Ncert Solutions Maths class 12th

Given,

$(M)2ta{n}^{-1}(cosx)=ta{n}^{-1}(2cosecx)$

$\Rightarrow ta{n}^{-1}\frac{2cosx}{1-co{s}^{2}x}=ta{n}^{-1}\frac{2}{sinx}$ $\left\{\begin{array}{l}using.\\ ta{n}^{-1}2x=\frac{2x}{1-x^2}\end{array}\right\}$

$\Rightarrow \frac{2cosx}{si{n}^{2}x}=\frac{2}{sinx}$ $\left\{?1-co{s}^{2}x=si{n}^{2}x\Rightarrow 1=si{n}^{2}x+co{s}^{2}x\right\}$

$\Rightarrow \frac{cosx}{sinx}=1$

$\Rightarrow cotx=cot\frac{x}{4}$

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

LH.S $=\left(ta{n}^{-1}\frac{1}{5}+ta{n}^{-1}\frac{1}{7}\right)+\left(ta{n}^{-1}\frac{1}{3}+ta{n}^{-1}\frac{1}{8}\right)$

$=ta{n}^{-1}\left[\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}·\frac{1}{7}}\right]+ta{n}^{-1}\left[\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3},\frac{1}{8}}\right]\left\{\begin{array}{cc}\hfill ?usingta{n}^{-1}x+ta{n}^{-1}y\hfill & \hfill \frac{x+y}{1-xy},xy<1\hfill \end{array}\right\}$

= $ta{n}^{-1}\left[\frac{\frac{7+5}{7*5}}{\frac{7*5-1}{7*5}}\right]+ta{n}^{-1}\left[\frac{\frac{8+3}{5*3}}{\frac{8*3-1}{8*3}}\right]$

$=ta{n}^{-1}\left(\frac{12}{35-1}\right)+ta{n}^{-1}\left(\frac{11}{24-1}\right)=ta{n}^{-1}\frac{12}{34}+ta{n}^{-1}\frac{11}{23}$

$=ta{n}^{-1}\frac{6}{17}+ta{n}^{-1}\frac{11}{23}$

$=ta{n}^{-1}\left(\frac{\frac{6}{17}+\frac{1}{23}}{1-\frac{6}{17}*\frac{3}{23}}\right)=ta{n}^{-1}\left(\frac{\frac{6*23+11*11}{17*23}}{\frac{?7*23-6*11}{17*23}}\right)$

$=ta{n}^{-1}\frac{138+187}{391-66}=ta{n}^{-1}\frac{325}{325}=ta{n}^{-1}1$

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

$=\frac{x}{4}=R.H.S$

Let $si{n}^{-1}\frac{5}{13}=x\text{\hspace{0.17em}and}co{s}^{-1}\frac{3}{5}=y.$

Then, $sinx=\frac{5}{13}andcos=\frac{3}{5}$

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

$=\frac{5/3}{12/1}=\frac{4/5}{3/5}.$

$=\frac{5}{12}=\frac{4}{3}.$

Using $tan(x+y)=\frac{lanx+lany}{1-tanxtany.}$

tan $\left(si{n}^{-1}\frac{5}{13}+co{s}^{-1}\frac{3}{5}\right)=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}*\frac{4}{3}}=$ $\frac{\frac{5*3+4*12}{12*3}}{\frac{12*3-5*4}{12*3}}$

$=\frac{15+48}{36-20}=\frac{63}{16}$

$\Rightarrow si{n}^{-1}\frac{5}{13}+co{s}^{-1}\frac{3}{5}=ta{n}^{-1}\frac{63}{16}.$

Hence proved.

$co{s}^{-1}\frac{12}{13}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\frac{56}{65}$

Let $co{s}^{-1}\frac{12}{13}=\mathrm{xandsin}{}^{-1}\frac{3}{5}=y.$

Thin, $cosx=\frac{12}{13}\text{\hspace{0.17em}and sin}y=\frac{3}{5}$

Using $sin(x+y)=sinxcosy+cosxsiny.$

$sin\left[co{s}^{-1}\frac{12}{13}+si{n}^{-1}\frac{3}{5}\right]=\frac{5}{13}*\frac{4}{5}+\frac{12}{13}*\frac{3}{5}=\frac{20+36}{65}=\frac{56}{65}.$

$co{s}^{-1}\frac{12}{13}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\frac{56}{65}.$

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

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 $cos(x+y)=cosxcosy-sinxsiny.$

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

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

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

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

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

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

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)}

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

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]

= 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