# All Inverse trigonometry functions and formulas or Identity list | pdf

## Inverse trigonometry functions and formulas

This is for all competitive exams like SSC, Railway, Bank, TGT, PGT and other competitive exams and academic exams like intermediate or High school exam CBSE, ICSE (class 10, 11, 12) or State board.

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### Negative Property

sin-1 (-x) = -sin-1 x

cos-1 (-x) = π - cos-1 x

tan-1 (-x) = -tan-1 x

cot-1 (-x) = π - cot-1 x

sec-1 (-x) = π - sec-1 x

cosec-1 (-x) = -cosec-1 x

### Complementary Property

sin-1 x + cos-1 x = $$\frac {π}{2}$$

tan-1 x + cot-1 x = $$\frac {π}{2}$$

sec-1 x + cosec-1 x = $$\frac {π}{2}$$

### Properties of Inverse Circular function

sin-1 x = cosec-1 $$\frac {1}{x}$$

cosec-1 x = sin-1 $$\frac {1}{x}$$

cos-1 x = sec-1 $$\frac {1}{x}$$

sec-1 x = cos-1 $$\frac {1}{x}$$

tan-1 x = cot-1 $$\frac {1}{x}$$

cot-1 x = tan-1 $$\frac {1}{x}$$

sin-1 (sin x) = x   :   x ∈ $$\left [ \frac {-π}{2}, \frac {π}{2} \right ]$$

cos-1 (cos x) = x   :   x ∈ $$\left [ 0, π \right ]$$

tan-1 (tan x) = x   :   x ∈ $$\left ( \frac {-π}{2}, \frac {π}{2} \right )$$

cot-1 (cot x) = x   :   x ∈ $$\left ( 0, π \right )$$

sec-1 (sec x) = x   :   x ∈ [0, π] - $$\left\{ \frac {π}{2} \right\}$$

cosec-1 (cosec x) = x   :   x ∈ $$\left [ \frac {-π}{2}, \frac {π}{2} \right ] - \left\{0\right\}$$

sin (sin-1 x) = x   :   x ∈ $$\left [ -1, 1 \right ]$$

cos (cos-1 x) = x   :   x ∈ $$\left [ -1, 1 \right ]$$

tan (tan-1 x) = x   :   x ∈ R

cot (cot-1 x) = x   :   x ∈ R

sec (sec-1 x) = x   :   x ∈ R - (-1, 1)

cosec (cosec-1 x) = x   :   x ∈ R - (-1, 1)

sin-1 x + sin-1 y = sin-1 $$x\sqrt {1 - y^2} + y\sqrt {1 - x^2}$$;  x ≥ 0, y ≥ 0 & x2 + y2 ≤ 1

sin-1 x + sin-1 y = π - sin-1 $$x\sqrt {1 - y^2} + y\sqrt {1 - x^2}$$;  x ≥ 0, y ≥ 0 & x2 + y2 ≥ 1

sin-1 x - sin-1 y = sin-1 $$x\sqrt {1 - y^2} - y\sqrt {1 - x^2}$$;  x,y ∈ [0, 1]

cos-1 x + cos-1 y = cos-1 $$xy - \sqrt {1 - x^2}\sqrt {1 - y^2}$$;  x,y ∈ [0, 1]

cos-1 x - cos-1 y = cos-1 $$xy + \sqrt {1 - x^2}\sqrt {1 - y^2}$$;  0 ≤ x < y ≤ 1

cos-1 x - cos-1 y = - cos-1 $$xy + \sqrt {1 - x^2}\sqrt {1 - y^2}$$;  0 ≤ y < x ≤ 1

tan-1 x + tan-1 y = tan-1 $$\frac{x+y}{1 - xy}$$;  x,y > 0 & xy < 1

tan-1 x + tan-1 y = π + tan-1 $$\frac{x+y}{1 - xy}$$;  x,y > 0 & xy > 1

tan-1 x + tan-1 y = π/2 if x,y > 0 & xy = 1

tan-1 x + tan-1 y = -π/2 if x,y < 0 & xy = 1

tan-1 x - tan-1 y = tan-1 $$\frac{x-y}{1 + xy}$$;  x,y ≥ 0 & xy > -1

2tan-1 x = sin-1 $$\frac {2x}{1+x^2}$$;  -1 ≤ x ≤ 1

2tan-1 x = tan-1 $$\frac {2x}{1-x^2}$$;   -1 < x < 1

2tan-1 x = cos-1 $$\frac {1-x^2}{1+x^2}$$;   x ≥ 0

2sin-1 x = sin-1 $$2x\sqrt {1 - x^2}$$

3sin-1 x = sin-1 (3x - 4x3 )

2cos-1 x = cos-1 (2x2 - 1)

3cos-1 x = cos-1 (4x3 - 3x)

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

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