Use these Inter 1st Year Maths 1A Formulas PDF Chapter 8 Inverse Trigonometric Functions to solve questions creatively.

## Intermediate 1st Year Maths 1A Inverse Trigonometric Functions Formulas

→ If sin θ = x and θ ∈ $$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$$, then sin-1(x) = θ,

→ If cos θ x and θ ∈ [0, π], then cos-1(x) = θ

→ If tan θ = x and θ ∈ $$\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$$. then tan-1 (y) = θ

→ If cot θ = x and θ ∈ (0, π), then cot-1(x) = θ.

→ If sec θ = x and θ ∈ [o, $$\frac{\pi}{2}$$) ∪ ($$\frac{\pi}{2}$$, π] then sec-1x = θ.

→ If cosec θ = x and θ ∈ [-$$\frac{\pi}{2}$$, 0) ∪ (o, $$\frac{\pi}{2}$$] then cosec-1x = θ.

→ If x ∈ [-1, 1] – {0}, then sin-1(x) = cosec-1$$\left(\frac{1}{x}\right)$$

→ If x ∈ [-1, 1] – {0}, then cos-1(x) = sec-1$$\left(\frac{1}{x}\right)$$

→ If x > 0, then tan-1(x) = cot-1$$\left(\frac{1}{x}\right)$$ and
x < 0, then tan-1(x) = cot-1$$\left(\frac{1}{x}\right)$$ – π

→ sin-1 x + cos-1x = $$\frac{\pi}{2}$$ (|x| ≤ 1) i.e., – 1 ≤ x ≤ 1

→ tan-1x + cot-1x = $$\frac{\pi}{2}$$, for any x ∈ R

→ sec-1x + cosec-1x = $$\frac{\pi}{2}$$, if (-∞, – 1] ∪ [1, ∞)

 Function y = f(x) Domain (x) Range (y) (i) sinh-1 (x) [-1, 1] $$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$$ (ii) cosh-1 (x) [1, 1] [0, π] (iii) tanh-1 (x) R $$\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$$ (iv) cot-1 (x) R (0, π) (v) sec-1 (x) (-∞, -1] ∪ [1, ∞) [0, $$\frac{\pi}{2}$$) ∪ ($$\frac{\pi}{2}$$, π] (vi) cosec-1 (x) (-∞, -1] ∪ [1, ∞) [-$$\frac{\pi}{2}$$, 0) ∪  (0, $$\frac{\pi}{2}$$, π]

→ Principal values:
For sin-1x, tan-1x, cot-1x, cosec-1x, principal values lies between –$$\frac{\pi}{2}$$ and $$\frac{\pi}{2}$$
For cos-1 x, sec-1 x, principal values lies between 0 and π.

→ tan-1x + tan-1y = tan-1$$\left(\frac{x+y}{1-x y}\right)$$ if (xy < 1), x > 0, y > 0
= $$\frac{\pi}{2}$$ if (xy = 1), x > 0, y > 0
= π +tan-1, if (xy > 1), x > 0, y > 0

→ If x < 0, y < 0 then tan-1x + tan-1y = tan-1$$\left(\frac{x+y}{1-x y}\right)$$ if xy > 1
= – $$\frac{\pi}{2}$$, if xy < 1
= –$$\frac{\pi}{2}$$, if xy = 1

→ If x, y ∈ [0, 1] and x2 + y2 ≤ 1, then sin-1 x + sin-1 y = sin-1$$\left(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right)$$

→ If x, y ∈ [0, 1] and x2 + y2 > 1 then sin-1 x + sin-1 (y) = π – sin-1$$\left(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right)$$

→ If x, y ∈ [0, 1], then sin-1 x + sin-1y) = cos-1($$\sqrt{1-x^{2}} \sqrt{1-y^{2}}$$ – xy)

→ If x, y ∈ [0, 1] then sin-1 y = sin-1(x$$\sqrt{1-y^{2}}$$ – y$$\sqrt{1-x^{2}}$$)

→ If x, y ∈ [0, 1], then cos-1 x + cos-1y = cos-1($$\sqrt{1-x^{2}} \sqrt{1-y^{2}}$$ + xy)

→ If x, y ∈ [0, 1] then cos-1 x + cos-1y = cos-1(xy – $$\sqrt{1-x^{2}} \sqrt{1-y^{2}}$$)

→ If x, y ∈ [0, 1] and x2 + y2 ≥ 1 then cos-1x + cos-1y = sin-1(y$$\sqrt{1-x^{2}}$$ + x$$\sqrt{1-y^{2}}$$)

