Inverse Trigonometric Functions Principal value ranges Because sin \sin sin cos \cos cos
Principal value branches Function Domain Range sin − 1 x \sin^{-1}x sin − 1 x [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] [ − π 2 , π 2 ] \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] [ − 2 π , 2 π ] cos − 1 x \cos^{-1}x cos − 1 x [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] [ 0 , π ] [0, \pi] [ 0 , π ] tan − 1 x \tan^{-1}x tan − 1 x R \mathbb{R} R ( − π 2 , π 2 ) \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right) ( − 2 π , 2 π ) cot − 1 x \cot^{-1}x cot − 1 x R \mathbb{R} R ( 0 , π ) (0, \pi) ( 0 , π ) sec − 1 x \sec^{-1}x sec − 1 x R − ( − 1 , 1 ) \mathbb{R} - (-1, 1) R − ( − 1 , 1 ) [ 0 , π ] − { π 2 } [0, \pi] - \left\{\tfrac{\pi}{2}\right\} [ 0 , π ] − { 2 π } csc − 1 x \csc^{-1}x csc − 1 x R − ( − 1 , 1 ) \mathbb{R} - (-1, 1) R − ( − 1 , 1 ) [ − π 2 , π 2 ] − { 0 } \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] - \{0\} [ − 2 π , 2 π ] − { 0 }
Worked example: evaluating a mixed expression With θ = cos − 1 1 5 \theta = \cos^{-1}\dfrac{1}{\sqrt5} θ = cos − 1 5 1 sin ( π 2 + θ ) = cos θ \sin\left(\tfrac{\pi}{2}+\theta\right) = \cos\theta sin ( 2 π + θ ) = cos θ sin ( π 2 + cos − 1 1 5 ) = 1 5 \sin\left(\dfrac{\pi}{2} + \cos^{-1}\dfrac{1}{\sqrt5}\right) = \dfrac{1}{\sqrt5} sin ( 2 π + cos − 1 5 1 ) = 5 1
Worked example: a principal value cot − 1 ( − 1 3 ) \cot^{-1}\left(-\dfrac{1}{\sqrt3}\right) cot − 1 ( − 3 1 ) ( 0 , π ) (0, \pi) ( 0 , π ) cot π 3 = 1 3 \cot\tfrac{\pi}{3} = \tfrac{1}{\sqrt3} cot 3 π = 3 1 π − π 3 = 2 π 3 \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3} π − 3 π = 3 2 π
Worked example: undoing sine Find sin − 1 ( sin 2 π 3 ) \sin^{-1}\left(\sin\dfrac{2\pi}{3}\right) sin − 1 ( sin 3 2 π ) 2 π 3 \dfrac{2\pi}{3} 3 2 π [ − π 2 , π 2 ] \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] [ − 2 π , 2 π ] sin 2 π 3 = sin π 3 \sin\tfrac{2\pi}{3} = \sin\tfrac{\pi}{3} sin 3 2 π = sin 3 π π 3 \tfrac{\pi}{3} 3 π π 3 \dfrac{\pi}{3} 3 π
Worked example: a domain fact sec θ \sec\theta sec θ ∣ sec θ ∣ ≥ 1 |\sec\theta| \ge 1 ∣ sec θ ∣ ≥ 1 sec − 1 x \sec^{-1}x sec − 1 x ∣ x ∣ ≥ 1 |x| \ge 1 ∣ x ∣ ≥ 1 R − ( − 1 , 1 ) \mathbb{R} - (-1, 1) R − ( − 1 , 1 )
Worked example: two identities tan − 1 x + cot − 1 x = π 2 \tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2} tan − 1 x + cot − 1 x = 2 π θ = tan − 1 x \theta = \tan^{-1}x θ = tan − 1 x θ ∈ ( − π 2 , π 2 ) \theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right) θ ∈ ( − 2 π , 2 π ) cot ( π 2 − θ ) = tan θ = x \cot\left(\tfrac{\pi}{2}-\theta\right) = \tan\theta = x cot ( 2 π − θ ) = tan θ = x π 2 − θ \tfrac{\pi}{2} - \theta 2 π − θ ( 0 , π ) (0, \pi) ( 0 , π ) cot − 1 x \cot^{-1}x cot − 1 x
tan − 1 1 + tan − 1 1 2 + tan − 1 1 3 = π 2 \tan^{-1}1 + \tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{3} = \dfrac{\pi}{2} tan − 1 1 + tan − 1 2 1 + tan − 1 3 1 = 2 π tan − 1 a + tan − 1 b = tan − 1 a + b 1 − a b \tan^{-1}a + \tan^{-1}b = \tan^{-1}\dfrac{a+b}{1-ab} tan − 1 a + tan − 1 b = tan − 1 1 − ab a + b a b < 1 ab < 1 ab < 1
tan − 1 1 2 + tan − 1 1 3 = tan − 1 5 6 1 − 1 6 = tan − 1 1 = π 4 , π 4 + π 4 = π 2 \tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{3} = \tan^{-1}\dfrac{\tfrac{5}{6}}{1-\tfrac{1}{6}} = \tan^{-1}1 = \dfrac{\pi}{4}, \qquad \dfrac{\pi}{4} + \dfrac{\pi}{4} = \dfrac{\pi}{2} tan − 1 2 1 + tan − 1 3 1 = tan − 1 1 − 6 1 6 5 = tan − 1 1 = 4 π , 4 π + 4 π = 2 π Common mistakes Reading sin − 1 x \sin^{-1}x sin − 1 x 1 sin x \dfrac{1}{\sin x} sin x 1 That's csc x \csc x csc x Cancelling sin − 1 ( sin θ ) = θ \sin^{-1}(\sin\theta) = \theta sin − 1 ( sin θ ) = θ Using the tangent sum formula when a b > 1 ab > 1 ab > 1 Then a correction of ± π \pm\pi ± π Mixing up ranges: cos − 1 \cos^{-1} cos − 1 cot − 1 \cot^{-1} cot − 1 π \pi π sin − 1 \sin^{-1} sin − 1 tan − 1 \tan^{-1} tan − 1 Where this comes up in exams The 2025-26 CBSE Class XII syllabus lists the definition, range, domain, principal value branch and graphs of inverse trigonometric functions. JEE Main 2026 lists inverse trigonometrical functions and their properties, and JEE Advanced 2026 lists inverse trigonometric functions (principal value only) and their elementary properties.
Checked against CBSE Mathematics curriculum, Classes XI-XII (2025-26) , JEE (Main) 2026 syllabus (NTA) and JEE (Advanced) 2026 syllabus . Syllabi can change from year to year, so confirm with the latest official notification for your exam.
Fit it into your week with the free timetable generator . Mark it off as you revise in the syllabus tracker . See how long you have left: JEE Main 2027 (Session 1) countdown and JEE Advanced 2027 countdown . Continue learning These rest on the standard trigonometric values and identities.