Inverse Trigonometric Function

In this article, we will discuss inverse trigonometric function. The use of the inverse function is seen in every branch of calculus.

If a function is bijective then there exists an inverse of that function.

Definition of Inverse of a Function

Let X and Y are two non-null set. If any function f : X → Y be such that f(x) = y is bijective, then there exists another function g : Y → X such that g(y) =x,  where x ∈ X and y = f(x), where y ∈ Y. here domain of g is the range of f and range of g is domain of f. Then g is called inverse function of f and it is denoted as f^-1. Again the function g is bijective and the inverse of g is f.

Then

\[{{g}^{-1}}={{\left( {{f}^{-1}} \right)}^{-1}}=f\]

\[\left( {{f}^{-1}}\circ f \right)\left( x \right)={{f}^{-1}}\left\{ f\left( x \right) \right\}={{f}^{-1}}\left( y \right)=x\]

\[\left( f\circ {{f}^{-1}} \right)\left( y \right)=f\left\{ {{f}^{-1}}\left( y \right) \right\}=f\left( x \right)=y\]

The primary six trigonometric functions sinx, cosx, tanx, cosecx, secx and cotx are not bijective because their values periodically repeat. The values of function sinx in the interval [-π/2, π/2 ] increases between -1 to 1. The sinx function is bijective in the interval [-π/2, π/2 ]. Therefore in this interval there exists an inverse function sin^-1x of sinx. The interval [-π/2, π/2 ] is called principal value region.

Definition of Principal Value Region

The region where any trigonometric function is one-one-onto i.e. bijective, then the region is called principal value region of that trigonometric function.

Domain, Range and Principal Value Region of various Inverse Functions

Properties of Inverse Trigonometric Functions and Formulas

A.

\[(i){{\sin }^{-1}}\left( \sin \theta  \right)=\theta ,where~~\theta \in \left[ -\frac{\pi }{2},\frac{\pi }{2} \right]\]

\[(ii){{\cos }^{-1}}\left( \cos \theta  \right)=\theta ,where~~\theta \in \left[ 0,\pi  \right]\]

\[(iii){{\tan }^{-1}}\left( \tan \theta  \right)=\theta ,where~~\theta \in \left( -\frac{\pi }{2},\frac{\pi }{2} \right)\]

\[(iv)\cos e{{c}^{-1}}\left( \cos ec\theta  \right)=\theta ,where~~\theta \in \left[ -\frac{\pi }{2},\frac{\pi }{2} \right],\theta \ne 0\]

\[(v){{\sec }^{-1}}\left( \sec \theta  \right)=\theta ,where~~\theta \in \left[ 0,\pi  \right],\theta \ne \frac{\pi }{2}\]

\[(vi){{\cot }^{-1}}\left( \cot \theta  \right)=\theta ,where~~\theta \in \left( 0,\pi  \right)\]

B.

\[(i)\sin \left( {{\sin }^{-1}}x \right)=x,where~~x\in \left[ -1,1 \right]\]

\[(ii)\cos \left( {{\cos }^{-1}}x \right)=x,where~~x\in \left[ -1,1 \right]\]

\[(iii)\tan \left( {{\tan }^{-1}}x \right)=x,where~~x\in R\]

\[(iv)\cos ec\left( \cos e{{c}^{-1}}x \right)=x,where~~x\in \left( -\infty ,\left. -1 \right]\cup \left[ 1,\infty  \right) \right.\]

\[(v)\sec \left( {{\sec }^{-1}}x \right)=x,where~~x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty  \right)\]

\[(vi)\cot \left( {{\cot }^{-1}}x \right)=x,where~~x\in R\]

C.

\[(i){{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x,where~~x\in \left[ -1,1 \right]\]

\[(ii){{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}x,where~~x\in \left[ -1,1 \right]\]

\[(iii){{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x,where~~x\in R\]

\[(iv)\cos e{{c}^{-1}}\left( -x \right)=-\cos e{{c}^{-1}}x,where~~x\in \left( -\infty ,\left. -1 \right]\cup \left[ 1,\infty  \right) \right.\]

\[(v){{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}x,where~~x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty  \right)\]

\[(vi){{\cot }^{-1}}\left( -x \right)=\pi -{{\cot }^{-1}}x,where~~x\in R\]

D.

\[(i){{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2},where~~x\in \left[ -1,1 \right]\]

\[(ii){{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2},where~~x\in R\]

\[(iii){{\sec }^{-1}}x+\cos e{{c}^{-1}}x=\frac{\pi }{2},where~~x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty  \right)\]