→ If 0 ≤ x ≤ y ≤ 1 then cos-1 x – cos-1y = cos-1(xy + $$\sqrt{1-x^{2}} \sqrt{1-y^{2}}$$)

→ If x, y ∈ [0, 1], then cos-1 x – cos-1y = sin-1(y$$\sqrt{1-x^{2}}$$ – x$$\sqrt{1-y^{2}}$$)

→ If x ∈ [-1, 1]- {0}, then sin-1 (x) = cosec-1$$\left(\frac{1}{x}\right)$$

→ If x ∈ [-1, 1] – {0}, then cos-1(x) = sec-1$$\left(\frac{1}{x}\right)$$

→ If x > 0, then tan-1x = cot-1$$\left(\frac{1}{x}\right)$$ and

→ If x < 0, then tan-1x = cot-1$$\left(\frac{1}{x}\right)$$ – π

→ sin-1(-x) = -sin-1(x), if x ∈ [-1, 1]

→ cos-1(-x) = π – cos-1(x), if x ∈ [-1, 1]

→ tan-1(-x) = -tan-1 (x), for any x ∈ R

→ For any x ∈ R, cot-7 (-x) = π – cot-1(x)

→ If x ∈ [-∞, -1] ∪ [1, ∞) then

• sec-1(-x) = π – sec-1(x)
• cosec-1(-x) = -cosec-1(x)

→ If θ ∈ $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$, then sin-1(sin θ) = θ and if x ∈ [-1, 1], then sin(sin-1x) = x.

→ If θ ∈ [0, π], then cos-1(cos θ) = θ and if x ∈ [-1, 1], then cos (cos-1x) = x.

→ If θ ∈ $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$, then tan-1(tan θ) = θ and for any x ∈ R, then tan(tan-1x) = x

→ If θ ∈ (0, $$\frac{\pi}{2}$$) ∪ ($$\frac{\pi}{2}$$, π) then cot-1(cot θ) = θ and for any x ∈ R, cot (cot-1x) = x.

→ If θ ∈ (0, $$\frac{\pi}{2}$$) ∪ ($$\frac{\pi}{2}$$, π) then sec-1(sec θ) = θ and if x ∈ (-∞, -1] ∪ [1, ∞) then sec (sec-1x) = x.

→ If θ ∈ [-$$\frac{\pi}{2}$$, 0) ∪ (0, $$\frac{\pi}{2}$$] then cosec-1 (cosec θ) = θ and if x ∈ (-∞, -1] ∪ [1, ∞), then cosec (cosec-1x) = x.

→ θ ∈ [0, π] sin-1(cos θ) = $$\frac{\pi}{2}$$ – θ

→ θ ∈ $$\left[\frac{\pi}{2}, \frac{\pi}{2}\right]$$ ⇒ cos-1(sin θ) = $$\frac{\pi}{2}$$ – θ

→ θ ∈ $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ ⇒ cot-1(tan θ) = $$\frac{\pi}{2}$$ – θ

→ θ ∈ (0, π) ⇒ tan-1(cot θ) = $$\frac{\pi}{2}$$ – θ

→ θ ∈ [-$$\frac{\pi}{2}$$, 0) ∪ (0, $$\frac{\pi}{2}$$] ⇒ sec-1(cosec θ) = $$\frac{\pi}{2}$$ – θ

→ θ ∈ [0, $$\frac{\pi}{2}$$) ∪ ($$\frac{\pi}{2}$$, π] ⇒ cosec-1(sec θ) = $$\frac{\pi}{2}$$ – θ

→ 0 ≤ x ≤ 1 ⇒ sin-1(x) = cos-1$$\left(\sqrt{1-x^{2}}\right)$$ and

→ If -1 ≤ x < 0 ⇒ sin-1(x) = -cos-1$$\left(\sqrt{1-x^{2}}\right)$$

→ -1 < x < 1 ⇒ sin-1(x) = tan-1$$\left(\frac{x}{\sqrt{1-x^{2}}}\right)$$

→ -1 ≤ x < 0 ⇒ cos-1(x) = π – sin-1$$\left(\sqrt{1-x^{2}}\right)$$ = π + tan-1$$\left(\frac{\sqrt{1-x^{2}}}{x}\right)$$

→ 0 ≤ x ≤ 7 ⇒ cos-1(x) = sin-1$$\left(\sqrt{1-x^{2}}\right)$$ = tan-1$$\left(\frac{\sqrt{1-x^{2}}}{x}\right)$$

→ x > 0 ⇒ tan-1(x) = sin-1$$\left(\frac{x}{\sqrt{1+x^{2}}}\right)$$ = cos-1$$\left(\frac{1}{\sqrt{1+x^{2}}}\right)$$

→ tan-1(x) + tan-1(y) + tan-1(z) = tan-1$$\left[\frac{x+y+z-x y z}{1-x y-y z-z x}\right]$$, ifx, y, z have the same sign and xy + yz + zx < 1.

