Inverse Trigonometric Functions

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

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

(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.}$ = $\frac{3}{4}$

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

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

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

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

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

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

(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 }

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

= $-\frac{\pi }{4}$

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

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

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

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

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

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

(E)Using $ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\frac{x+y}{1-xy}.$

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

$\Rightarrow \frac{\frac{(x-1)(x+2)-1(x+1)(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2)-(x-1)(x+1)}{(x-2)(x+2)}}=tan\frac{\pi }{4}$

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

$\Rightarrow \frac{x^2+2x-x-2+x^2-2x+x-2}{\left(x^2-4\right)-\left(x^2-1^2\right)}=1.$ $\left\{\begin{array}{cc}\hfill ?(a-b)(a+b)=\hfill & \hfill a^2-b^2\hfill \end{array}\right\}$

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

$sin\left(si{n}^{-1}\frac{1}{5}+co{s}^{-1}x\right)=1.$

(E) $\Rightarrow sin\left(si{n}^{-1}\frac{1}{5}+co{s}^{-1}x\right)=sin\frac{\pi }{2}\left\{?sin\frac{\pi }{2}=1\right\}$

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

$\Rightarrow co{s}^{-1}x=\frac{\pi }{2}-si{n}^{-1}\frac{1}{5}$

$\Rightarrow co{s}^{-1}x=co{s}^{-1}\frac{1}{5}$ $\left\{?\frac{\pi }{2}=si{n}^{-1}x+co{s}^{-1}x\right\}$

$=\frac{\pi }{2}-si{n}^{-1}\frac{1}{5}=co{s}^{-1}\frac{1}{5}$ $\{\begin{array}{l}\Rightarrow \frac{\pi }{2}-si{n}^{-1}x=co{s}^{-1}x\\ x=\frac{1}{5}\\ \Rightarrow \frac{\pi }{2}-si{n}^{-1}\frac{1}{5}=co{s}^{-1}\frac{1}{5}\end{array}.$

= x = $\frac{1}{5}.$

 $tan\frac{1}{2}\left[si{n}^{-1}\frac{2x}{1+x^2}+co{s}^{-1}\frac{1-y^2}{1+y^2}\right],\left|x\right|<1,y>$

(M) Let x = tanØ . then tan^-1x= 0 and y = tan ω then tan^-1y = ω. we have,

tan $\frac{1}{2}$ $\{sin{}^{-1}\frac{2tan\theta }{1+ta{n}^2\theta }+\frac{1-ta{n}^2\omega }{1+ta{n}^2\omega }\}$

$=tan\frac{1}{2}\left\{si{n}^{-1}(sin2\theta )+co{s}^{-1}(cos2\omega )\right\}$ $\left\{\therefore sin2\theta \frac{2tan\theta }{1+ta{n}^2\theta }cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }tanx+y=\frac{tanx+tany}{1-tanxtany}\right\}$

$=tan\frac{1}{2}\{2\theta +2\omega \}=tan(\theta +\omega )$

$=\frac{tan\theta +tan\omega }{1-tan\theta tan\omega }$

= $\frac{x=y}{1-xy}$

cot $\left(ta{n}^{-1}a+co{t}^{-1}a\right)=cot\frac{\pi }{2}\left\{?ta{n}^{-1}x+co{t}^{-1}x=\frac{\pi }{2}\right\}$

(E) $=0$

$ta{n}^{-1}\left[2cos\left(2si{n}^{-1}\frac{1}{2}\right)\right]=ta{n}^{-1}\left[2cos\left(2si{n}^{-1}sin\frac{\pi }{6}\right)\right]$

(E) $=ta{n}^{-1}\left[2cos\left(2\times \frac{\pi }{6}\right)\right]$

$=ta{n}^{-1}\left[2cos\frac{\pi }{3}\right]$

$=ta{n}^{-1}2\times \frac{1}{2}$

$=ta{n}^{-1}1$

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

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

Given tan ^-1 $\frac{cosx-sinx}{cosx+sinx}$ , $\frac{-\pi }{4},$ x< $\frac{3x}{4}$

(M) Dividing numerator & denominator cos x we get,

tan ^-1 $\frac{\frac{cosx-sinx}{cosx}}{\frac{cosx+sinx}{cosx}}=ta{n}^{-1}\frac{\frac{cosx}{cosx}-\frac{sinx}{cosx}}{\frac{cosx}{cosx}+\frac{sinx}{cosx}}=ta{n}^{-1}\frac{1-tanx}{1+tanx.}$

We know that $tan\frac{\pi }{4}=1$ = 1 so,

$=ta{n}^{-1}\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}tanx}=ta{n}^{-1}\left[tan\left(\frac{\pi }{4}-x\right)\right]$ $\{?tan(x-y)=\frac{tanx-tany}{1+tanx·tany}\}$

$=\frac{\pi }{4}-x$ .