In Engineering, Construction, and Architecture, inverse trigonometric ratios are frequently used. Inverse trigonometric ratios are the most straightforward approach to obtaining an unknown angle. Therefore, we utilise them in locations where we need to find the angle and rapidly acquire the desired result. They are also used to measure a right-angled triangle’s unknown angles, the depth of a hole and the angle of inclination.

Architects use inverse trigonometric functions to determine the angle of a bridge and its supports. Carpenters use this tool to achieve a specific cut angle. Let us learn more about inverse trigonometric functions in the following article.

Inverses of Trigonometric Functions

The concepts of inverse functions and their types are used to define the inverses of trigonometric functions and study their properties. An inverse trigonometric function performs the inverse operation of a trigonometric function. We already know that trigonometric functions are based on right-angle triangles. When two sides of a right triangle’s measure are known, the six main functions are employed to obtain the angle measure.

The inverse trigonometric functions are denoted by using the prefix ‘arc’ since, for a given value of trigonometric functions, the length of arc of a unit circle is produced which is needed to obtain that particular value. Hence, the functions are \(\arcsin \,x,\,\arccos \,x,\,\arctan \,x,\,{\text{arccosec x, arcsec x}}\) and \(\operatorname{arccot} \,x\). They are also denoted as \({\sin ^{ – 1}}\,x,\,{\cos ^{ – 1}}\,x,\,{\tan ^{ – 1}}\,x,\,{\cot ^{ – 1}}\,x,\,{\sec ^{ – 1}}\,x\) and \({{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x\).

As trigonometric functions are periodic, in general, we can say that all trigonometric functions are not bijections. Consequently, their inverses do not exist. However, if we restrict their domains and co-domains, they become bijections, and we can obtain their inverses.

Domain, Ranges and Principal Values

Domains, ranges and principal value branches of all inverse trigonometric functions are tabulated below

Function	Domain	Range	Principal Value Branch
\({\sin ^{ – 1}}\)	\({\rm{[ – 1,}}\,{\rm{1]}}\)	\(\left[ { – \frac{\pi }{2},\,\frac{\pi }{2}} \right]\)	\( – \frac{\pi }{2} \le y \le \frac{\pi }{2}\), where \(y = {\sin ^{ – 1}}\,x\)
\({\cos ^{ – 1}}\)	\({\rm{[ – 1,}}\,{\rm{1]}}\)	\([0,\,\pi ]\)	\(0 \le y \le \pi \), where \(y = {\cos ^{ – 1}}\,x\)
\({{{\tan }^{ – 1}}}\)	\(R\)	\(\left( { – \frac{\pi }{2},\,\frac{\pi }{2}} \right)\)	\({ – \frac{\pi }{2} < y < \frac{\pi }{2}}\), where \({y = {{\tan }^{ – 1}}\,x}\)
\({{\mathop{\rm cosec}\nolimits} ^{ – 1}}\)	\(( – \infty ,\, – 1] \cup [1,\,\infty )\)	\(\left[ { – \frac{\pi }{2},\,\frac{\pi }{2}} \right] – \left\{ {0} \right\}\)	\( – \frac{\pi }{2} \le y \le \frac{\pi }{2}\), where \(y = {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x,\,y \ne 0\)
\({\sec ^{ – 1}}\)	\(( – \infty ,\, – 1] \cup [1,\,\infty )\)	\([0,\,\pi ] – \left\{ {\frac{\pi }{2}} \right\}\)	\(0 \le y \le \pi \), where \(y = {\sec ^{ – 1}}\,x,\,y \ne \frac{\pi }{2}\)
\({\cot ^{ – 1}}\)	\(R\)	\((0,\,\pi )\)	\(0 < y < \pi \), where \(y = {\cot ^{ – 1}}\,x\)

Properties of Inverse Trigonometric Functions

 Some important properties of inverse trigonometric functions will be demonstrated. These conclusions are valid within the principal value branches of the inverse trigonometric functions and wherever they are defined. Some inverse trigonometric function results may not be valid for all domain values. They will only be valid for a subset of \(x\) values for which inverse trigonometric functions exist.

Property 1: Properties of Inverse Trigonometric Functions of the Form \({f^{ – 1}}(f(x))\)

• \({\sin ^{ – 1}}(\sin \,x) = x\) for all \(x \in \left[ { – \frac{\pi }{2},\,\frac{\pi }{2}} \right]\)
• \({\cos ^{ – 1}}(\cos \,x) = x\) for all \(x \in [0,\,\pi ]\)
• \({\tan ^{ – 1}}(\tan \,x) = x\) for all \(x \in \left( { – \frac{\pi }{2},\,\frac{\pi }{2}} \right)\)
• \({{\mathop{\rm cosec}\nolimits} ^{ – 1}}({\mathop{\rm cosec}\nolimits} \,x) = x\) for all \(x \in \left[ { – \frac{\pi }{2},\,0} \right) \cup \left( {0,\,\frac{\pi }{2}} \right]\)
• \({\sec ^{ – 1}}(\sec \,x) = x\) for all \(x \in \left[ {0,\,\frac{\pi }{2}} \right) \cup \left( {\frac{\pi }{2},\,\pi } \right]\)
• \({\cot ^{ – 1}}(\cot \,x) = x\) for all \(x \in (0,\,\pi )\)

Here, we observe that the relations between trigonometric functions and their inverses hold for specific values of \(x\). If \(x\) does not lie in the domain of a trigonometric function in which it is not a bijection, then the above relations do not hold good. 