Find the principal value of the following inverse trigonometric functions

\[(i){{\cos }^{-1}}\left( -\frac{1}{2} \right)~~~~~(ii)\cos ec\left( -\sqrt{2} \right)~~~~~~~(iii){{\tan }^{-1}}\left( \tan \frac{3\pi }{4} \right)\]

Solution:

\[(i)Let~~~{{\cos }^{-1}}\left( -\frac{1}{2} \right)~=\theta ,~~~\theta \in \left[ 0,\pi  \right]~~\]

\[\therefore \cos \theta =-\frac{1}{2}=\cos \left( \frac{2\pi }{3} \right)\]

\[\therefore \theta =\frac{2\pi }{3}\in \left[ 0,\pi  \right]\]

\[\therefore P\text{rincipal Value}~~of{{\cos }^{-1}}\left( -\frac{1}{2} \right)~=\frac{2\pi }{3}\]

\[(ii)Let~~~\cos ec\left( -\sqrt{2} \right)=\theta ,~~~\theta \in \left[ -\frac{\pi }{2},\frac{\pi }{2} \right]-\left\{ 0 \right\}~~\]

\[\Rightarrow \cos ec\theta =-\sqrt{2}=\cos ec\left( -\frac{\pi }{4} \right)\]

\[\therefore \theta =-\frac{\pi }{4}\in \left[ -\frac{\pi }{2},\frac{\pi }{2} \right]\]

\[\therefore P\text{rincipal Value}~~of\cos ec\left( -\sqrt{2} \right)=-\frac{\pi }{4}\]

\[~(iii){{\tan }^{-1}}\left( \tan \frac{3\pi }{4} \right)\ne \frac{3\pi }{4}~~\left[ \because ~~it~~not~~lies~~between~~-\frac{\pi }{2}~~and~~\frac{\pi }{2} \right]\]

\[\therefore {{\tan }^{-1}}\left( \tan \frac{3\pi }{4} \right)={{\tan }^{-1}}\left[ \tan \left( \pi -\frac{\pi }{4} \right) \right]\]

\[\Rightarrow {{\tan }^{-1}}\left( \tan \frac{3\pi }{4} \right)={{\tan }^{-1}}\left( -\tan \frac{\pi }{4} \right)\]

\[\Rightarrow {{\tan }^{-1}}\left( \tan \frac{3\pi }{4} \right)={{\tan }^{-1}}\left[ \tan \left( -\frac{\pi }{4} \right) \right]=-\frac{\pi }{4}\]

\[\therefore P\text{rincipal Value}~~of{{\tan }^{-1}}\left( \tan \frac{3\pi }{4} \right)=-\frac{\pi }{4}\]

Prove that

\[\tan \left[ \frac{1}{2}.{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+\frac{1}{2}.{{\cos }^{-1}}\frac{1-{{y}^{2}}}{1+{{y}^{2}}} \right]=\frac{x+y}{1-xy}\]

\[where~~\left| x \right|<1,y>0,xy<1\]

Solution:

\[Let~~x=\tan \theta ~~and~~y=\tan \phi \]

\[LHS=\tan \left[ \frac{1}{2}.{{\sin }^{-1}}\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }+\frac{1}{2}.{{\cos }^{-1}}\frac{1-{{\tan }^{2}}\phi }{1+{{\tan }^{2}}\phi } \right]\]

\[=\tan \left[ \frac{1}{2}.{{\sin }^{-1}}\left( \sin 2\theta  \right)+\frac{1}{2}.{{\cos }^{-1}}\left( \cos 2\phi  \right) \right]\]

\[\left[ \because \sin 2\theta =\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }~~and~~\cos 2\phi =\frac{1-{{\tan }^{2}}\phi }{1+{{\tan }^{2}}\phi } \right]\]

\[=\tan \left[ \frac{1}{2}\left( 2\theta  \right)+\frac{1}{2}\left( 2\phi  \right) \right]\]

\[=\tan \left( \theta +\phi  \right)=\frac{\tan \theta +\tan \phi }{1-\tan \theta .\tan \phi }\]

\[\therefore \tan \left[ \frac{1}{2}.{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+\frac{1}{2}.{{\cos }^{-1}}\frac{1-{{y}^{2}}}{1+{{y}^{2}}} \right]=\frac{x+y}{1-xy}\]

