# Inverse Trigonometry Function: Concept and Formulas

In basic trigonometry, trigonometric functions takes the angle and provide the ratio of length of sides of triangle.

In the above illustration, angle = 30 degree.
With the help of trigonometry function we can say that;

Sin\ 30\ =\ \frac{Perpendicular}{Base}\\\ \\ \frac{1}{2} \ =\ \frac{Perpendicular\ }{Base} \\\ \\

Hence, we got the ratio of sides with the help of sin function.

But what if ratio of sides are given and we have to find the measure of corresponding angle?
In this case, we will use the concept of inverse trigonometry functions.

## Introduction to Inverse Trigonometry

Consider the following illustration:

From the above figure you can observe that the ratio of length is given and we need to find the corresponding angle.
This can easily been done with the help of inverse trigonometry functions.

we\ know\ that\\\ \\ sin\theta \ =\ \frac{Perpendicular}{Hypotenuse}\\\ \\ \theta \ =\ sin^{-1}\frac{1}{\sqrt{2}} \\\ \\ \theta =\ 45\ degree\\\ \\

Now you have understood the basic concepts of inverse trigonometry functions, let us now move on to the rules and formulas for this concept.

### Calculation of the Principal Value

To solve questions related to calculation of angle, try to understand the following table

Given below are some of questions related to Principal Value. I will suggest you to solve the questions to understand this concept fully.

#### Questions related to Principal Value

(01) Find the principle value of cos^{-1}\left(\frac{-1}{2}\right) \\\ \\ Read Solution

Solution
cos^{-1}\left(\frac{-1}{2}\right) =y \\\ \\ \frac{-1}{2} \ =\ cos\ y\ \ \ \ \ \ \ \ -\ -\ -eq( 1)\\\ \\ we\ know\ that\\ \\ cos\ 60\ =\ \frac{1}{2}\\\ \\ Angle=60\ degree\\\ \\ Calculating\ angle\ in\ Radian\\\ \\ \theta \ =\ 60\ \times \frac{\pi }{180}\\\ \\ \theta =\frac{\pi }{3}\\\ \\ Since\ value\ is\ in\ negative\\\ \\ \theta =\pi -\frac{\pi }{3}\\\ \\ \theta \ =\frac{2\pi }{3}\\\ \\ Putting\ the\ value\ in\ eq( 1)\\\ \\ cos\left(\frac{2\pi }{3}\right) \ =\ cos\ y\\\ \\ hence\ y\ =\left(\frac{2\pi }{3}\right)\\\ \\ Hence\ \left(\frac{2\pi }{3}\right) \ is\ the\ required\ angle

(02) Find the Principal Value of cot^{-1}\left(\frac{-1}{\sqrt{3}}\right) \\\ \\ Read Solution

Solution
cot^{-1}(\frac{-1}{\sqrt{3}})=y \\\ \\ (\frac{-1}{\sqrt{3}})=cot y - -eq(1)\\\ \\ we\ know\ that\\ \\ Cot 60 = \frac{1}{\sqrt{3}}\\\ \\ converting\ 60\ degree\ into\ radian\\\ \\ Angle = 60 ×\frac{\pi}{180}\\\ \\ Angle =\frac{\pi}{3}\\\ \\ Since\ the\ value\ is\ negative \\ \\ So Angle = \pi -\frac{\pi}{3}\\\ \\ Angle= \frac{2\pi}{3}\\\ \\ Putting\ the\ value\ in\ eq(1)\\ \\ cot (\frac{2\pi}{3}) = cot y\\\ \\ hence\ y = \frac{2\pi}{3}\\\ \\

(03) Find the principal value of sin^{-1}\left(\frac{-1}{2}\right) \\\ \\ Read Solution