Property 2: Properties of Inverse Trigonometric Functions of the Form \(f\left( {{f^{ – 1}}(x)} \right)\)

If \(f:A \to B\) is a bijection, then \({f^{ – 1}}:B \to A\) exists such that \(f \circ {f^{ – 1}}(x) = x\) or \(f\left( {{f^{ – 1}}(x)} \right) = x\) for all \(x \in R\). Applying this property to the trigonometric functions and their inverses, we obtain the following.

• \(\sin \left( {{{\sin }^{ – 1}}\,x} \right) = x\) for all \(x \in [ – 1,\,1]\).
• \(\cos \left( {{{\cos }^{ – 1}}\,x} \right) = x\) for all \(x \in [ – 1,\,1]\)
• \(\tan \left( {{{\tan }^{ – 1}}\,x} \right) = x\) for all \(x \in R\)
• \({\mathop{\rm cosec}\nolimits} \left( {{{{\mathop{\rm cosec}\nolimits} }^{ – 1}}\,x} \right) = x\) for all \(x \in (x,\, – 1] \cup [1,\,\infty )\)
• \(\sec \left( {{{\sec }^{ – 1}}\,x} \right) = x\) for all \(x \in ( – \infty ,\, – 1] \cup [1,\,\infty )\)
• \(\cot \left( {{{\cot }^{ – 1}}\,x} \right) = x\) for all \(x \in R\)

This property helps us in finding the values of expression of the form \(f\left( {{g^{ – 1}}(x)} \right)\) where \(f\) and \(g\) are trigonometric functions.

Property 3: Reflection Identities

1. \({\sin ^{ – 1}}( – x) = – {\sin ^{ – 1}}\,x\) for all \(x \in [ – 1,\,1]\)

Proof: Clearly, \( – x \in [ – 1,\,1]\) for all \(x \in [ – 1,\,1]\)

Let \({\sin ^{ – 1}}( – x) = \theta \,……..(i)\)

\( \Rightarrow – x = \sin \,\theta \)

\( \Rightarrow x = – \sin \,\theta \)

\( \Rightarrow x = \sin ( – \theta )\)

\( \Rightarrow – \theta = {\sin ^{ – 1}}\,x \) \( [ \because \,x \in [ – 1,\,1]\) and \(- \theta \in [ – \frac{\pi }{2},\,\frac{\pi }{2} ]\) for all \(\theta \in [ – \frac{\pi }{2},\,\frac{\pi }{2} ] ]\)

\(\therefore \,\theta = – {\sin ^{ – 1}}\,x\,……..(ii)\)

From \((i)\) and \((ii)\), we have

\({\sin ^{ – 1}}( – x) = – {\sin ^{ – 1}}\,x\)

2. \({\cos ^{ – 1}}( – x) = \pi – {\cos ^{ – 1}}\,x\) for all \(x \in [ – 1,\,1]\)

Proof: Clearly, \( – x \in [ – 1,\,1]\) for all \(x \in [ – 1,\,1]\)

Let \({\cos ^{ – 1}}( – x) = \theta \,……..(i)\)

\( \Rightarrow – x = \cos \,\theta \)

\( \Rightarrow x = – \cos \,\theta \)

\( \Rightarrow x = \cos (\pi – \theta )\)

\( \Rightarrow {\cos ^{ – 1}}\,x = \pi – \theta \) \([\therefore \,x \in [ – 1,\,1]\) and \(\pi – \theta \in [0,\,\pi ]\) for all \(\theta \in [0,\,\pi ]]\)

\(\therefore \,\theta = \pi – {\cos ^{ – 1}}x\,……..(ii)\)

From \((i)\) and \((ii)\), we get

\({\cos ^{ – 1}}( – x) = \pi – {\cos ^{ – 1}}\,x\)

Similarly, other results can also be stated as shown below.

3. \({\tan ^{ – 1}}( – x) = – {\tan ^{ – 1}}\,x\) for all \(x \in R\)
4. \({{\mathop{\rm cosec}\nolimits} ^{ – 1}}( – x) = – {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x\), for all \(x \in ( – \infty ,\, – 1] \cup [1,\,\infty )\)
5. \({\sec ^{ – 1}}( – x) = \pi – {\sec ^{ – 1}}\,x\), for all \(x \in ( – \infty ,\, – 1] \cup [1,\,\infty )\)
6. \({\cot ^{ – 1}}( – x) = \pi – {\cot ^{ – 1}}\,x\) for all \(x \in R\)

Property 4: Reciprocal Inverse Identities

i. \({\sin ^{ – 1}}\left( {\frac{1}{x}} \right) = {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x\) for all \(x \in ( – \infty ,\, – 1] \cup [1,\,\infty )\)

Proof: Let \({{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x = \theta \,……..(i)\)

\( \Rightarrow x = {\mathop{\rm cosec}\nolimits} \,\theta \)

\( \Rightarrow \theta = {\sin ^{ – 1}}\frac{1}{x}……..(ii)\) \(\left[ {\therefore \,x \in ( – \infty ,\, – 1] \cup [1,\,\infty ) \Rightarrow \frac{1}{x} \in [ – 1,\,1] – { 0}\;{{{\mathop{\rm cosec}\nolimits} }^{ – 1}}x = \theta \Rightarrow \theta \in \left[ { – \frac{\pi }{2},\,\frac{\pi }{2}} \right] – { 0} } \right]\)

From \((i)\) and \((ii)\), we get

\({\sin ^{ – 1}}\left( {\frac{1}{x}} \right) = {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x\)

ii. \({\cos ^{ – 1}}\left( {\frac{1}{x}} \right) = {\sec ^{ – 1}}\,x\) for all \(x \in ( – \infty ,\, – 1] \cup [1,\,\infty )\)