Prove that

\[{{\tan }^{-1}}\left[ \frac{3{{a}^{2}}x-{{x}^{3}}}{{{a}^{3}}-3a{{x}^{2}}} \right]=3{{\tan }^{-1}}\left( \frac{x}{a} \right)\]

Solution:

\[Let~~x=a\tan \theta \]

\[\Rightarrow \tan \theta =\frac{x}{a}\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{x}{a} \right)\]

\[\therefore {{\tan }^{-1}}\left[ \frac{3{{a}^{2}}x-{{x}^{3}}}{{{a}^{3}}-3a{{x}^{2}}} \right]={{\tan }^{-1}}\left[ \frac{3{{a}^{2}}a\tan \theta -{{a}^{3}}{{\tan }^{3}}\theta }{{{a}^{3}}-3a.{{a}^{2}}{{\tan }^{2}}\theta } \right]\]

\[={{\tan }^{-1}}\left[ \frac{{{a}^{3}}\left( 3\tan \theta -{{\tan }^{3}}\theta  \right)}{{{a}^{3}}\left( 1-3{{\tan }^{2}}\theta  \right)} \right]={{\tan }^{-1}}\left[ \tan 3\theta  \right]=3\theta \]

\[\therefore {{\tan }^{-1}}\left[ \frac{3{{a}^{2}}x-{{x}^{3}}}{{{a}^{3}}-3a{{x}^{2}}} \right]=3{{\tan }^{-1}}\left( \frac{x}{a} \right)\]

Prove that

\[{{\tan }^{-1}}\left[ \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]=\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x,~~x\in \left( 0,\frac{\pi }{4} \right)\]

Solution:

\[Let~~x=\cos 2\theta \]

\[\Rightarrow 2\theta ={{\cos }^{-1}}x\Rightarrow \theta =\frac{1}{2}.{{\cos }^{-1}}x\]

Now,

\[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}\]

\[=\frac{\sqrt{2{{\cos }^{2}}\theta }-\sqrt{2{{\sin }^{2}}\theta }}{\sqrt{2{{\cos }^{2}}\theta }+\sqrt{2{{\sin }^{2}}\theta }}~\]

\[\left[ \because 1+\cos 2\theta =2{{\cos }^{2}}\theta ~~and~~1-\cos 2\theta =2{{\sin }^{2}}\theta  \right]\]

\[=\frac{\sqrt{2}\left( \cos \theta -\sin \theta  \right)}{\sqrt{2}\left( \cos \theta +\sin \theta  \right)}=\frac{1-\tan \theta }{1+\tan \theta }\]

\[=\frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}.\tan \theta }=\tan \left( \frac{\pi }{4}-\theta  \right)\]

\[\therefore {{\tan }^{-1}}\left[ \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]=\frac{\pi }{4}-\theta \]

\[\left[ \because 0<\theta <\frac{\pi }{4}\Rightarrow 0\le \frac{\pi }{4}-\theta <\frac{\pi }{4} \right]\]

\[\therefore {{\tan }^{-1}}\left[ \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]=\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x\]

Some More Important Formulas about Inverse Trigonometric Function

A.

\[(i){{\sin }^{-1}}x+{{\sin }^{-1}}y={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right),\]

\[where~~-1\le x,y\le 1~~and~~{{x}^{2}}+{{y}^{2}}\le 1\]

\[(ii){{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right),\]

\[where~~-1\le x,y\le 1~~and~~{{x}^{2}}+{{y}^{2}}\le 1\]

B.

\[(i){{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)\]

\[where~~-1\le x,y\le 1~~and~~x+y\ge 0\]

\[(ii){{\cos }^{-1}}x-{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)\]

\[where~~-1\le x,y\le 1~~and~~x\le y\]

C.

\[(i){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right),~~if~~xy<1\]

\[(ii){{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right),\]

\[~if~~x>0,y>0~~and~~xy>1\]

\[(iii){{\tan }^{-1}}x+{{\tan }^{-1}}y=-\pi +{{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right),\]

\[~if~~x<0,y<0~~and~~xy>1\]

\[(iv){{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right),~~~if~~xy>-1\]

\[(v){{\tan }^{-1}}x-{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right),~\]

\[if~~x>0,y<0~~and~~xy<-1\]

D.

\[(i){{\cot }^{-1}}x+{{\cot }^{-1}}y={{\cot }^{-1}}\left( \frac{xy-1}{y+x} \right)\]

\[(ii){{\cot }^{-1}}x-{{\cot }^{-1}}y={{\cot }^{-1}}\left( \frac{xy+1}{y-x} \right)\]

