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
\(=\frac{\pi }{4}+\frac{2\pi }{3}-\frac{\pi }{6}\),
\(=\frac{3\pi +8\pi -2\pi }{12}=\frac{9\pi }{12}=\frac{3\pi }{4}\).