Proof: Let \({\sec ^{ – 1}}\,x = \theta ……..(i)\)

Then, \(x \in ( – \infty ,\, – 1] \cup [1,\,\infty )\) and \(\theta \in [0,\,\pi ] – \left[ { – \frac{\pi }{2}} \right]\)

Now, \({\sec ^{ – 1}}\,x = \theta \)

\( \Rightarrow x = \sec \,\theta \)

\( \Rightarrow \frac{1}{x} = \cos \,\theta \)

\( \Rightarrow \theta = {\cos ^{ – 1}}\frac{1}{x}\, \ldots \ldots (ii)\) \([ \because \,x \in ( – \infty ,\, – 1] \cup [1,\,\infty ) \Rightarrow \frac{1}{x} \in [ – 1,\,1] – { 0} \) and \(\theta \in [0,\,\pi ] ]\)

From \((i)\) and \((ii)\), we get

\({\cos ^{ – 1}}\left( {\frac{1}{x}} \right) = {\sec ^{ – 1}}\,x\)

iii. \({\tan ^{ – 1}}\left( {\frac{1}{x}} \right) = \left\{ {\begin{array}{*{20}{c}}{{{\cot }^{ – 1}}\,x\quad {\rm{ for }}\,x > 0}\\{ – \pi  + {{\cot }^{ – 1}}\,x,\quad {\rm{ for }}\,x < 0}\end{array}} \right.\)

Proof: Let \({\cot ^{ – 1}}\,x = \theta \ldots \ldots (i)\)

Then, \(x \in R,\,x \ne 0\) and \(\theta \in (0,\,\pi )\)

Now, two cases arise.

Case 1: When \(x > 0\)

In this case, \(\theta \in \left( {0,\,\frac{\pi }{2}} \right)\)

\(\therefore \,{\cot ^{ – 1}}\,x = \theta \)

\( \Rightarrow x = \cot \,\theta \)

\( \Rightarrow \frac{1}{x} = \tan \,\theta \)

\( \Rightarrow \theta = {\tan ^{ – 1}}\left( {\frac{1}{x}} \right)\, \ldots \ldots (ii)\) \(\left[ {\because \,\theta \in \left( {0,\,\frac{\pi }{2}} \right)} \right]\)

From \((i)\) and \((ii)\), we get

\({\tan ^{ – 1}}\left( {\frac{1}{x}} \right) = {\cot ^{ – 1}}\,x\) for all \(x > 0\)

Case 2: When \(x < 0\)

In this case, \(\theta \in \left( {\frac{\pi }{2},\,\pi } \right)\)

Now, \(\frac{\pi }{2} < \theta < \pi \)

\( \Rightarrow – \frac{\pi }{2} < \theta – \pi < 0\)

\( \Rightarrow \theta – \pi \in \left( { – \frac{\pi }{2},\,0} \right)\)

\(\therefore \,{\cot ^{ – 1}}\,x = \theta \)

\( \Rightarrow x = \cot \,\theta \)

\( \Rightarrow \frac{1}{x} = \tan \,\theta \)

\( \Rightarrow \frac{1}{x} = – \tan (\pi – \theta )\)

\( \Rightarrow \frac{1}{x} = \tan (\theta – \pi )\)

\( \Rightarrow \theta – \pi = {\tan ^{ – 1}}\left( {\frac{1}{x}} \right)\) \(\left[ {\because \,\theta – \pi \in \left( { – \frac{\pi }{2},\,0} \right)} \right]\)

\( \Rightarrow {\tan ^{ – 1}}\left( {\frac{1}{x}} \right) = – \pi + \theta \,……..(iii)\)

From equations \((i)\), and \((iii)\), we have

\({\tan ^{ – 1}}\left( {\frac{1}{x}} \right) = – \pi + {\cot ^{ – 1}}\,x\), if \(x < 0\)

Hence,

\({\tan ^{ – 1}}\left( {\frac{1}{x}} \right) = \left\{ {\begin{array}{*{20}{c}}{{{\cot }^{ – 1}}\,x,}&{{\rm{ for }}\;x > 0}\\{ – \pi  + {{\cot }^{ – 1}}\,x,}&{{\rm{ for }}\;x < 0}\end{array}} \right.\)

Property 5: Cofunction Inverse Identities

i. \({\sin ^{ – 1}}\,x + {\cos ^{ – 1}}\,x = \frac{\pi }{2}\), for all \(x \in [ – 1,\,1]\).

Proof :  Let \({\sin ^{ – 1}}\,x = \theta \,……..(i)\)

Then, \(\theta \in \left[ { – \frac{\pi }{2},\,\frac{\pi }{2}} \right]\)

\( \Rightarrow – \frac{\pi }{2} \le \theta \le \frac{\pi }{2} \Rightarrow – \frac{\pi }{2} \le – \theta \le \frac{\pi }{2} \Rightarrow 0 \le \frac{\pi }{2} – \theta \le \pi \Rightarrow \frac{\pi }{2} – \theta \in [0,\,\pi ]\)

Now, \({\sin ^{ – 1}}\,x = \theta \)

\( \Rightarrow x = \cos \left( {\frac{\pi }{2} – \theta } \right)\)

\( \Rightarrow {\cos ^{ – 1}}\,x = \frac{\pi }{2} – \theta \) \([ \because \,x \in [ – 1,\,1]\) and \(( {\frac{\pi }{2} – \theta } ) \in [0,\,\pi ] ]\)

