Also, read |

Basics of Trigonometry |

Trigonometry for class 10 |

Trigonometric Formulas |

Applications of Trigonometry |

# Inverse Trigonometric Function:

While studying calculus we see that Inverse trigonometric function plays a very important role. For every section of trigonometry with limited inputs in function, we use inverse trigonometric function formula to solve various types of problems. In this article, we have listed all the important formulas and identities related to inverse trigonometric functions.

**Domain and range of simple trigonometric functions:**

Functions |
Domain |
Range |

sinϴ | R | [-1, 1] |

cosϴ | R | [-1, 1] |

tanϴ | R- { x : x = (2n + 1)\(\frac{π}{2}\), n ∈ Z} | R |

cotϴ | R – { x : x = nπ, n ∈ Z} |
R |

secϴ | R- { x : x = (2n + 1)\(\frac{π}{2}\), n ∈ Z} | R – (-1, 1) |

cosecϴ | R – { x : x = nπ, n ∈ Z} |
R –(-1, 1) |

**Domain and range of inverse trigonometric function formula:**

Functions |
Domain |
Range(principal value) |

sin^{-1}ϴ |
[–1, 1] | [-\(\frac{π}{2}\), \(\frac{π}{2}\)] |

cos^{-1}ϴ |
[–1, 1] | [0, π] |

tan^{-1}ϴ |
R | [-\(\frac{π}{2}\), \(\frac{π}{2}\)] |

cot^{-1}ϴ |
R | [0, π] |

sec^{-1}ϴ |
R – (–1, 1) | [0, π] – {\(\frac{π}{2}\)} |

cosec^{-1}ϴ |
R – (–1, 1) | [-\(\frac{π}{2}\), \(\frac{π}{2}\)] – {0} |

**Properties of inverse trigonometric function formula:**

Functions and values |
Domain |

sin^{-1}\(\frac{1}{x}\) = cosec^{-1}x |
|x| ≥ 1 |

cos^{-1}\(\frac{1}{x}\) = sec^{-1}x |
|x| ≥ 1 |

tan^{-1}\(\frac{1}{x}\) = cot^{‑1}x |
X > 0 |

Functions and values |
Domain |

sin^{–1}(–x) = – sin^{–1}x |
x ∈ [– 1, 1] |

tan–1 (–x) = – tan–1 x | x ∈ R |

cosec^{–1}(–x) = – cosec^{–1}x |
|x| ≥ 1 |

Functions and values |
Domain |

cos^{–1}(–x) = π – cos^{–1}x |
x ∈ [– 1, 1] |

sec^{–1}(–x) = π – sec^{–1}x |
|x| ≥ 1 |

cot^{–1}(–x) = π – cot^{–1} x |
x ∈ R |

**Identities of inverse trigonometric function formula:**

Identities |
Domain |

sin^{-1}x + cos^{–1}x = \(\frac{π}{2}\) |
x ∈ [– 1, 1] |

tan^{–1}x + cot^{–1}x = \(\frac{π}{2}\) |
x ∈ R |

cosec^{–1}x + sec^{–1}x = \(\frac{π}{2}\) |
|x| ≥ 1 |

**Common inverse trigonometric function formula:**

Formulas |
Domain |

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 |

Formulas |
Domain |

2 tan^{–1}x = sin^{-1}\(\frac{2x}{1+{x}^2}\) |
|x| ≤ 1 |

2 tan^{–1}x = cos^{-1}\(\frac{1-{x}^2}{1+{x}^2}\) |
x ≥ 0 |

2 tan^{–1}x = tan^{-1}\(\frac{2x}{1-{x}^2}\) |
– 1 < x < 1 |

**Some special inverse trigonometric function formula:**

- sin
^{-1}x + sin^{-1}y = sin^{-1}( x\(\sqrt{1-{y}^2}\) + y\(\sqrt{1-{x}^2}\) )

**if x, y ≥ 0 and x**^{2}+ y^{2}≤ 1. - sin
^{-1}x + sin^{-1}y = π – sin^{-1}( x\(\sqrt{1-{y}^2}\) + y\(\sqrt{1-{x}^2}\) )

