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$co{s}^{-1}\frac{4}{5}+co{s}^{-1}\frac{12}{13}=co{s}^{-1}\frac{33}{65^{°}}.$

Let cos^-1 $\frac{4}{5}$ and cos^-1 $\frac{12}{13}$ = y.

Then, $cosx=\frac{4}{5}\text{\hspace{0.17em}and}\text{\hspace{0.17em}cos}y=\frac{12}{13}.$

Using $cos(x+y)=cosxcosy-sinxsiny.$

$\Rightarrow cos\left[co{s}^{-1}\frac{4}{5}+co{s}^{-1}\frac{12}{13}\right]=\frac{4}{5}\frac{12}{13}-\frac{3}{5}\times \frac{5}{13}$

$\Rightarrow cos\left[co{s}^{-1}\frac{4}{5}+co{t}^{-1}\frac{12}{13}\right]=\frac{48-15}{65}=\frac{33}{65}$

$\Rightarrow co{s}^{-1}\frac{4}{5}+co{s}^{-1}\frac{12}{13}=co{s}^{-1}\frac{33}{65}.?$

Let $si{n}^{-1}\frac{8}{17}=xsi{n}^{-1}\frac{3}{5}=y.$

(N. then, $sinx=\frac{8}{17}siny=\frac{3}{5}.$

So, $tanx=\frac{sinx}{cosx}$ and $tany=\frac{siny}{cosy}$

$=\frac{8/17}{15/17}=\frac{3/4}{4/5}$

$=\frac{8}{15}=\frac{3}{4}$

Using. $\text{\hspace{0.17em}tan}(x+y)=\frac{tanx+tany}{1-tanxtany}$

$tan\left(si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{3}{5}\right)=\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}$

tan $(si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{3}{5}=\frac{\frac{8\times 4+3\times 15}{15\times 4}}{\frac{15\times 4-8\times 3}{15\times 4}}=\frac{32+45}{60-24}\Rightarrow sin$

sin^-1 $\frac{8}{17}+si{n}^{-1}\frac{3}{5}=$ tan^-1 $\frac{77}{36}$

Hence proved.

$ta{n}^{-1}\left(tan\frac{7\pi }{6}\right)=ta{n}^1\left(tan\frac{6\pi +\pi }{6}\right)$

$=ta{n}^{-1}\left(tan\frac{6\pi }{6}+\frac{\pi }{6}\right)$

$=ta{n}^{-1}\left(tan\pi +\frac{\pi }{6}\right)$

$=ta{n}^{-1}\left(tan\frac{\pi }{6}\right)$ { Ø tan ($\frac{\pi }{}$+ Ø ) = tan Ø as tan b (+) we in 3rd quadrant)}

$=\frac{\pi }{6}$ ∈$\left(\frac{-\pi }{2},\frac{\pi }{2}\right)$

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Cos^-1 $\left(cos\frac{13\pi }{6.}\right)$ = $co{s}^{-1}$ $\left(cos\frac{12\pi +\pi }{6}\right)$

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

= $co{s}^{-1}$ $\left[co{s}^{-1}\left(2\pi +\frac{\pi }{6}\right)\right]$  {Øcor2$\frac{\pi }{}$+=ØcosØ}

$=co{s}^{-1}\left(cos\frac{\pi }{6}\right)$

$=\frac{\pi }{6}$ ∈ [0, x]

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= sin $\left[\frac{\pi }{3}+si{n}^{-1}\left(sin\frac{\pi }{6}\right)\right]$

= sin $\left[\frac{\pi }{3}+\frac{\pi }{6}\right]$ = sin $\left[\frac{2\pi +\pi }{6}\right]$ = sin $\left(\frac{3\pi }{6}\right)$

= sin $\frac{\pi }{2}=1$ 

Cos^-1 cos $\frac{7\pi }{6}$

(M) As $\frac{7\pi }{6}$  [0, $\frac{\pi }{}$] ;principal value branch of cos^-1

cos^-1 $\left(cos\frac{7\pi }{6}\right)$ = cos^-1 $\left(cos2x-\frac{7\pi }{6}\right)$ $\{?cos(2\pi -\theta )=cos\theta \}$

= cos^-1$\left(sis\frac{12\pi -7\pi }{6}\right)$

$=co{s}^{-1}cos\frac{5\pi }{6}\frac{5\pi }{6}\in [0,\pi ]$

= $\frac{5\pi }{6}$

So, option B is correct

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$tan\left(si{n}^{-1}\frac{3}{5}+co{t}^{-1}\frac{3}{2}\right)$

(M) Let $si{n}^{-1}\frac{3}{5}=x$ and cot^-1 $\frac{3}{2,}$ = y .

Then $sinx=\frac{3}{5}coty=\frac{3}{2}$

Hence, tan x = $\frac{sinx}{cosx}$ = $\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ = $\frac{3}{4}$

$tan\left(si{n}^{-1}\frac{3}{5}+co{t}^{-1}\frac{3}{2}\right)=tan(x+y)$

$=\frac{tanx+tany}{1-tanxtany}$ .$\frac{\frac{\mathrm{}}{}}{}$

$\frac{\frac{3\times 3+2\times 4}{4\times 3}}{\frac{4\times 3-3\times 2}{4\times 3}}$

$\frac{4\times 3-3\times 2}{4\times 3}$

$\frac{9+8}{12-6}=\frac{17}{6}$

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$ta{n}^{-1}\left(tan\frac{3\pi }{4}\right).$

(M). As $\frac{3\pi }{4}\notin \left(-\frac{\pi }{2},\frac{\pi }{2}\right);$ principal value branch of tan ^-1we can write,

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

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

$=ta{n}^{-1}\left(-tan\frac{\pi }{4}\right)\{?tanis(-)wein{I}^{-1}$ quadent }

$=-ta{n}^{-1}\left(tan\frac{1}{4}\right)\left\{?ta{n}^{-1}(-x)=-ta{n}^{-1}x\right\}$

= $-\frac{\pi }{4}$

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$si{n}^{-1}\left(sin\frac{2\pi }{3}\right).$

(M) As $\frac{2\pi }{3}\notin \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ principal value branch of sin-1 we can write,

$si{n}^{-1}\left(sin\frac{3\pi -\pi }{3}\right)=si{n}^{-1}\left(sin\frac{3\pi }{3}-\frac{\pi }{3}\right)=si{n}^{-1}\left(sin\pi -\frac{\pi }{3}\right)$

$=si{n}^{-1}\left(sin\frac{\pi }{3}\right)\{?si\mathrm{nis}(+)ve\; in\; ind\; quadrat$
${}^{}\}$

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