Inverse Trigonometric Functions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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,

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

$=\frac{\pi }{4}-x$ .

L.H.S= 2 tan ^-1 $\frac{1}{2}$ + tan ^-1 $\frac{1}{7}$

(E) Using2 tan ^-1x= tan ^-1 $\frac{2a}{1-x^2}$ we can write.

L.H.S = tan ^-1 $\frac{2*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{1-{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{1}{1\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$ + tan ^-1$\frac{1}{7}$

= tan ^-1 $\frac{1}{4\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$ + tan ^-1$\frac{1}{7}$ = tan ^-1 $\frac{4}{3}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1$\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}*\frac{1}{7}}$ { Ø tan -1 x + tan -1 y = tan - 1 $\frac{x+y}{1-xy}$ }

= tan ^-1 $\frac{\frac{7*4+1*3}{3*7}}{\frac{3*7-4*1}{3*7}}$

= tan -1 $\frac{28+3}{21-4}$ = tan ^-1 $\frac{31}{17}$ = R H S

L.H.S = tan ^-1$\frac{2}{11}$ + tan ^-1 $\frac{7}{24}$

Using tan ^-1x+ tan ^-1y= tan ^-1 $\frac{x+y}{1-xy}$ , xy<1

L.H.S =tan ^-1 $\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}*\frac{7}{24}}$ = tan ^-1 $\frac{\frac{2*24+7*2}{11*24}}{\frac{11*24-7*2}{11*24}}$

tan ^-1 $\frac{48+14}{264-14}$ = tan ^-1 $\frac{125}{250}$ = tan ^-1 $\frac{1}{2}$ = R.H.S

We know that,

cos3θ= 4cos³θ - 3cosθ

Letx = cosθ Then θ = cos^-1x. We have,

Cos3 (cos^-1x) = 4x³-3x

3cos^-1x = cos^-1 (4x³- 3x)

Hence Proved

We know that.

sin 3θ =3 sin θ 4sin³θ (identity).

(E) Let x = sinθ. Then, sin ⁻¹x=θ . We have,

Sin3 (sin ⁻¹x) = 3x−4x³

3sin ⁻¹x =sin^-1 (3x−4x³)

Hence proved.