E.

\[(i)2{{\sin }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right),where~~-\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}\]

\[(ii)3{{\sin }^{-1}}x={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right),where~~-\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}\]

F.

\[(i)2{{\cos }^{-1}}x={{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right),where~~~0\le x\le 1\]

\[(ii)3{{\cos }^{-1}}x={{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right),where~~~\frac{1}{2}\le x\le 1\]

G.

\[(i)2{{\tan }^{-1}}x={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right),where~~~-1\le x\le 1\]

\[(ii)2{{\tan }^{-1}}x={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right),where~~~0<x<\infty \]

\[(iii)2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right),where~~~-1<x<1\]

\[(iv)3{{\tan }^{-1}}x={{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right),where~~~-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\]

H.

\[(i){{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z={{\tan }^{-1}}\left( \frac{x+y+z-xyz}{1-xy-yz-zx} \right)\]

Prove that

\[{{\sin }^{-1}}\frac{12}{13}+{{\cos }^{-1}}\frac{4}{5}+{{\tan }^{-1}}\frac{63}{16}=\pi \]

Solution:

\[Let~~{{\sin }^{-1}}\frac{12}{13}=\alpha \Rightarrow \sin \alpha =\frac{12}{13}\]

\[\therefore \cos \alpha =\sqrt{1-{{\sin }^{2}}\alpha }=\sqrt{1-\frac{144}{169}}=\frac{5}{13}\]

\[\therefore \tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{12}{5}\]

\[Let~~{{\cos }^{-1}}\frac{4}{5}=\beta \Rightarrow \cos \beta =\frac{4}{5}\]

\[\therefore \sin \beta =\sqrt{1-{{\cos }^{2}}\beta }=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\]

\[\therefore \tan \beta =\frac{\sin \beta }{\cos \beta }=\frac{3}{4}\]

\[And~~let~~{{\tan }^{-1}}\frac{63}{16}=\gamma \Rightarrow \tan \gamma =\frac{63}{16}\]

Now

\[\tan \left( \alpha +\beta  \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }\]

\[=\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}.\frac{3}{4}}=\frac{48+15}{20-36}=-\frac{63}{16}\]

\[\therefore \tan \left( \alpha +\beta  \right)=-\tan \gamma \]

\[\Rightarrow \tan \left( \alpha +\beta  \right)=\tan \left( -\gamma  \right)=\tan \left( \pi -\gamma  \right)\]

\[\therefore \alpha +\beta =-\gamma ~~else~~\alpha +\beta =\pi -\gamma \Rightarrow \alpha +\beta +\gamma =\pi \]

\[But~~\alpha ,\beta ,\gamma ~~are~~positive~~as~~\sin \alpha =\frac{12}{13},\cos \beta =\frac{4}{5},\tan \gamma =\frac{63}{16}\]

\[\therefore \alpha +\beta \ne -\gamma ~\]

\[\therefore \alpha +\beta +\gamma =\pi \]

\[i.e.,{{\sin }^{-1}}\frac{12}{13}+{{\cos }^{-1}}\frac{4}{5}+{{\tan }^{-1}}\frac{63}{16}=\pi \]

Show that

\[{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}}\tan \frac{x}{2} \right)=\frac{1}{2}{{\cos }^{-1}}\left( \frac{1+2\cos x}{2+\cos x} \right)\]

Solution:

\[{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}}\tan \frac{x}{2} \right)\]

\[=\frac{1}{2}\left[ 2{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}}\tan \frac{x}{2} \right) \right]\]

\[=\frac{1}{2}{{\cos }^{-1}}\left[ \frac{1-\frac{1}{3}{{\tan }^{2}}\frac{x}{2}}{1+\frac{1}{3}{{\tan }^{2}}\frac{x}{2}} \right]~~\left[ \because ~~2{{\tan }^{-1}}A={{\cos }^{-1}}\frac{1-{{A}^{2}}}{1+{{A}^{2}}} \right]\]

\[=\frac{1}{2}{{\cos }^{-1}}\left[ \frac{3-{{\tan }^{2}}\frac{x}{2}}{3+{{\tan }^{2}}\frac{x}{2}} \right]\]

\[=\frac{1}{2}{{\cos }^{-1}}\left[ \frac{3-\frac{1-\cos x}{1+\cos x}}{3+\frac{1-\cos x}{1+\cos x}} \right]~~\left[ \because ~~{{\tan }^{2}}\frac{x}{2}=\frac{1-\cos x}{1+\cos x} \right]\]

