# arcsin (x) - arcsin(y) = arcsin (x $$\sqrt{1 - y^{2}}$$ - y$$\sqrt{1 - x^{2}}$$)

We will learn how to prove the property of the inverse trigonometric function arcsin (x) - arcsin(y) = arcsin (x $$\sqrt{1 - y^{2}}$$ - y$$\sqrt{1 - x^{2}}$$)

Proof:

Let, sin$$^{-1}$$ x = α and sin$$^{-1}$$ y = β

From sin$$^{-1}$$ x = α we get,

x = sin α

and from sin$$^{-1}$$ y = β we get,

y = sin β

Now, sin (α - β) = sin α cos β - cos α sin β

⇒ sin (α - β) = sin α $$\sqrt{1 - sin^{2} β}$$ - $$\sqrt{1 - sin^{2} α}$$ sin β

⇒ sin (α - β) = x ∙ $$\sqrt{1 - y^{2}}$$ - $$\sqrt{1 - x^{2}}$$ ∙ y

Therefore, α - β = sin$$^{-1}$$ (x $$\sqrt{1 - y^{2}}$$ - y$$\sqrt{1 - x^{2}}$$)

or, sin$$^{-1}$$ x - sin$$^{-1}$$ y = sin$$^{-1}$$ (x $$\sqrt{1 - y^{2}}$$ - y$$\sqrt{1 - x^{2}}$$).       Proved.

Note: If x > 0, y > 0 and x$$^{2}$$ + y$$^{2}$$ > 1, then the sin$$^{-1}$$ x + sin$$^{-1}$$ y may be an angle more than π/2 while sin$$^{-1}$$ (x $$\sqrt{1 - y^{2}}$$ + y$$\sqrt{1 - x^{2}}$$), is an angle between – π/2 and π/2.

Therefore, sin$$^{-1}$$ x - sin$$^{-1}$$ y = π - sin$$^{-1}$$ (x $$\sqrt{1 - y^{2}}$$ + y$$\sqrt{1 - x^{2}}$$)