Sum and Difference of Inverse Trigonometric Functions

Inverse trigonometric functions can be defined as the inverses of the basic trigonometric functions - sine, cosine, tangent, cotangent, secant, and cosecant. We know that trigonometric functions are periodic and hence, man-one in their actual domain. So, to define an inverse trigonometric function, we have to restrict its actual domain to make the function injective. In real life, we use the inverse trigonometric function for determining the depth of the hole or the angle of inclination.

In this article, we will cover the concept of Sum and the difference of angles in terms of arctan. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of fourteen questions have been asked on this topic in JEE MAINS (2013 - 2023) including three in 2019, four in 2021, and five in 2022, two in 2023.

What is an Inverse Trigonometric Function?

In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function and vice versa.

Sum of the Angle of Inverse Trigonometric Function

The sum and difference of trigonometry formulas convert the sum and difference of inverse trigonometric functions into single inverse trigonometric functions.

Sum of angles in terms of arctan

We have the following formulas for the sum of angles when angles are in terms of arctan

1. $\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\left\{\begin{array}{cc}\tan ^{-1}\left(\frac{x+y}{1-x y}\right), & \text { If } \mathrm{x}>0, y>0, x y<1 \\ \pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right), & \text { If } \mathrm{x}>0, \mathrm{y}>0 \text { and } \mathrm{xy}>1 \\ -\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right), & \text { If } \mathrm{x}<0, \mathrm{y}<0 \text { and } \mathrm{xy}>1\end{array}\right.$

Difference of angles in terms of arctan

We have the following formulas for the difference of angles when angles are in terms of arctan

1. $\tan ^{-1} \mathrm{x}-\tan ^{-1} \mathrm{y}=\left\{\begin{array}{cc}\tan ^{-1}\left(\frac{x-y}{1+x y}\right), & \text { If } x y>-1 \\ \pi+\tan ^{-1}\left(\frac{x-y}{1+x y}\right), & \text { If } \mathrm{x}>0, \mathrm{y}<0 \text { and } \mathrm{xy}<-1 \\ -\pi+\tan ^{-1}\left(\frac{x-y}{1+x y}\right), & \text { If } \mathrm{x}<0, \mathrm{y}>0 \text { and } \mathrm{xy}<-1\end{array}\right.$

Sum and difference of angles in terms of arcsin

We have the following formulas for the sum of angles when angles are in terms of arcsin

1. $\sin ^{-1} \mathrm{x}+\sin ^{-1} \mathrm{y}= \begin{cases}\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\} & \text { if }-1 \leq x, y \leq 1 \text { and } x^2+y^2 \leq 1 \\ \pi-\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\} & \text { or, if } x y<0 \text { and } x^2+y^2>1 \\ & \text { if } 0<x, y \leq 1 \text { and } x^2+y^2>1 \\ -\pi-\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\} & \text { if }-1 \leq x, y<0 \text { and } x^2+y^2>1\end{cases}$

We have the following formulas for the difference of angles when angles are in terms of arcsin

2. $\sin ^{-1} \mathrm{x}-\sin ^{-1} \mathrm{y}= \begin{cases}\sin ^{-1}\left\{x \sqrt{1-y^2}-y \sqrt{1-x^2}\right\} & \text { if }-1 \leq x, y \leq 1 \text { and } x^2+y^2 \leq 1 \\ \pi-\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\} & \text { or, if } x y>0 \text { and } x^2+y^2>1 \\ & \text { if } 0<x \leq 1 ;-1 \leq y<0 \text { and } x^2+y^2>1 \\ -\pi-\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\} & \text { if }-1 \leq x<0 ; 0<y \leq 1 \text { and } x^2+y^2>1\end{cases}$

Sum and difference of angles in terms of arccos

We have the following formulas for the sum and difference of angles when angles are in terms of arc

