Inverse Trigonometric Function formulas

Every formula JEE Advanced expects you to know for Inverse Trigonometric Function, with the variables spelled out. Free, no sign-up. Read it before a mock, not during one.

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Formulas: 36
Subject: Mathematics
Exam: JEE Advanced
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Inverse Trigonometric Function, every formula

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Range of sin⁻¹
$sin^{- 1} x \in [- \frac{π}{2}, \frac{π}{2}], x \in [- 1, 1]$
Range of cos⁻¹
$cos^{- 1} x \in [0, π], x \in [- 1, 1]$
Range of tan⁻¹
$tan^{- 1} x \in (- \frac{π}{2}, \frac{π}{2}), x \in R$
Range of cot⁻¹
$cot^{- 1} x \in (0, π), x \in R$
Range of sec⁻¹
$sec^{- 1} x \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π]$
Range of cosec⁻¹
$csc^{- 1} x \in [- \frac{π}{2}, 0) \cup (0, \frac{π}{2}]$
Reciprocal: sin⁻¹ ↔ cosec⁻¹
$sin^{- 1} x = csc^{- 1} \frac{1}{x}, x \in [- 1, 1], x \neq = 0$
Reciprocal: cos⁻¹ ↔ sec⁻¹
$cos^{- 1} x = sec^{- 1} \frac{1}{x}, x \in [- 1, 1], x \neq = 0$
Reciprocal: tan⁻¹ ↔ cot⁻¹
$tan^{- 1} x = cot^{- 1} \frac{1}{x} (x > 0)$
Negative argument: sin⁻¹
$sin^{- 1} (- x) = - sin^{- 1} x$
Negative argument: cos⁻¹
$cos^{- 1} (- x) = π - cos^{- 1} x$
Negative argument: tan⁻¹
$tan^{- 1} (- x) = - tan^{- 1} x$
Negative argument: cot⁻¹
$cot^{- 1} (- x) = π - cot^{- 1} x$
Negative argument: sec⁻¹
$sec^{- 1} (- x) = π - sec^{- 1} x$
sin⁻¹x + cos⁻¹x
$sin^{- 1} x + cos^{- 1} x = \frac{π}{2}, x \in [- 1, 1]$
tan⁻¹x + cot⁻¹x
$tan^{- 1} x + cot^{- 1} x = \frac{π}{2}, x \in R$
sec⁻¹x + cosec⁻¹x
$sec^{- 1} x + csc^{- 1} x = \frac{π}{2}$
sin(sin⁻¹x)
$sin (sin^{- 1} x) = x, x \in [- 1, 1]$
sin⁻¹(sin θ)
$sin^{- 1} (sin θ) = θ, θ \in [- \frac{π}{2}, \frac{π}{2}]$
tan⁻¹x + tan⁻¹y
$tan^{- 1} x + tan^{- 1} y = tan^{- 1} \frac{x + y}{1 - x y}, x y < 1$
tan⁻¹x − tan⁻¹y
$tan^{- 1} x - tan^{- 1} y = tan^{- 1} \frac{x - y}{1 + x y}$
sin⁻¹x + sin⁻¹y
$sin^{- 1} x + sin^{- 1} y = sin^{- 1} (x 1 - y^{2} + y 1 - x^{2})$
for x²+y²≤1
cos⁻¹x + cos⁻¹y
$cos^{- 1} x + cos^{- 1} y = cos^{- 1} (x y - 1 - x^{2} 1 - y^{2})$
for x+y≥0
2 tan⁻¹x (sin form)
$2 tan^{- 1} x = sin^{- 1} \frac{2 x}{1 + x ^{2}}, ∣ x ∣ \leq 1$
2 tan⁻¹x (cos form)
$2 tan^{- 1} x = cos^{- 1} \frac{1 - x ^{2}}{1 + x ^{2}}, x \geq 0$
2 tan⁻¹x (tan form)
$2 tan^{- 1} x = tan^{- 1} \frac{2 x}{1 - x ^{2}}, ∣ x ∣ < 1$
2 sin⁻¹x
$2 sin^{- 1} x = sin^{- 1} (2 x 1 - x^{2}), ∣ x ∣ \leq \frac{1}{2}$
3 sin⁻¹x
$3 sin^{- 1} x = sin^{- 1} (3 x - 4 x^{3}), ∣ x ∣ \leq \frac{1}{2}$
3 cos⁻¹x
$3 cos^{- 1} x = cos^{- 1} (4 x^{3} - 3 x), \frac{1}{2} \leq x \leq 1$
3 tan⁻¹x
$3 tan^{- 1} x = tan^{- 1} \frac{3 x - x ^{3}}{1 - 3 x ^{2}}, ∣ x ∣ < \frac{1}{3}$
tan⁻¹ chain (=π/2)
$tan^{- 1} x + tan^{- 1} y + tan^{- 1} z = \frac{π}{2} if x y + y z + z x = 1$
tan⁻¹ chain (=π)
$tan^{- 1} x + tan^{- 1} y + tan^{- 1} z = π if x + y + z = x y z$
θ∈[0,π]: sin⁻¹(cosθ)
$sin^{- 1} (cos θ) = \frac{π}{2} - θ, θ \in [0, π]$
Trig substitution √(a²−x²)
$a^{2} - x^{2} : put x = a sin θ$
Trig substitution √(a²+x²)
$a^{2} + x^{2} : put x = a tan θ$
Range of (sin⁻¹x)²+(cos⁻¹x)²
$\frac{π ^{2}}{8} \leq (sin^{- 1} x)^{2} + (cos^{- 1} x)^{2} \leq \frac{5 π ^{2}}{4}$

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Inverse Trigonometric Function formulas

Inverse Trigonometric Function, every formula

Range of sin⁻¹

Range of cos⁻¹

Range of tan⁻¹

Range of cot⁻¹

Range of sec⁻¹

Range of cosec⁻¹

Reciprocal: sin⁻¹ ↔ cosec⁻¹

Reciprocal: cos⁻¹ ↔ sec⁻¹

Reciprocal: tan⁻¹ ↔ cot⁻¹

Negative argument: sin⁻¹

Negative argument: cos⁻¹

Negative argument: tan⁻¹

Negative argument: cot⁻¹

Negative argument: sec⁻¹

sin⁻¹x + cos⁻¹x

tan⁻¹x + cot⁻¹x

sec⁻¹x + cosec⁻¹x

sin(sin⁻¹x)

sin⁻¹(sin θ)

tan⁻¹x + tan⁻¹y

tan⁻¹x − tan⁻¹y

sin⁻¹x + sin⁻¹y

cos⁻¹x + cos⁻¹y

2 tan⁻¹x (sin form)

2 tan⁻¹x (cos form)

2 tan⁻¹x (tan form)

2 sin⁻¹x

3 sin⁻¹x

3 cos⁻¹x

3 tan⁻¹x

tan⁻¹ chain (=π/2)

tan⁻¹ chain (=π)

θ∈[0,π]: sin⁻¹(cosθ)

Trig substitution √(a²−x²)

Trig substitution √(a²+x²)

Range of (sin⁻¹x)²+(cos⁻¹x)²

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