\( \Rightarrow \theta + {\cos ^{ – 1}}\,x = \frac{\pi }{2}\,……..(ii)\)

From \((i)\) and \((ii)\), we get

\({\sin ^{ – 1}}\,x + {\cos ^{ – 1}}\,x = \frac{\pi }{2}\)

ii. \({\tan ^{ – 1}}\,x + {\cot ^{ – 1}}\,x = \frac{\pi }{2}\) for all \(x \in R\)

Proof: Let \(\tan^{ – 1}\,x = \theta \,……..(i)\)

Then, \(\theta \in \left( { – \frac{\pi }{2},\,\frac{\pi }{2}} \right)\)

\( \Rightarrow – \frac{\pi }{2} < \theta < \frac{\pi }{2} \Rightarrow – \frac{\pi }{2} < – \theta < \frac{\pi }{2} \Rightarrow 0 < \frac{\pi }{2} – \theta < \pi \Rightarrow \left( {\frac{\pi }{2} – \theta } \right) \in (0,\,\pi )\)

Now, \({\tan ^{ – 1}}\,x = \theta \)

\( \Rightarrow x = \tan \,\theta \)

\( \Rightarrow x = \cot \left( {\frac{\pi }{2} – \theta } \right)\)

\( \Rightarrow {\cot ^{ – 1}}\,x = \frac{\pi }{2} – \theta \) \(\left[ {\because \,\frac{\pi }{2} – \theta \in (0,\,\pi )} \right]\)

\( \Rightarrow \theta + {\cot ^{ – 1}}\,x = \frac{\pi }{2}\,……..(ii)\)

From \((i)\) and \((ii)\), we have

\({\tan ^{ – 1}}\,x + {\cot ^{ – 1}}\,x = \frac{\pi }{2}\)

iii. \({\sec ^{ – 1}}\,x + {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x = \frac{\pi }{2}\) for all \(x \in ( – \infty ,\, – 1] \cup [1,\,\infty )\)

Proof: Let \(\sec^{ – 1}\,x = \theta \,……..(i)\)

Then, \(\theta  \in [0,\,\pi ] – \left\{ {\frac{\pi }{2}} \right\}\) \([\because \,x \in ( – \infty ,\, – 1] \cup [1,\,\infty )]\)

\( \Rightarrow 0 \le \theta \le \pi ,\,\theta \ne \frac{\pi }{2}\)

\( \Rightarrow – \pi \le – \theta \le 0,\,\theta \ne \frac{\pi }{2}\)

\( \Rightarrow – \frac{\pi }{2} \le \frac{\pi }{2} – \theta \le \frac{\pi }{2},\,\frac{\pi }{2} – \theta \ne 0\)

\( \Rightarrow \left( {\frac{\pi }{2} – \theta } \right) \in \left[ { – \frac{\pi }{2},\,\frac{\pi }{2}} \right]\) and \(\frac{\pi }{2} – \theta \ne 0\)

Now, \({\sec ^{ – 1}}\,x = \theta \)

\( \Rightarrow x = \sec \,\theta \)

\( \Rightarrow x = {\mathop{\rm cosec}\nolimits} \,\left( {\frac{\pi }{2} – \theta } \right)\)

\( \Rightarrow {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x = \frac{\pi }{2} – \theta \) \([ \because \,( {\frac{\pi }{2} – \theta } ) \in [ { – \frac{\pi }{2},\,\frac{\pi }{2}} ]\) and \(\frac{\pi }{2} – \theta \ne 0 ]\)

\( \Rightarrow \theta + {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x = \frac{\pi }{2}\,……..(ii)\)

From \((i)\) and \((ii)\), we have

\({\sec ^{ – 1}}\,x + {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x = \frac{\pi }{2}\)

Property 6: Formulae to find sum or difference of \({\tan ^{ – 1}}\,x\) and \({\tan ^{ – 1}}\,y\)

i. \({\tan ^{ – 1}}\,x + {\tan ^{ – 1}}\,y = \left\{ {\begin{array}{*{20}{c}}{{{\tan }^{ – 1}}\left( {\frac{{x + y}}{{1 – xy}}} \right),{\rm{ if }}\,xy < 1}\\{\pi  + {{\tan }^{ – 1}}\left( {\frac{{x + y}}{{1 – xy}}} \right),{\rm{ if }}\,x > 0,\,y > 0\,{\rm{ and }}\,xy > 1}\\{ – \pi  + {{\tan }^{ – 1}}\left( {\frac{{x + y}}{{1 – xy}}} \right),{\rm{ if }}\,x < 0,\,y < 0\,{\rm{ and }}\,xy > 1}\end{array}} \right.\)

Proof: Let \({\tan ^{ – 1}}\,x = \theta \) and \({\tan ^{ – 1}}\,y = \phi \). Then, \(x = \tan \,\theta ,\,y = \tan \,\phi \)

Now,
\(\tan (\theta + \phi ) = \frac{{\tan \,\theta + \tan \,\phi }}{{1 – \tan \,\theta \,\tan \,\phi }} = \frac{{x + y}}{{1 – xy}}\)

\( \Rightarrow \theta + \phi = {\tan ^{ – 1}}\frac{{x + y}}{{1 – xy}}\)

Hence, \({\tan ^{ – 1}}\,x + {\tan ^{ – 1}}\,y = {\tan ^{ – 1}}\frac{{x + y}}{{1 – xy}}\)

In the above result, if we replace \(y\) by \(-y\), we get the second result that is given below.

