Properties of Inverse Trigonometric Functions

A strong understanding of inverse trigonometric identities helps in simplifying expressions, proving relations, and solving board as well as entrance-oriented problems based on principal value and transformation rules.

• Fundamental Relationship: If \[y = \sin^{-1}x\], then this implies \[x = \sin y\], and vice versa.

• Self-Canceling Property: A trigonometric function and its inverse cancel each other out when applied sequentially, provided the values are within the defined domains:

  • \[\sin(\sin^{-1}x) = x\], where \[x \in [-1, 1]\]

  • \[\sin^{-1}(\sin x) = x\], where \[x \in [-\pi/2, \pi/2]\]

• Substitution Method: Many complex inverse trigonometric expressions can be simplified by substituting algebraic variables (like x) with trigonometric variables (like \[\sin\theta\] or \[\sec\theta\]) to utilize standard trigonometric identities.

Express tan⁻¹\[\frac{\cos x}{1-\sin x}\], \[-\frac{3\pi}{2}<x<\frac{\pi}{2}\] in the simplest form.

Solution: We write

tan⁻¹\[\left(\frac{\cos x}{1-\sin x}\right)\]
= tan⁻¹\[\left[\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}\right]\]

= tan⁻¹ \[\left[\frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2}\right]\]

= tan⁻¹ \[\left[\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}\right]\]

= tan⁻¹ \[\left[\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right]\]

= tan⁻¹ \[\left[\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right]\]

\[=\frac{\pi}{4}+\frac{x}{2}\]

Write cot⁻¹\[\left(\frac{1}{\sqrt{x^{2}-1}}\right)\], x > 1 in the simplest form.

Solution: Let x = sec θ, then \[\sqrt{x^2-1}=\sqrt{\sec^2\theta-1}\] = tan θ

Therefore, \[\cot^{-1}\frac{1}{\sqrt{x^{2}-1}}\] = cot⁻¹(cot θ) = θ = sec⁻¹x, which is the simplest form.

i. \[\sin^{-1}\frac{1}{x}=\mathrm{cosec}^{-1}\] if x ≥ 1 or x ≤ −1
\[\cos^{-1}\frac{1}{x}=\sec^{-1}x\] if x ≥ 1 or x ≤ −1
\[\tan^{-1}\frac{1}{x}=\cot^{-1}x\] if x > 0

ii. sin⁻¹(−x) = −sin⁻¹x, for x ∈ [−1, 1]
tan⁻¹(−x) = −tan⁻¹x, for x ∈ R
cosec⁻¹(−x) = −cosec⁻¹x, for x ≥ 1
cos⁻¹(−x) = π − cos⁻¹x, for x ∈ [−1, 1]
sec⁻¹(−x) = π − sec⁻¹x, for x ≥ 1
cot⁻¹(−x) = π − cot⁻¹x, for x ∈ R

\[\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\] for x ∈ [−1, 1]

\[\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2},\] for x ∈ R

\[\sec^{-1}x+\cos\sec^{-1}x=\frac{\pi}{2},\] for |x| ≥ 1

\[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right),\] for x > 0, y > 0 and xy < 1

\[\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right),\] for x, y > 0 and xy > 1

\[\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right),\] for x, y > 0

\[2\tan^{-1}x=\sin^{-1}\left(\frac{2x}{1+x^{2}}\right),\] if −1 ≤ x ≤ 1

\[2\tan^{-1}x=\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right),\] if x > 0

\[2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^{2}}\right),\] if −1 < x < 1