Solution
\theta =\ sin^{-1}\left(\frac{-1}{2}\right)\\\ \\ sin\theta \ =\left(\frac{-1}{2}\right) \ \ --\ eq( 1) \\\ \\ We\ know\ that\\\ \\ Sin\ 30=\frac{1}{2}\\\ \\ converting\ 30\ degree\ into\ radian\\\ \\ Angle\ =\ 30\ \times \frac{\pi }{180}\\\ \\ Angle\ =\ \frac{\pi }{6}\\\ \\ Observe\ that\ here\ angle\ is\ going\ negative\\ \\ so\ putting\ angle\ value\ with\ negative\ sign\\\ \\ sin\theta \ =sin\ \left(\frac{-\pi }{6}\right) \\\ \\ Hence\ \theta =\frac{-\pi }{6}\\\ \\

(04) Find the principal value of tan^{-1}( 1) \\ \\

Solution
\theta =\ tan^{-1}( 1)\\\ \\ tan\theta \ =\ 1\\\ \\ tan\theta \ =\ tan\ 45\\\ \\ \theta =45\\\ \\ Converting\ angle\ into\ radian\\\ \\ angle\ =\ 45\ \times \frac{\pi }{180}\\\ \\ angle\ =\frac{\pi }{4}\\\ \\ Hence\ \frac{\pi }{4\ } \ is\ the\ right\ answer\

(05) Find the principal value of tan^{-1}( -1) \\\ \\ Read Solution

Solution
\theta= tan^{-1}(-1) \\\ \\ tan\theta= -1 \\\ \\ This\ can\ be\ written\ as\\ \\ tan\theta = tan (-45) - - -eq(1)\\\ \\ converting\ the\ degree\ into\ radian\\ \\ angle = -45 ×\frac{\pi}{180}\\\ \\ angle = \frac{-\pi} {4}\\\ \\ Putting\ the\ value\ in\ eq(1)\\\ \\ tan\theta = tan (-\frac{\pi}{4})\\\ \\hence\\ \\ \theta =\frac{-\pi}{4}

## Formulas for Inverse Trigonometry

### (01)Formulas with negative variable

(01)\ \ sin^{-1}( -x) =-sin^{-1}( x)\\\ \\ (02)\ \ cos^{-1}( -x) =\pi -cos^{-1}( x)\\\ \\ (03)\ \ tan^{-1}( -x) =-tan^{-1}( x)\\\ \\ (04)\ \ cot^{-1}( -x) =\pi -cot^{-1}( x)\\\ \\ (05)\ \ sec^{-1}( -x) =\pi -sec^{-1}( x)\\\ \\ (06)\ \ cosec^{-1}( -x) =-cosec^{-1}( x) \\ \\

Note:
(a) In Sin, Tan and Cosec inverse function there is simple relocation of negative sign
(b) In Cos, Cot and Sec inverse function the function is subtracted with /pi

### (02)Formulas with variable\frac{1}{x}

( 01) \ \ sin^{-1}\left(\frac{1}{x}\right) =cosec^{-1}( x)\\\ \\ ( 02) \ \ cos^{-1}\left(\frac{1}{x}\right) =sec^{-1}( x)\\\ \\ ( 03) \ \ tan^{-1}\left(\frac{1}{x}\right) =cot^{-1}( x)\\\ \\ ( 04) \ \ cot^{-1}\left(\frac{1}{x}\right) =tan^{-1}( x)\\\ \\ ( 05) \ \ sec^{-1}\left(\frac{1}{x}\right) =cos^{-1}( x)\\\ \\ ( 06) \ \ cosec^{-1}\left(\frac{1}{x}\right) =sin^{-1}( x) \\\ \\

Note:
(a) Sin inverse is converted into cosec inverse and vice-versa
(b) Cos inverse is converted into Sec inverse and vice versa
(c) Tan inverse is converted into Cot Inverse and vice-versa

### (03) Inverse function with Trigonometry variable

( 01) \ \ sin^{-1}( sin\theta ) =\ \theta \\\ \\ ( 02) \ \ cos^{-1}( cos\theta ) =\theta \\\ \\ ( 03) \ \ tan^{-1}( tan\theta ) =\ \theta \\\ \\ ( 04) \ \ cot^{-1}( cot\theta ) =\theta \\\ \\ ( 05) \ \ sec^{-1}( sec\theta ) =\theta \\\ \\ ( 06) \ \ cosec^{-1}( cosec\theta ) =\theta