ii. \({\tan ^{ – 1}}\,x – {\tan ^{ – 1}}\,y = \left\{ {\begin{array}{*{20}{c}}{{{\tan }^{ – 1}}\left( {\frac{{x – y}}{{1 + xy}}} \right),{\rm{ if }}\,xy >  – 1}\\{\pi  + {{\tan }^{ – 1}}\left( {\frac{{x – y}}{{1 + xy}}} \right),{\rm{ if }}\,x > 0,\,\,y < 0\,{\rm{ and }}\,xy <  – 1}\\{ – \pi  + {{\tan }^{ – 1}}\left( {\frac{{x – y}}{{1 + xy}}} \right),{\rm{ if }}\,x < 0,\,y > 0\,{\rm{ and }}\,xy <  – 1}\end{array}} \right.\)

Property 7: Formulae of \(2\,{\tan ^{ – 1}}\,x\)

i. \(2\,{\tan ^{ – 1}}\,x = {\sin ^{ – 1}}\frac{{2x}}{{1 + {x^2}}},\,|x| \le 1\)

Proof: Let \({\tan ^{ – 1}}\,x = y\)

\( \Rightarrow x = \tan \,y\)

Now, \({\sin ^{ – 1}}\frac{{2x}}{{1 + {x^2}}} = {\sin ^{ – 1}}\frac{{2\,\tan \,y}}{{1 + {{\tan }^2}\,y}}\)

\( = {\sin ^{ – 1}}(\sin \,2y)\)

\({\rm{ = 2}}y\)

\( = 2\,{\tan ^{ – 1}}\,x\)

Hence, \(2\,{\tan ^{ – 1}}\,x = {\sin ^{ – 1}}\frac{{2x}}{{1 + {x^2}}},\,|x| \le 1\)

ii. \(2\,{\tan ^{ – 1}}\,x = {\cos ^{ – 1}}\frac{{1 – {x^2}}}{{1 + {x^2}}},\,x \ge 0\)

Proof: Let \({\tan ^{ – 1}}\,x = y\)

\( \Rightarrow x = \tan \,y\)

Now, \({\cos ^{ – 1}}\frac{{1 – {x^2}}}{{1 + {x^2}}} = {\cos ^{ – 1}}\frac{{1 – {{\tan }^2}\,y}}{{1 + {{\tan }^2}\,y}}\)

\( = {\cos ^{ – 1}}(\cos \,2y)\)

\( = 2y\)

\( = 2\,{\tan ^{ – 1}}\,x\)

Hence, \(2\,{\tan ^{ – 1}}\,x = {\cos ^{ – 1}}\frac{{1 – {x^2}}}{{1 + {x^2}}},\,x \ge 0\)

iii. \(2\,{\tan ^{ – 1}}\,x = {\tan ^{ – 1}}\frac{{2x}}{{1 – {x^2}}},\, – 1 < x < 1\)

Proof: Let \({\tan ^{ – 1}}\,x = y\)

\( \Rightarrow x = \tan \,y\)

Now, \({\tan ^{ – 1}}\frac{{2x}}{{1 – {x^2}}} = {\tan ^{ – 1}}\frac{{2\,\tan \,y}}{{1 – {{\tan }^2}\,y}}\)

\( = {\tan ^{ – 1}}\,\tan \,2y\)

\( = 2y\)

\(= 2\,{\tan ^{ – 1}}\,x\)

Hence, \(2\,{\tan ^{ – 1}}\,x = {\tan ^{ – 1}}\frac{{2x}}{{1 – {x^2}}},\, – 1 < x < 1\)

Solved Examples – Properties of Inverse Trigonometric Functions

Q.1. Express \({\tan ^{ – 1}}\left( {\frac{{\cos \,x}}{{1 – \sin \,x}}} \right),\, – \frac{{3\pi }}{2} < x < \frac{\pi }{2}\) in the simplest form.
Ans:
\({\tan ^{ – 1}}\left( {\frac{{\cos \,x}}{{1 – \sin \,x}}} \right) = {\tan ^{ – 1}}\left[ {\frac{{{{\cos }^2}\frac{x}{2} – {{\sin }^2}\frac{x}{2}}}{{{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} – 2\,\sin \frac{x}{2}\cos \frac{x}{2}}}} \right]\)
\( = {\tan ^{ – 1}}\left[ {\frac{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)\left( {\cos \frac{x}{2} – \sin \frac{x}{2}} \right)}}{{{{\left( {\cos \frac{x}{2} – \sin \frac{x}{2}} \right)}^2}}}} \right]\)
\( = {\tan ^{ – 1}}\left[ {\frac{{\cos \frac{x}{2} + \sin \frac{x}{2}}}{{\cos \frac{x}{2} – \sin \frac{x}{2}}}} \right]\)
\( = {\tan ^{ – 1}}\left[ {\frac{{1 + \tan \frac{x}{2}}}{{1 – \tan \frac{x}{2}}}} \right]\) [Dividing numerator and denominator by \(\cos \frac{x}{2}\).]
\( = {\tan ^{ – 1}}\left[ {\frac{{\tan \frac{\pi }{4} + \tan \frac{x}{2}}}{{1 – \tan \frac{\pi }{4}\tan \frac{x}{2}}}} \right]\)
\( = {\tan ^{ – 1}}\left[ {\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)} \right]\)
\( = \frac{\pi }{4} + \frac{x}{2}\)
Hence,
\({\tan ^{ – 1}}\left( {\frac{{\cos \,x}}{{1 – \sin \,x}}} \right) = \frac{\pi }{4} + \frac{x}{2}\), where \( – \frac{{3\pi }}{2} < x < \frac{\pi }{2}\)