**if x, y ≥ 0 and x**^{2}+ y^{2}> 1. - sin
^{-1}x – sin^{-1}y = sin^{-1}( x\(\sqrt{1-{y}^2}\) – y\(\sqrt{1-{x}^2}\) )

**if x, y ≥ 0 and x**^{2}+ y^{2}≤ 1. - sin
^{-1}x + sin^{-1}y = π – sin^{-1}( x\(\sqrt{1-{y}^2}\) – y\(\sqrt{1-{x}^2}\) )

**if x, y ≥ 0 and x**^{2}+ y^{2}> 1. - cos
^{−}^{1}x + cos^{−}^{1}y = cos^{−}^{1}(xy – \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )

**if x, y ≥ 0 and x**^{2}+ y^{2}≤ 1. - cos
^{−}^{1}x + cos^{−}^{1}y = π – cos^{−}^{1}(xy – \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )

**if x, y ≥ 0 and x**^{2}+ y^{2}> 1. - cos
^{−}^{1}x – cos^{−}^{1}y = cos^{−}^{1}(xy + \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )

**if x, y ≥ 0 and x**^{2}+ y^{2}≤ 1. - cos
^{−}^{1}x – cos^{−}^{1}y = π – cos^{−}^{1}(xy + \(\sqrt{1-{y}^2}\)\(\sqrt{1-{x}^2}\) )

**if x, y ≥ 0 and x**^{2}+ y^{2}> 1. - tan
^{−}^{1}x + tan^{−}^{1}y = tan^{−}^{1}(\(\frac{x+y}{1-xy}\) )

**if x > 0, y > 0 and xy < 1.** - tan
^{−}^{1}x + tan^{−}^{1}y = π + tan^{−}^{1}(\(\frac{x+y}{1-xy}\) )

**if x > 0, y > 0 and xy > 1.** - tan
^{−}^{1}x + tan^{−}^{1}y = tan^{−}^{1}(\(\frac{x+y}{1-xy}\) ) – π

**if x < 0, y > 0 and xy > 1.** - tan
^{−}^{1}x + tan^{−}^{1}y + tan^{−}^{1}z = tan^{−}^{1}(\(\frac{x+y+z-xyz}{1-xy-yz-zx}\) ) - tan
^{−}^{1}x – tan^{−}^{1}y = tan^{−}^{1}(\(\frac{x-y}{1+xy}\) ) – π

**if x < 0, y > 0 and xy > 1.** - 2 sin
^{−}^{1}x = sin^{−}^{1}(2x\(\sqrt{1-{x}^2}\) - 2 cos
^{−}^{1}x = cos^{−}^{1}(2x^{2}– 1) - 2tan
^{-1}x = tan^{-1}( \(\frac{2x}{1-{x}^2}\) ) = sin^{-1}( \(\frac{2x}{1+{x}^2}\) ) = cos^{-1 }(\(\frac{1-{x}^2}{1+{x}^2}\)) - 3sin
^{-1}x = sin^{-1}(3x – 4x^{3}) - 3 cos
^{-1}x = cos^{-1}(4x^{3 }– 3x) - 3tan
^{-1}x = tan^{-1}( \(\frac{3x – {x}^3}{1 – 3{x}^2}\) )

## Examples based on inverse trigonometric function formula:

Let sin*Find the principal value of sin*Solution:^{–1}(\(\frac{1}{\sqrt{2}}\) ).

^{–1}(\(\frac{1}{\sqrt{2}}\) ) = y. Then, sin y = ( \(\frac{1}{\sqrt{2}}\) )

We know that the range of the principal value branch of sin^{–1}is [-\(\frac{π}{2}\), \(\frac{π}{2}\)].

Also, sin( \(\frac{π}{4}\) ) = \(\frac{1}{\sqrt{2}}\)

so, principal value of sin^{-1}( \(\frac{1}{\sqrt{2}}\) ) is \(\frac{π}{4}\).Let x = sec θ, then \(\sqrt{{x}^2-1}\) = \(\sqrt{{sec}^2θ-1}\) = tan θ*Write cot*Solution:^{-1}( \(\frac{1}{\sqrt{{x}^2-1}}\) ), | x | > 1 in the simplest form.

Therefore, cot^{–1}= \(\frac{1}{\sqrt{x^{2}–1}}\) = cot^{–1}(cot θ) = θ = sec^{–1}x, which is the simplest form.**More from Calculus****Relation and Functions****Limits Formula**Continuity Rules Differentiability Rules Derivative Formula **Integral Formula****Application of Integrals****Logarithm Formulas**

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