### (04)Trigonometry function with inverse variable

( 01) \ \ sin\left( sin^{-1} x\right) =\ x\\\ \\ ( 02) \ \ cos\left( cos^{-1} x\right) =x\\\ \\ ( 03) \ \ tan\left( tan^{-1} x\right) =\ x\\\ \\ ( 04) \ \ cot\left( cot^{-1} x\right) =x\\\ \\ ( 05) \ \ sec\left( sec^{-1} x\right) =x\\\ \\ ( 06) \ \ cosec\left( cosec^{-1} x\right) =x \\\ \\

### (05)Addition of different Inverse Trigonometry Function

sin^{-1} x+\ cos^{-1} x\ =\ \frac{\pi }{2}\\\ \\ tan^{-1} x+\ cot^{-1} x\ =\ \frac{\pi }{2}\\\ \\ sec^{-1} x+\ cosec^{-1} x\ =\ \frac{\pi }{2}

### (06)Addition and Subtraction of same type of Inverse Trigonometry Function

#### (A) Formula for Sin Inverse

(01)\ sin^{-1} x+\ sin^{-1} y\ =\ sin^{-1} [ x\sqrt{1-y^{2}} +y\sqrt{1-x^{2}} \ ]\\\ \\ (02)\ sin^{-1} x-\ sin^{-1} y\ =\ sin^{-1}[ x\sqrt{1-y^{2}} -y\sqrt{1-x^{2}} \ ]

#### (B) Formula for Cos Inverse

( 01) \ \ cos^{-1} x+\ cos^{-1} y\ =\ cos^{-1}\left[ xy-\sqrt{1-x^{2}} .\sqrt{1-y^{2}} \ \right]\\\ \\ ( 02) \ \ cos^{-1} x-\ cos^{-1} y\ =\ cos^{-1}\left[ xy+\sqrt{1-x^{2}} .\sqrt{1-y^{2}} \ \right]

#### (C) Formula for Tan Inverse

( 01) \ \ tan^{-1} x+\ tan^{-1} y\ =\ tan^{-1}\frac{x+y}{1-xy} \ ,\ where\ xy< 1\\\ \\ ( 02) \ \ tan^{-1} x+\ tan^{-1} y\ =\ \pi +\ tan^{-1}\frac{x+y}{1-xy} \ ,\ where\ xy >1\\\ \\ ( 03) \ \ tan^{-1} x-\ tan^{-1} y\ =\ tan^{-1}\frac{x-y}{1+xy} \ ,\ where\ xy >-1\

#### (D) 2 * Inverse Trigonometry Functions

( 01) \ \ 2\ sin^{-1} x\ =\ sin^{-1}\left( 2x\sqrt{1-x^{2}}\right)\\\ \\ ( 02) \ 2\ cos^{-1} x\ =\ cos^{-1}\left( 2x^{2} -1\right)\\\ \\ ( 03) \ 2\ tan^{-1} x\ =\ tan^{-1}\left(\frac{2x}{1-x^{2}}\right)\\ \\ \ OR\\ \\ 2\ tan^{-1} x\ =\ sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\\ \\ OR\ \\ \\ 2\ tan^{-1} x\ =\ cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\

#### (E) 3 * Inverse Trigonometry Functions

(01) \ \ 3\ sin^{-1} x\ =\ sin^{-1}\left( 3x-4x^{3}\right)\\\ \\ ( 02) \ 3\ cos^{-1} x\ =\ cos^{-1}\left( 4x^{3} -3x\right)\\\ \\ ( 03) \ 3\ tan^{-1} x\ =\ tan^{-1}\left(\frac{3x-x^{2}}{1-3x^{2}}\right)\\\ \\