Q.2. Evaluate: \(\cos \left( {{{\sin }^{ – 1}}\frac{1}{4} + {{\sec }^{ – 1}}\frac{4}{3}} \right)\).
Ans: Given:\(\cos \left( {{{\sin }^{ – 1}}\frac{1}{4} + {{\sec }^{ – 1}}\frac{4}{3}} \right)\)
\( = \cos \left( {{{\sin }^{ – 1}}\frac{1}{4} + {{\cos }^{ – 1}}\frac{3}{4}} \right)\left[ {{{\sec }^{ – 1}}\frac{4}{3} = {{\cos }^{ – 1}}\frac{3}{4}} \right]\)
\( = \cos \left( {{{\sin }^{ – 1}}\frac{1}{4}} \right)\cos \left( {{{\cos }^{ – 1}}\frac{3}{4}} \right) – \sin \left( {{{\sin }^{ – 1}}\frac{1}{4}} \right)\sin \left( {{{\cos }^{ – 1}}\frac{3}{4}} \right)\)
\( = \cos \left( {{{\cos }^{ – 1}}\frac{{\sqrt {15} }}{4}} \right)\cos \left( {{{\cos }^{ – 1}}\frac{3}{4}} \right) – \sin \left( {{{\sin }^{ – 1}}\frac{1}{4}} \right)\sin \left( {{{\sin }^{ – 1}}\frac{{\sqrt 7 }}{4}} \right)\) \(\,\left[ {\because \,{{\sin }^{ – 1}}\frac{1}{4} = {{\cos }^{ – 1}}\frac{{\sqrt {15} }}{4}\,{\rm{and }}\,{{\cos }^{ – 1}}\frac{3}{4} = {{\sin }^{ – 1}}\frac{{\sqrt 7 }}{4}} \right]\)
\( = \frac{{\sqrt {15} }}{4} \times \frac{3}{4} – \frac{1}{4} \times \frac{{\sqrt 7 }}{4}\)
\( = \frac{{3\sqrt {15} – \sqrt 7 }}{{16}}\)
Hence, \(\cos \left( {{{\sin }^{ – 1}}\frac{1}{4} + {{\sec }^{ – 1}}\frac{4}{3}} \right) = \frac{{3\sqrt {15} – \sqrt 7 }}{{16}}\)

Q.3. If \({\sin ^{ – 1}}\,x + {\sin ^{ – 1}}\,y + {\sin ^{ – 1}}\,z = \pi \), prove that
\(x\sqrt {1 – {x^2}} + y\sqrt {1 – {y^2}} + z\sqrt {1 – {z^2}} = 2xyz\)
Ans: Let \({\sin ^{ – 1}}\,x = A,\,{\sin ^{ – 1}}\,y = B\), and \({\sin ^{ – 1}}\,z = C\).
Then, \(x = \sin \,A,\,y = \sin \,B\), and \(z = \sin \,C\)
Now,
\({\sin ^{ – 1}}\,x + {\sin ^{ – 1}}\,y + {\sin ^{ – 1}}\,z = \pi \)
\( \Rightarrow A + B + C = \pi \)
\( \Rightarrow \sin \,2A + \sin \,2B + \sin \,2C = 4\,\sin \,A\,\sin \,B\,\sin \,C\)
\( \Rightarrow 2\,\sin \,A\,\cos \,A + 2\,\sin \,B\,\cos \,B + 2\,\sin \,C\,\cos \,C = 4\,\sin \,A\,\sin \,B\,\sin \,C\)
\( \Rightarrow \sin \,A\sqrt {1 – {{\sin }^2}\,A} + \sin \,B\sqrt {1 – {{\sin }^2}\,B} + \sin \,C\sqrt {1 – {{\sin }^2}\,C} = 2\,\sin \,A\,\sin \,B\,\sin \,C\)
\( \Rightarrow x\sqrt {1 – {x^2}} + y\sqrt {1 – {y^2}} + z\sqrt {1 – {z^2}} = 2xyz\)
Hence, proved.