→ 2 sin-1(x) = sin-12x$$\sqrt{1-x^{2}}$$, if x ≤ $$\frac{1}{\sqrt{2}}$$
= π – sin-12x$$\sqrt{1-x^{2}}$$, if x > $$\frac{1}{\sqrt{2}}$$

→ 2 cos-1(x) = cos-1(2x2 – 1), if x ≥ $$\frac{1}{\sqrt{2}}$$
= π – cos-1(1 – 2x2), if x < $$\frac{1}{\sqrt{2}}$$

→ 2 tan-1(x) = tan-1$$\left(\frac{2 x}{1-x^{2}}\right)$$, if |x| < 1
= π – tan-1$$\left(\frac{2 x}{1-x^{2}}\right)$$, if |x| ≥ 1

→ 2tan-1(x) = sin-1$$\left(\frac{2 x}{1+x^{2}}\right)$$, ∀ x ∈ R
= cos-1$$\left(\frac{1-x^{2}}{1+x^{2}}\right)$$, if x ≥ 0
= -cos-1$$\left(\frac{1-x^{2}}{1+x^{2}}\right)$$, if x < 0
= tan-1$$\left(\frac{2 x}{1-x^{2}}\right)$$, ∀ x ∈ R

→ 3 sin-1(x) = sin-1(3x – 4x3) for 0 ≤ x < $$\frac{1}{2}$$
3 cos-1(x) = cos-1(4x3 – 3x) for $$\frac{\sqrt{3}}{2}$$ ≤ x < 1
3 tan-1(x) = tan-1$$\left\{\frac{3 x-x^{3}}{1-3 x^{2}}\right\}$$ for 0 ≤ x < $$\frac{1}{\sqrt{3}}$$

→ If sin θ = x, we write θ = sin-1x.

→ sin(sin-1x) = x, sin-1(sin θ) = θ if ‘θ‘ ∈ $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

→ cos(cos-1x)=x, cos-1(cos θ) = θ if θ ∈ [0, n]

→ tan (tan-1x) = x, tan-1(tan θ) = θ if θ ∈ $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

→ That value of sin-1x lying between – $$\frac{\pi}{2}$$ and$$\frac{\pi}{2}$$ is called the principal value of sin-1x.

→ That value of cos-1x lying between 0 and π is called the principal value of cos-1x.

→ That value of tan-1x lying between – $$\frac{\pi}{2}$$ and $$\frac{\pi}{2}$$ is called the principal value of tan-1x.

→ If -1 ≤ x ≤ 1, then

• sin-1(- x) = – sin-1x
• cos-1(-x) = π – cos-1x

→ If x ∈ R , then

• tan-1(-x) = -tan-1x
• cot-1(-x) = π – cot-1x

→ If x ≤ -1 or x ≥ 1, then

• cosec-1(-x) = -cosec-1x
• sec-1(-x) = π – sec-1x

→ cosec-1x = sin-1$$\frac{1}{x}$$ (if x ≠ 0)

→ sec-1x = cos-1$$\frac{1}{x}$$ (if x ≠ 0)

→ cot-1x = tan-1$$\frac{1}{x}$$ (if x > 0)
= π + tan-1$$\frac{1}{x}$$ (if x < 0)

→ sin-1x + cos-1x = π/2 ,

→ tan-1x + cot-1x = π/2,

→ sec-1x + cosec-1x = π/2.

→ If sin-1x + sin-1y = π/2, then x2 + y2 = 1.

→ sin(cos-1x) = $$\sqrt{1-x^{2}}$$, cos(sin-1x) = $$\sqrt{1-x^{2}}$$

→ sin-1 = cos-1$$\sqrt{1-x^{2}}$$ for 0 ≤ x ≤ 1
= -cos-1$$\sqrt{1-x^{2}}$$ for – 1 ≤ x ≤ 0