1. $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left\{x y-\sqrt{1-x^2} \sqrt{1-y^2}\right\} \quad$ if $0<x, y \leq 1$
2. $\cos ^{-1} x-\cos ^{-1} y= \begin{cases}\cos ^{-1}\left\{x y+\sqrt{1-x^2} \sqrt{1-y^2}\right\} & \text { if } 0 \leq x, y \leq 1 \text { and } x \leq y \\ -\cos ^{-1}\left\{x y+\sqrt{1-x^2} \sqrt{1-y^2}\right\} & \text { if } 0<x, y \leq 1 \text { and } x>y\end{cases}$

Important notes

1) $x<0$ and $y<0$,these identities can be used with the help of property $\sin ^{-1}(-x)=-\sin ^{-1} x$, i.e., change x and y to-x and -y. respectively. Also, replace y by -y in the $\sin ^{-1} x-\sin ^{-1} y$ above results.

2) $\mathrm{x}<0$ and $\mathrm{y}<0$$\cos (\alpha+\beta)=\frac{3}{5}, \sin (\alpha-\beta)=\frac{5}{13}$ these identities can be used with the help of property $\cos ^{-1}(-x)=\pi-\cos ^{-1} x$, i.e., change x and y to-x and -y. respectively.

Summary

The sum and difference of inverse trigonometric functions provide us with a relationship between the sum and difference of the inverse trigonometric function of an angle and the sum and difference of angles in the inverse trigonometric function. These formulas are derived from trigonometric identities and are useful in simplifying the expression involving inverse trigonometric functions. These formulas make our calculation easy and are helpful for solving various trigonometric problems.

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Solved Examples Based on the Sum and difference of angles

Example 1: If $\cos (\alpha+\beta)=\frac{3}{5}, \sin (\alpha-\beta)=\frac{5}{13}$ and $0<\alpha, \beta<\frac{\pi}{4}$ then $\tan (2 \alpha)$ is equal to : [JEE MAINS 2023]

Solution:
$
\begin{aligned}
& \cos (\alpha+\beta)=\frac{5}{5} \\
& \sin (\alpha-\beta)=\frac{5}{13}
\end{aligned}
$
$
0<\alpha, \beta<\frac{4}{4} \quad \text { then } \quad \tan (2 \alpha)=\text { ? }
$

$\begin{aligned} & \cos (\alpha+\beta)=\frac{3}{5} \\ & \tan (\alpha+\beta)=\frac{4}{3}\end{aligned}$

$\begin{aligned} & \sin (\alpha-\beta)=\frac{5}{13} \\ & \tan (\alpha-\beta)=\frac{5}{12}\end{aligned}$

$\begin{aligned} & \alpha+\beta=\tan ^{-1}\left(\frac{4}{3}\right) \\ & \alpha-\beta=\tan ^{-1}\left(\frac{5}{12}\right)\end{aligned}$

$=2 \alpha=\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{5}{12}\right)$

$\begin{aligned} & =\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3} \times \frac{5}{12}}\right) \\ & =\tan ^{-1}\left(\frac{\frac{21}{12}}{\frac{36-20}{36}}\right)=\frac{21}{\frac{12}{36}} \\ & =\tan ^{-1}\left(\frac{3 \times 21}{16}\right) \\ & =\tan ^{-1}\left(\frac{63}{16}\right)\end{aligned}$

Hence, the answer is $\frac{63}{16}$

Example 2: If $S=\left\{x \in R: \sin ^{-1}\left(\frac{x+1}{\sqrt{x^2+2 x+2}}\right)-\sin ^{-1}\left(\frac{x}{\sqrt{x^2+11}}\right)=\frac{\pi}{4}\right\}$,then is equal to ____________. [JEE MAINS 2023]

Solution: $\sin ^{-1}\left(\frac{(x+1)}{\sqrt{(x+1)^2+1}}\right)-\sin ^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)=\frac{\pi}{4}$