Q.4. If \({\tan ^{ – 1}}\left\{ {\frac{{\sqrt {1 + {x^2}}  – \sqrt {1 – {x^2}} }}{{\sqrt {1 + {x^2}}  + \sqrt {1 – {x^2}} }}} \right\} = \alpha \), then prove that \({x^2} = \sin \,2\alpha \).
Ans: Given: \({\tan ^{ – 1}}\left\{ {\frac{{\sqrt {1 + {x^2}}  – \sqrt {1 – {x^2}} }}{{\sqrt {1 + {x^2}}  + \sqrt {1 – {x^2}} }}} \right\} = \alpha \)
\( \Rightarrow \frac{{\sqrt {1 + {x^2}} – \sqrt {1 – {x^2}} }}{{\sqrt {1 + {x^2}} + \sqrt {1 – {x^2}} }} = \tan \,\alpha \)
\( \Rightarrow \frac{{\left( {\sqrt {1 + {x^2}} – \sqrt {1 – {x^2}} } \right) + \left( {\sqrt {1 + {x^2}} + \sqrt {1 – {x^2}} } \right)}}{{\left( {\sqrt {1 + {x^2}} – \sqrt {1 – {x^2}} } \right) – \left( {\sqrt {1 + {x^2}} + \sqrt {1 – {x^2}} } \right)}} = \frac{{\tan \,\alpha + 1}}{{\tan \,\alpha – 1}}\)
\( \Rightarrow \frac{{2\sqrt {1 + {x^2}} }}{{ – 2\sqrt {1 – {x^2}} }} = \frac{{\tan \,\alpha + 1}}{{\tan \,\alpha – 1}}\)
\( \Rightarrow \frac{{\sqrt {1 – {x^2}} }}{{\sqrt {1 + {x^2}} }} = \frac{{1 – \tan \,\alpha }}{{1 + \tan \,\alpha }}\)
\( \Rightarrow \sqrt {\frac{{1 – {x^2}}}{{1 + {x^2}}}} = \frac{{\cos \,\alpha – \sin \,\alpha }}{{\cos \,\alpha + \sin \,\alpha }}\)
\( \Rightarrow \frac{{1 – {x^2}}}{{1 + {x^2}}} = {\left( {\frac{{\cos \,\alpha – \sin \,\alpha }}{{\cos \,\alpha + \sin \,\alpha }}} \right)^2}\)
\( \Rightarrow \frac{{1 – {x^2}}}{{1 + {x^2}}} = \frac{{{{\cos }^2}\,\alpha + {{\sin }^2}\,\alpha – 2\,\sin \,\alpha \,\cos \,\alpha }}{{{{\cos }^2}\,\alpha + {{\sin }^2}\,\alpha + 2\,\sin \,\alpha \,\cos \,\alpha }}\)
\( \Rightarrow \frac{{1 – {x^2}}}{{1 + {x^2}}} = \frac{{1 – \sin \,2\alpha }}{{1 + \sin \,2\alpha }}\)
\( \Rightarrow {x^2} = \sin \,2\alpha \)
Hence, proved.

Q.5. If \({\left( {{{\tan }^{ – 1}}\,x} \right)^2} + {\left( {{{\cot }^{ – 1}}\,x} \right)^2} = \frac{{5{\pi ^2}}}{8}\) then find \(x\).
Ans: Given: \({\left( {{{\tan }^{ – 1}}\,x} \right)^2} + {\left( {{{\cot }^{ – 1}}\,x} \right)^2} = \frac{{5{\pi ^2}}}{8}\)
\( \Rightarrow {\left( {{{\tan }^{ – 1}}\,x} \right)^2} + {\left( {{{\cot }^{ – 1}}\,x} \right)^2} + 2\,{\tan ^{ – 1}}\,x\,{\cot ^{ – 1}}\,x – 2\,{\tan ^{ – 1}}\,x\,{\cot ^{ – 1}}\,x = \frac{{5{\pi ^2}}}{8}\)
\( \Rightarrow {\left( {{{\tan }^{ – 1}}\,x + {{\cot }^{ – 1}}\,x} \right)^2} – 2\,{\tan ^{ – 1}}x\,{\cot ^{ – 1}}\,x = \frac{{5{\pi ^2}}}{8}\)
\( \Rightarrow \frac{{{\pi ^2}}}{4} – 2\,{\tan ^{ – 1}}\,x\left( {\frac{\pi }{2} – {{\tan }^{ – 1}}\,x} \right) = \frac{{5{\pi ^2}}}{8}\) \(\left[ {\because \,{{\tan }^{ – 1}}\,x + {{\cot }^{ – 1}}\,x = \frac{\pi }{2}} \right]\)
\( \Rightarrow \frac{{{\pi ^2}}}{4} – \pi \,{\tan ^{ – 1}}\,x + 2{\left( {{{\tan }^{ – 1}}\,x} \right)^2} = \frac{{5{\pi ^2}}}{8}\)
\( \Rightarrow 2{\left( {{{\tan }^{ – 1}}\,x} \right)^2} – \pi \,{\tan ^{ – 1}}\,x – \frac{{3{\pi ^2}}}{8} = 0\)
\( \Rightarrow 16{\left( {{{\tan }^{ – 1}}\,x} \right)^2} – 8\pi \left( {{{\tan }^{ – 1}}\,x} \right) – 3{\pi ^2} = 0\)
\( \Rightarrow 16{\left( {{{\tan }^{ – 1}}\,x} \right)^2} – 12\pi \,{\tan ^{ – 1}}\,x + 4\pi \,{\tan ^{ – 1}}\,x – 3{\pi ^2} = 0\)
\( \Rightarrow 4\,{\tan ^{ – 1}}\,x\left( {4\,{{\tan }^{ – 1}}\,x – 3\pi } \right) + \pi \left( {4\,{{\tan }^{ – 1}}\,x – 3\pi } \right) = 0\)
\( \Rightarrow \left( {4\,{{\tan }^{ – 1}}\,x – 3\pi } \right)\left( {4\,{{\tan }^{ – 1}}\,x + \pi } \right) = 0\)
\( \Rightarrow 16\left( {{{\tan }^{ – 1}}\,x – \frac{{3\pi }}{4}} \right)\left( {{{\tan }^{ – 1}}\,x + \frac{\pi }{4}} \right) = 0\)
\( \Rightarrow {\tan ^{ – 1}}\,x + \frac{\pi }{4} = 0\) \(\left[ {\because \, – \frac{\pi }{2} < {{\tan }^{ – 1}}\,x < \frac{\pi }{2}\,\therefore \,{{\tan }^{ – 1}}\,x – \frac{{3\pi }}{4} \ne 0} \right]\)

\( \Rightarrow {\tan ^{ – 1}}\,x = – \frac{\pi }{4}\)

\( \Rightarrow x = \tan \left( { – \frac{\pi }{4}} \right)\)

Hence, \(x = – 1\)

