Surjective is not a type of relation. It is a type of function. Reflexive, Symmetric and Transitive are type of relations.

A binary operation ‘*’ defined on a set A is said to be commutative only if a * b = b *a, ∀ a, b ∈ A.

If (a * b) * c = a * (b * c), then the operation is said to associative ∀ a, b∈ A.

If (b ο c) * a = (b * a) ο (c * a), then the operation is said to be distributive ∀ a, b, c ∈ A.

tan⁻¹ √3 = π/3, sec⁻¹2 = π/3, cos⁻¹1 = 0

tan⁻¹√3 + sec⁻¹2 – cos−¹1 = π/3 + π/3

= 2π/3

Let sin^-1⁡x = y

⇒ x = sin⁡y

⇒ x = √1 - cos²y

⇒ x² = 1 - cos²y

⇒ cos²y = 1 - x²

∴ y = cos^-1⁡ √1 - x² = sin^-1⁡x

A relation in a set A is said to be symmetric if (a₁, a₂)∈R implies that (a₁, a₂)∈R,for every a₁, a₂∈R.

Hence, for the given set A={1, 2, 3}, R={(1, 2), (2, 1)} is symmetric. It is not reflexive since every element is not related to itself and neither transitive as it does not satisfy the condition that for a given relation R in a set A if (a₁, a₂)∈R and (a₂, a₃)∈R implies that (a₁, a₃)∈ R for every a₁, a₂, a₃∈R.

Given that, g(x) = 3 x 2 + 7 and f(x) = √x

∴ gοf(x) = g(f(x)) = g(√x) = 3(√x)2 + 7 = 3x + 7.

Hence, gοf(x) = 3x + 7.

This is an equivalence relation. A relation R is said to be an equivalence relation when it is reflexive, transitive and symmetric.

Reflexive: We know that a line is always parallel to itself. This implies that I₁ is parallel to I₁ i.e. (I₁, I₂)∈R. Hence, it is a reflexive relation.

Symmetric: Now if a line I₁ || I₂ then the line I₂ || I₁. Therefore, (I₁, I₂)∈R implies that (I₂, I₁)∈R. Hence, it is a symmetric relation.

Transitive: If two lines (I₁, I₃) are parallel to a third line (I₂) then they will be parallel to each other i.e. if (I₁, I₂) ∈R and (I₂, I₃) ∈R implies that (I₁, I₃) ∈R.

Let sec^-1⁡x = y

⇒ x = sec⁡y

⇒ x = √ 1 + tan²y

⇒ x² - 1 = tan²y

∴ y = tan^-1√x² - 1 = sec^-1⁡x