$\begin{aligned} & \because \frac{t}{\sqrt{t^2+1}} \in(-1,1) \\ & \sin ^{-1}\left(\frac{(\mathrm{x}+1)}{\sqrt{(x+1)^2+1}}\right)=\sin ^{-1}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^2+1}}\right)+\frac{\pi}{4} \\ & \frac{(x+1)}{\sqrt{(x+1)^2+1}}=\left(\frac{1}{\sqrt{2}}\right) \cos \left(\sin ^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)\right)+\frac{1}{\sqrt{2}}\left(\frac{x}{\sqrt{x^2+1}}\right) \\ & =\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{x^2+1}}+\frac{x}{\sqrt{x^2+1}}\right) \\ & \frac{(x+1)}{\sqrt{(x+1)^2+1}}=\frac{1}{\sqrt{2}}\left(\frac{1+x}{\sqrt{x^2+1}}\right)\end{aligned}$
After solving this equation, we get$\begin{aligned} & \mathrm{x}=-1 \text { or } \mathrm{x}=0 \\ & S=\{-1,0\}\end{aligned}$

$\begin{aligned} & \sum_{x \in R}\left(\sin \left(\left(x^2+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^2+x+5\right) \pi\right)\right) \\ & =\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right]+\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right]=4\end{aligned}$

Hence, the answer is 4.

Example 3: The value of $\cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right)$ is [JEE MAINS 2022]

Solution

$\sum_{\mathrm{n}=1}^{50} \tan ^{-1}\left(\frac{(\mathrm{n}+1)-(\mathrm{n})}{1+(\mathrm{n}+1)(\mathrm{n})}\right)=\sum_{\mathrm{n}=1}^{50}\left(\tan ^{-1}(\mathrm{n}+1)-\tan ^{-1}(\mathrm{n})\right]$

$\begin{aligned} & =\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+\ldots-\left(\tan ^{-1} 51-\tan ^{-1} 50\right) \\ & =\tan ^{-1}(51)-\tan ^{-1}(1) \\ & =\tan ^{-1} \frac{51-1}{1+51 \times 1}=\tan ^{-1} \frac{50}{52}=\tan ^{-1} \frac{25}{26} \\ & \text { so } \cot \left(\tan ^{-1}\left(\frac{25}{26}\right)\right)=\frac{26}{25}\end{aligned}$

Hence, the answer is $\frac{26}{25}$

Example 4: $\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)$ is equal to [JEE MAINS 2022]

Solution: $\tan \left(2\left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)\right)$

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

$=\tan \left[\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{2}\right]$

$=\tan \left(\tan ^{-1} 2\right)$

$=2$

Hence, the answer is 2

Example 5: If S is the sum of the first 10 terms of the series $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots \ldots$

then $\tan (S)$ is equal to [JEE MAINS 2020]

Solution

$\begin{aligned} & \mathrm{S}=\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\ldots \\ & \mathrm{S}=\tan ^{-1}\left(\frac{2-1}{1+1.2}\right)+\tan ^{-1}\left(\frac{3-2}{1+2 \times 3}\right)+\tan ^{-1}\left(\frac{4-3}{1+3 \times 4}\right)+\ldots+ \\ & \tan ^{-1}\left(\frac{11-10}{1+10 \times 11}\right)\end{aligned}$

$\begin{aligned} & S=\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+\left(\tan ^{-1} 4-\tan ^{-1} 3\right)+\ldots \ldots+ \\ & \left(\tan ^{-1}(11)-\tan ^{-1}(10)\right)\end{aligned}$$\begin{aligned} & S=\tan ^{-1} 11-\tan ^{-1} 1=\tan ^{-1}\left(\frac{11-1}{1+11}\right) \\ & \tan (S)=\frac{11-1}{1+11 \times 1}=\frac{10}{12}=\frac{5}{6}\end{aligned}$

Hence, the answer is $\frac{5}{6}$