Summary

We have learned about the various properties of the six inverse trigonometric functions. These properties are useful in simplifying expressions and solving equations involving inverse trigonometric functions. The properties of these functions are based on the concepts of the inverse of functions, types of functions and bijective functions such as if \(f:A \to B\) is a bijection, then \({f^{ – 1}}:B \to A\) exists such that \({f^{ – 1}}(f(x)) = x\) for all \(x \in A\). We have also learned the reflection identities such as \({\sin ^{ – 1}}( – x) = – {\sin ^{ – 1}}\,x\) for all \(x \in [ – 1,\,1],\,{\cos ^{ – 1}}( – x) = \pi – {\cos ^{ – 1}}\,x\) for all \(x \in [ – 1,\,1]\), etc. The reciprocal inverse identities are \({\sin ^{ – 1}}\left( {\frac{1}{x}} \right) = {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x\) for all \(x \in ( – \infty ,\, – 1] \cup [1,\,\infty ),\,{\cos ^{ – 1}}\left( {\frac{1}{x}} \right) = {\sec ^{ – 1}}\,x\) for all \(x \in \left( { – \infty ,\, – 1} \right] \cup [1,\,\infty ),\,{\tan ^{ – 1}}\left( {\frac{1}{x}} \right) =\left\{ {\begin{array}{*{20}{c}}{{{\cot }^{ – 1}}\,x,}&{x > 0}\\{ – \pi  + {{\cot }^{ – 1}}\,x,}&{x < 0}\end{array}} \right.\). We have also learned the formulae to find the sum and difference of \({\tan ^{ – 1}}\,x\) and \({\tan ^{ – 1}}\,y\) and \(2\,{\tan ^{ – 1}}\,x\). We have also provided the solved examples based on all these properties.

Frequently Asked Questions (FAQs)

Q.1. What are the elementary properties of inverse trigonometric functions?
Ans: The elementary properties of inverse trigonometric functions are as follows:

• \({\sin ^{ – 1}}\frac{1}{x} = {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x,\,x \ge 1\) or \(x \le – 1\)
• \({\cos ^{ – 1}}\frac{1}{x} = {\sec ^{ – 1}}\,x,\,x \ge 1\) or \(x \le – 1\)
• \({\tan ^{ – 1}}\frac{1}{x} = {\cot ^{ – 1}}\,x,\,x > 0\)
• \({\sin ^{ – 1}}( – x) = – {\sin ^{ – 1}}\,x,\,x \in [ – 1,\,1]\)
• \({\tan ^{ – 1}}( – x) = – {\tan ^{ – 1}}\,x,\,x \in R\)
• \({{\mathop{\rm cosec}\nolimits} ^{ – 1}}( – x) = – {{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x,\,|x| \ge 1\)
• \({\cos ^{ – 1}}( – x) = \pi – {\cos ^{ – 1}}\,x,\,x \in [ – 1,\,1]\)
• \({\sec ^{ – 1}}( – x) = \pi – {\sec ^{ – 1}}\,x,\,|x| \ge 1\)
• \({\cot ^{ – 1}}( – x) = \pi – {\cot ^{ – 1}}\,x,\,x \in R\)
• \({\sin ^{ – 1}}\,x + {\cos ^{ – 1}}\,x = \frac{\pi }{2},\,x \in [ – 1,\,1]\)
• \({\tan ^{ – 1}}\,x + {\cot ^{ – 1}}\,x = \frac{\pi }{2},\,x \in R\)
• \({{\mathop{\rm cosec}\nolimits} ^{ – 1}}\,x + {\sec ^{ – 1}}\,x = \frac{\pi }{2},\,|x| \ge 1\)
• \({\tan ^{ – 1}}\,x + {\tan ^{ – 1}}\,y = {\tan ^{ – 1}}\frac{{x + y}}{{1 – xy}};\,xy < 1\)
• \({\tan ^{ – 1}}\,x – {\tan ^{ – 1}}\,y = {\tan ^{ – 1}}\frac{{x – y}}{{1 – xy}};\,xy > – 1\)
• \(2\,{\tan ^{ – 1}}\,x = {\tan ^{ – 1}}\frac{{2x}}{{1 – {x^2}}};\,|x| < 1\)
• \(2\,{\tan ^{ – 1}}\,x = {\sin ^{ – 1}}\frac{{2x}}{{1 + {x^2}}},\,|x| \le 1\)
• \(2\,{\tan ^{ – 1}}\,x = {\cos ^{ – 1}}\frac{{1 – {x^2}}}{{1 + {x^2}}},\,x \ge 0\)

Q.2. What is the principal value of inverse trigonometric functions?
Ans: The principal value of an inverse trigonometric function at a point \(x\) is the value of the inverse function at the point \(x\) that is within the principal range of the branch. When two numerically equal values have the opposite signs, the primary value of the inverse trigonometric function is the positive number.

Q.3. What is the importance of inverse trigonometric functions?
Ans: Sometimes known as the arc functions or anti-trigonometric functions, these are the inverses of the trigonometric functions. Any trigonometric ratios can be used to find the unknown angles of a right-angle triangle.

Q.4. What is the use of inverse trigonometric functions in real life?
Ans: The height of the Sun from the ground is calculated using inverse trigonometric functions. Inverse trigonometric functions can be used to calculate the tilt angle. They can also help in the identification of bridge angles in scale models.

Q.5. Why are inverse trigonometric functions called arc?
Ans: The inverse trigonometric functions are often called arc functions, because they give the length of the arc required to calculate a trigonometric function.

Learn All Trigonometric Functions

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