# Inverse Trigonometric Functions – Problems

## Inverse Trigonometric Functions – Problems

If a function is one – one and onto from A to B, then function g which associates each elements y ϵ B to one and only one element x ϵ A such that y = f(x), then g is called the inverse function of f, denoted by x = g(y).

g = f⁻¹

∴ x = f⁻¹ (y)

Problems:

1. tan⁻¹ √3 – sec⁻¹ (-2) is equal

Solution: Given that tan⁻¹ √3 – sec⁻¹ (-2)

Let tan⁻¹ √3 = x

tan x = √3

$$\tan x=\tan \left( \frac{\pi }{3} \right)$$,

$$x=\frac{\pi }{3}\in \left( \frac{-\pi }{2},\frac{\pi }{2} \right)$$ … (1)

sec⁻¹ (-2) = y

sec y = -2

$$\sec y=-\sec \left( \frac{\pi }{3} \right)$$,

$$\sec y=\sec \left( \pi -\frac{\pi }{3} \right)$$,

$$\sec y=\sec \left( \frac{2\pi }{3} \right)$$,

$$y=\frac{2\pi }{3}\in [0,\pi ]-\left( \frac{\pi }{2} \right)$$ … (2)

Equation (1) and (2) substitute in given equation

∴ $${{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}(-2)=x-y=\frac{\pi }{3}-\frac{2\pi }{3}=-\frac{\pi }{3}$$,

2. $${{\tan }^{-1}}(1)+{{\cos }^{-1}}\left( \frac{-1}{2} \right)+{{\sin }^{-1}}\left( \frac{-1}{2} \right)$$ equal to

Solution: Let tan⁻¹ (1) = x

tan x = 1

$$\tan x=\tan \left( \frac{\pi }{4} \right)$$,

$$x=\frac{\pi }{4}$$ … (1)

cos⁻¹ (-½) = y

cos y = -½

$$\cos y=-\cos \left( \frac{\pi }{3} \right)$$,

$$\cos y=-\cos \left( \pi -\frac{\pi }{3} \right)$$,

$$\cos y=\cos \left( \frac{2\pi }{3} \right)$$,

$$y=\left( \frac{2\pi }{3} \right)$$ … (2)

sin⁻¹ (-½) = z

sin z = -½

$$\sin z=-\sin \left( \frac{\pi }{3} \right)$$,

$$z=-\frac{\pi }{3}$$ … (3)

$$z\in \left[ \frac{-\pi }{2},\frac{\pi }{2} \right]$$,

Equation (1), (2), and (3) substitute in given equation

tan⁻¹ (1) + cos⁻¹ (-½) + sin⁻¹ (-½) = x + y + z

$$=\frac{\pi }{4}+\frac{2\pi }{3}-\frac{\pi }{6}$$,

$$=\frac{3\pi +8\pi -2\pi }{12}=\frac{9\pi }{12}=\frac{3\pi }{4}$$.