### (07)Addition of Three Inverse Trigonometry Functions

#### (A) Addition of Sin Inverse Function

( 01) \ If\ \ sin^{-1} x+sin^{-1} y+sin^{-1} z\ =\ \frac{\pi }{2}\\\ \\ then\ x^{2} +y^{2} +z^{2} +2xyz=1\\\ \\ \\\ \\ ( 02) \ If\ \ sin^{-1} x+sin^{-1} y+sin^{-1} z\ =\pi \\\ \\ then\ x\sqrt{1-x^{2}} \ +y\sqrt{1-y^{2}} +z\sqrt{1-z^{2}} =2xyz\\\ \\ \\\ \\ ( 03) \ \ If\ \ sin^{-1} x+sin^{-1} y+sin^{-1} z\ =\frac{3\pi }{2}\\\ \\ then\ \ xy+yz+zx=3

#### (B) Addition of Cos Inverse Function

( 01) \ If\ \ cos^{-1} x+cos^{-1} y+cos^{-1} z\ =\ 3\pi \\\ \\ then\ xy\ +yz+zx=3\\\ \\ \\\ \\ ( 02) \ If\ \ cos^{-1} x+cos^{-1} y+cos^{-1} z\ =\pi \\\ \\ then\ x^{2} +y^{2} +z^{2} +2xyz=1\

#### (C) Addition of Tan Inverse Function

( 01) \ \ tan^{-1} x+tan^{-1} y+tan^{-1} z\ \\\ \\ \Longrightarrow \ tan^{-1\ }\left[\frac{x+y+z-xyz}{1-( xy+yz+zx)}\right]\\\ \\ \\\ \\ ( 02) \ If\ tan^{-1} x+tan^{-1} y+tan^{-1} z\ \ =\frac{\pi }{2}\\\ \\ then\ xy+yz+zx=1\\\ \\ \\\ \\ ( 03) \ If\ tan^{-1} x+tan^{-1} y+tan^{-1} z\ \ =\pi \\\ \\ then\ x+y+z=xyz\

## Questions on Inverse Trigonometry Function

(01) If\ sin^{-1}\left(\frac{3}{x}\right) +sin^{-1}\left(\frac{4}{x}\right) =\frac{\pi }{2}\\\ \\ Then\ find\ value\ of\ x\ \\\ \\ Read Solution

Solution
\Longrightarrow \ sin^{-1}\left(\frac{3}{x}\right) +sin^{-1}\left(\frac{4}{x}\right) =\frac{\pi }{2}\\\ \\ \Longrightarrow sin^{-1}\left(\frac{3}{x}\right) =\frac{\pi }{2} \ -\ sin^{-1}\left(\frac{4}{x}\right)\\\ \\ Using\ Formula\\\ \\ cos^{-1} \alpha \ =\frac{\pi }{2} \ -sin^{-1} \alpha \\\ \\ \Longrightarrow sin^{-1}\left(\frac{3}{x}\right) \ =\ cos^{-1}\left(\frac{4}{x}\right)\\\ \\ Using\ another\ formula\\ \\ cos^{-1} \alpha =sin^{-1}\sqrt{1-\alpha ^{2}}\\\ \\ \Longrightarrow \ sin^{-1}\left(\frac{3}{x}\right) \ \ =\ sin^{-1}\sqrt{1-\frac{16}{x^{2}}}\\\ \\ LHS=\ RHS\\ \\ \Longrightarrow \ \left(\frac{3}{x}\right) \ =\sqrt{1-\frac{16}{x^{2}}}\\\ \\ On\ solving\ the\ equation\ we\ get\\\ \\ \Longrightarrow \ x=5\\\ \\

(02) Find the value of

tan^{-1}\left(\frac{m}{n}\right) -tan^{-1}\left(\frac{m-n}{m+n}\right)\\\ \\

Solution
\Longrightarrow \ tan^{-1}\left(\frac{m}{n}\right) -tan^{-1}\left(\frac{m-n}{m+n}\right)\\\ \\ Using\ formula\\ \\ tan^{-1} x\ -\ tan^{-1} y\ =\ tan^{-1}\left(\frac{x-y}{1+xy}\right)\\\ \\ \Longrightarrow \ tan^{-1}\frac{\left(\frac{m}{n} -\frac{m-n}{m+n}\right)}{1+\frac{m}{n} \times \frac{m-n}{m+n}}\\\ \\ \Longrightarrow \ tan^{-1}\frac{\frac{m^{2} +n^{2}}{n( m+n)}}{\frac{m^{2} +n^{2}}{n( m+n)}}\\\ \\ \Longrightarrow \ tan^{-1\ }( 1)\\\ \\ \Longrightarrow \ 45\ degree\