\[=\frac{1}{2}{{\cos }^{-1}}\left[ \frac{3\left( 1+\cos x \right)-\left( 1-\cos x \right)}{3\left( 1+\cos x \right)+\left( 1-\cos x \right)} \right]\]

\[=\frac{1}{2}{{\cos }^{-1}}\left[ \frac{2+4\cos x}{4+2\cos x} \right]\]

\[=\frac{1}{2}{{\cos }^{-1}}\left[ \frac{2\left( 1+2\cos x \right)}{2\left( 2+\cos x \right)} \right]\]

\[=\frac{1}{2}{{\cos }^{-1}}\left( \frac{1+2\cos x}{2+\cos x} \right)\]

Prove that

\[2{{\tan }^{-1}}\left[ \tan \frac{\alpha }{2}.\tan \left( \frac{\pi }{4}-\frac{\beta }{2} \right) \right]={{\tan }^{-1}}\left( \frac{\sin \alpha \cos \beta }{\cos \alpha +\sin \beta } \right)\]

Solution:

\[2{{\tan }^{-1}}\left[ \tan \frac{\alpha }{2}.\tan \left( \frac{\pi }{4}-\frac{\beta }{2} \right) \right]\]

\[={{\tan }^{-1}}\left[ \frac{2\tan \frac{\alpha }{2}.\tan \left( \frac{\pi }{4}-\frac{\beta }{2} \right)}{1-{{\tan }^{2}}\frac{\alpha }{2}.{{\tan }^{2}}\left( \frac{\pi }{4}-\frac{\beta }{2} \right)} \right]~~\left[ \because ~~2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right) \right]\]

\[={{\tan }^{-1}}\left[ \frac{2\frac{\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}}.\frac{\sin \left( \frac{\pi }{4}-\frac{\beta }{2} \right)}{\cos \left( \frac{\pi }{4}-\frac{\beta }{2} \right)}}{1-\frac{{{\sin }^{2}}\frac{\alpha }{2}}{{{\cos }^{2}}\frac{\alpha }{2}}.\frac{{{\sin }^{2}}\left( \frac{\pi }{4}-\frac{\beta }{2} \right)}{{{\cos }^{2}}\left( \frac{\pi }{4}-\frac{\beta }{2} \right)}} \right]\]

\[={{\tan }^{-1}}\left[ \frac{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\sin \left( \frac{\pi }{4}-\frac{\beta }{2} \right)\cos \left( \frac{\pi }{4}-\frac{\beta }{2} \right)}{{{\cos }^{2}}\frac{\alpha }{2}{{\cos }^{2}}\left( \frac{\pi }{4}-\frac{\beta }{2} \right)-{{\sin }^{2}}\frac{\alpha }{2}{{\sin }^{2}}\left( \frac{\pi }{4}-\frac{\beta }{2} \right)} \right]\]

\[={{\tan }^{-1}}\left[ \frac{1}{2}.\frac{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}2\sin \left( \frac{\pi }{4}-\frac{\beta }{2} \right)\cos \left( \frac{\pi }{4}-\frac{\beta }{2} \right)}{\left\{ \cos \frac{\alpha }{2}\cos \left( \frac{\pi }{4}-\frac{\beta }{2} \right)+\sin \frac{\alpha }{2}\sin \left( \frac{\pi }{4}-\frac{\beta }{2} \right) \right\}\left\{ \cos \frac{\alpha }{2}\cos \left( \frac{\pi }{4}-\frac{\beta }{2} \right)-\sin \frac{\alpha }{2}\sin \left( \frac{\pi }{4}-\frac{\beta }{2} \right) \right\}} \right]\]

\[={{\tan }^{-1}}\left[ \frac{\sin \alpha .\sin \left( \frac{\pi }{2}-\beta  \right)}{2\cos \left( \frac{\alpha }{2}+\frac{\pi }{4}-\frac{\beta }{2} \right)\cos \left( \frac{\alpha }{2}-\frac{\pi }{4}+\frac{\beta }{2} \right)} \right]\]

\[={{\tan }^{-1}}\left[ \frac{\sin \alpha .\cos \beta }{\cos \alpha +\cos \left( \frac{\pi }{2}-\beta  \right)} \right]={{\tan }^{-1}}\left( \frac{\sin \alpha \cos \beta }{\cos \alpha +\sin \beta } \right)\]

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