Inverse Trigonometric Functions

$si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}\to (1)$

(M) Let $x=sin\theta .Then,\theta =si{n}^{-1}x.$

Putting this in $q^n(1)$ we get

$si{n}^{-1}(1-x)-2·\theta =\frac{\pi }{2}$

$\Rightarrow si{n}^{-1}(1-x)=\frac{\pi }{2}+2\theta$

$\Rightarrow 1-x=sin\left(\frac{\pi }{2}+2\theta \right)\left\{sin\left(\frac{\pi }{2}+x\right)=cosx\right\}$

$\Rightarrow 1-x=cos2\theta$

$\Rightarrow 1-x=1-2si{n}^2\theta ·\left\{cos2x=1-2si{n}^{2}x\right\}$

$\Rightarrow 1-x=1-2x^2·\{sin\theta =x\}$

$\begin{array}{l}\Rightarrow 2x^2-x=0\\ \Rightarrow x(2x-1)=0\end{array}$

$\Rightarrow so,x=0x2x-1=0x2x=1\to x=\frac{1}{2}.$

Putting $x=0$ in qⁿ (1) .

L.H.S $=si{n}^{-1}(1-0)-2si{n}^{-1}0=si{n}^{-1}sin\frac{x}{2}-0=\frac{\pi }{2}=R.H.S.$

$x=\frac{1}{2}\mathrm{q(1)}$

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

$=si{n}^{-1}\frac{1}{2}-2si{n}^{-1}\frac{1}{2}$

$=-si{n}^{-1}\frac{1}{2}=-si{n}^{-1}\left(sin\frac{\pi }{6}\right)=-\frac{\pi }{6}$ $\ne$

So, $=0.$

Option (c) is correct.

Given, (M)

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

$x=tan\mathrm{\theta .Then}\theta =ta{n}^{-1}\mathrm{x\; Bo\; we\; have,}$

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

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

$\Rightarrow ta{n}^{-1}\left\{tan\left(\frac{x}{4}-\theta \right)\right\}=\frac{\theta }{2}\{\frac{?tanx-tany}{1+tanxtan}=tan(x-y)$

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

$\Rightarrow \frac{\theta }{2}+\theta =\frac{x}{4}$

$\Rightarrow \frac{3\theta }{2}=\frac{\pi }{4}$

$\Rightarrow \theta =\frac{\pi }{4}*\frac{2}{3}=\frac{\pi }{6}$

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