(03) Show that
sin^{-1}\left( 2x\sqrt{1-x^{2}}\right) \ =\ 2\ sin^{-1} x\\\ \\ Read Solution

Solution
\Longrightarrow \ sin^{-1}\left( 2x\sqrt{1-x^{2}}\right) \\\ \\ Let\ x=\ sin\theta \\\ \\ \Longrightarrow \ sin\ ^{-1}\left( 2\ sin\theta \sqrt{1-sin^{2} \theta }\right)\\\ \\ \Longrightarrow \ sin^{-1}( 2\ sin\theta \ cos\theta )\\\ \\ \Longrightarrow \ sin^{-1}( sin\ 2\theta )\\\ \\ \Longrightarrow \ 2\theta \

(04) If\ 2\ tan^{-1} x=sin^{-1}\frac{2a}{1+a^{2}} +\ cos^{-1}\frac{1-b^{2}}{1+b^{2}}\\\ \\ Find\ the\ value\ of\ x\\\ \\ Read Solution

Solution
2\ tan^{-1} x=sin^{-1}\frac{2a}{1+a^{2}} +\ cos^{-1}\frac{1-b^{2}}{1+b^{2}}\\\ \\ We\ know\ that\\ \\ 2\ tan^{-1} a=sin^{-1}\frac{2a}{1+a^{2}}\\\ \\ also\\ \\ 2\ tan^{-1} b=cos^{-1}\frac{1-b^{2}}{1+b^{2}}\\\ \\ Putting\ these\ values\ in\ main\ question\\\ \\ 2tan^{-1} x=\ 2tan^{-1} a+2tan^{-1} b\\\ \\ tan^{-1} x=\ tan^{-1} a+tan^{-1} b\\\ \\ tan^{-1} x=\ tan^{-1}\frac{a+b}{1-ab}\\\ \\ hence\\ \\ x\ =\frac{a+b}{1-ab}\

(05) If\ \left( tan^{-1} x\right)^{2} +\ \left( cot^{-1} x\right)^{2} =\frac{5\pi ^{2}}{8}\\\ \\ Then\ find\ value\ of\ x\\\ \\ Read Solution

Solution
If\ \left( tan^{-1} x\right)^{2} +\ \left( cot^{-1} x\right)^{2} =\frac{5\pi ^{2}}{8}\\\ \\ we\ know\ that\\ \\ a^{2} +b^{2} =( a+b)^{2} -2ab \\\ \\ Putting\ the\ values\\\ \\ \left( tan^{-1} x\ +\ cot^{-1} x\right)^{2} -2.\ tan^{-1} x.\ cot^{-1} x=\frac{5\pi ^{2}}{8}\\\ \\ we\ know\ that\\\ \\ tan^{-1} x\ +\ cot^{-1} x\ =\ \frac{\pi }{2}\\\ \\ Putting\ the\ values\\\ \\ \left(\frac{\pi }{2}\right)^{2} -2tan^{-1} x\ \left(\frac{\pi }{2} -tan^{-1} x\right) =\frac{5\pi ^{2}}{8}\\\ \\ \frac{\pi }{4}^{2} -\pi tan^{-1} x+2\ \left( tan^{-1} x\right)^{2} =\frac{5\pi ^{2}}{8}\\\ \\ 2\ \left( tan^{-1} x\right)^{2} -\pi tan^{-1} x-\frac{3\pi ^{2}}{8} =0\\\ \\ This\ is\ a\ Quadratic\ Equation\\\ \\ On\ solving\ we\ get\\\ \\ tan^{-1} x\ =\ \frac{-\pi }{4}\\\ \\ hence\ x=1