution } $

$ m=0$

$|x| = 3 \Rightarrow cos^{-1}cos9 = 9 -2 \pi$

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

Let $\tan ^{-1} \mathrm{x}=\theta \Rightarrow \mathrm{x}=\tan \theta$

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

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

Now

$\sin \left[\cot ^{-1}(1+x)\right]$

$\cot ^{-1}(1+x)=\phi \Rightarrow 1+x=\cot \phi$

$\sin \cot ^{-1}(1+x)=\sin \phi=\frac{1}{\csc \phi}=\frac{1}{\sqrt{1+(1+x)^{2}}}$

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

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

$1+2 \mathrm{x}=0 \Rightarrow \mathrm{x}=-\frac{1}{2}$

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

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

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

$\Rightarrow 4 \alpha=\pi$

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

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

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

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

$\Rightarrow 3 \beta=\pi \quad \ldots \ldots$

From (1) and (2) we have

$\therefore 4 \alpha=3 \beta$

$\Rightarrow \frac{\mathrm{n}}{\pi}<\cot \frac{\pi}{6}$

[as ${{{\cot }^{ - 1}}x}$ is a decreasing function]

$\Rightarrow \quad \frac{\mathrm{n}}{\pi}<\sqrt{3}$

$\Rightarrow \quad n<\sqrt{3} \pi$

$\Rightarrow \quad n<5.46$

$\Rightarrow \quad$ maximum value of $n$ is $5$

$ \sin ^{-1} \theta =5-2 \pi $

$ \theta =\sin (5-2 \pi) $

Also $\cos ^{-1} \sqrt{x^{2}-1} \Rightarrow 0 \leq x^{2}-1 \leq 1$

$\mathrm{x}^{2} \in[1,2]$

$\therefore $ Possible value of $x=1$

$\therefore $ Equation becomes $\frac{\pi}{2}+\frac{\pi}{2}+\tan ^{-1} \tan \mathrm{y}=\mathrm{a}$

$\therefore $ for solution $a \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$

$\therefore $ integral values are $2,3,4$

$\mathrm{T}_{\mathrm{r}}=\tan ^{-1}\left(\frac{\mathrm{r}}{3}\right)-\tan ^{-1}\left(\frac{(\mathrm{r}-1)}{3}\right)$

$\mathrm{T}_{1}=\tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}(0)$

$\mathrm{T}_{2}=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{1}{3}\right)$

$\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{\mathrm{n}}{3}\right)-\tan ^{-1}\left(\frac{(\mathrm{n}-1)}{3}\right)$

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

$=\tan ^{-1}(\mathrm{n}+1)-\tan ^{-1}(\mathrm{n})$

so that $L.H.S.$ of the given eq uation is

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

$+\tan ^{-1}(\mathrm{n}+1)-\tan ^{-1} \mathrm{n}$

$=\tan ^{-1}(\mathrm{n}+1)-\tan ^{-1} 1$

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

so that $\tan ^{-1} \frac{n}{n+2}=\tan ^{-1} \theta \Rightarrow \theta=\frac{n}{n+2}$

$=2\left(2 \tan ^{-1} \frac{1}{5}\right)-\tan ^{-1} \frac{1}{239}$

$=2 \tan ^{-1} \frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}-\tan ^{-1} \frac{1}{239} \quad\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$

$=2 \tan ^{-1} \frac{2 / 5}{24 / 25}-\tan ^{-1} \frac{1}{239}$

$=2 \tan ^{-1} \frac{5}{12}-\tan ^{-1} \frac{1}{239}$

$=2 \tan ^{-1} \frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}-\tan ^{-1} \frac{1}{239} \quad\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$

$=2 \tan ^{-1} \frac{2 / 5}{24 / 25}-\tan ^{-1} \frac{1}{239}$

$=2 \tan ^{-1} \frac{5}{12}-\tan ^{-1} \frac{1}{239}$

$=\tan ^{-1} \frac{2 \cdot \frac{5}{12}}{1-\left(\frac{5}{12}\right)^{2}}-\tan ^{-1} \frac{1}{239} \quad\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$

$=\tan ^{-1} \frac{144 \times 5}{119 \times 6}-\tan ^{-1} \frac{1}{239}$

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

$=\tan ^{-1} \frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119} \cdot \frac{1}{239}} \quad\left(\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}\right)$

$=\tan ^{-1} \frac{120 \times 239-119}{119 \times 239+120}=\tan ^{-1} \frac{28680-119}{28441+120}$

$=\tan ^{-1} \frac{28561}{28561}=\tan ^{-1} 1=\frac{\pi}{4}$

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

$\Rightarrow \cos ^{-1}(2 x \cdot 3 x-\sqrt{1-4 x^{2}} \sqrt{1-9 x^{2}})=\cos ^{-1}(-x)$

$\Rightarrow \quad 6 x^{2}-\sqrt{1-4 x^{2}} \sqrt{1-9 x^{2}}=-x$

$\Rightarrow \quad\left(6 x^{2}+x\right)=\sqrt{1-4 x^{2}} \sqrt{1-9 x^{2}}$

$\Rightarrow \quad 36 x^{4}+x^{2}+12 x^{3}=\left(1-4 x^{2}\right)\left(1-9 x^{2}\right)$

$\Rightarrow \quad 36 x^{4}+x^{2}+12 x^{3}=1-13 x^{2}+36 x^{4}$

$\Rightarrow 12 \mathrm{x}^{3}+14 \mathrm{x}^{2}-1=0$

$\therefore \quad a=12, b=14, c=0$

$\sum\limits_{n = 1}^\infty  {{{\tan }^{ - 1}}\left( {\frac{1}{{{n^2} + n + 1}}} \right)} $

$\sum\limits_{n = 1}^\infty  {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)} $

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

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

$ = {\tan ^{ - 1}}\left( \infty  \right) - {\tan ^{ - 1}}\left( 1 \right)$

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

$ = \sum\limits_{n = 1}^5 {{{\sin }^{ - 1}}\left( { - \sin \frac{\pi }{6}} \right) + \sum\limits_{n = 1}^5 {{{\sin }^{ - 1}}\left( {\sin \frac{\pi }{6}} \right)} } $

$ =  - \frac{{5\pi }}{6} + \frac{{5\pi }}{6} = 0$

$=\tan ^{-1}\left(1+\frac{2}{25}\right)>\frac{\pi}{4}$