Class 12 Maths NCERT Solutions Chapter 2 Ex 2.2 – Properties of Inverse Trigonometric Functions

Let x=\sin\theta. Since x\in\left[-\frac12,\frac12\right], \theta\in\left[-\frac{\pi}{6},\frac{\pi}{6}\right] and \sin^{-1}x=\theta.

By the triple-angle identity, \sin3\theta=3\sin\theta-4\sin^3\theta=3x-4x^3.

Since \theta\in\left[-\frac{\pi}{6},\frac{\pi}{6}\right], 3\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] — the principal range of \sin^{-1}.

So \sin^{-1}(3x-4x^3)=\sin^{-1}(\sin3\theta)=3\theta.

Let x=\cos\theta. Since x\in\left[\frac12,1\right], \theta\in\left[0,\frac{\pi}{3}\right] and \cos^{-1}x=\theta.

By the triple-angle identity, \cos3\theta=4\cos^3\theta-3\cos\theta=4x^3-3x.

Since \theta\in\left[0,\frac{\pi}{3}\right], 3\theta\in[0,\pi] — the principal range of \cos^{-1} — so \cos^{-1}(4x^3-3x)=\cos^{-1}(\cos3\theta)=3\theta.

Let x=\tan\theta, \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right), \theta\neq0, so \sqrt{1+x^2}=\sec\theta (positive throughout this range).

The expression becomes \dfrac{\sec\theta-1}{\tan\theta}=\dfrac{1-\cos\theta}{\sin\theta} after multiplying through by \cos\theta.

Using the half-angle identities 1-\cos\theta=2\sin^2\frac{\theta}{2} and \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, this simplifies to \tan\dfrac{\theta}{2}.

Using the half-angle identities 1-\cos x=2\sin^2\frac{x}{2} and 1+\cos x=2\cos^2\frac{x}{2}, the fraction under the root is \tan^2\frac{x}{2}.

Since 0 \lt x \lt \pi puts \frac{x}{2} in the first quadrant, the square root simplifies to \tan\frac{x}{2} without a sign issue.

Dividing numerator and denominator by \cos x gives \dfrac{1-\tan x}{1+\tan x}, which is exactly the tangent-subtraction pattern \tan\left(\frac{\pi}{4}-x\right).

For the given domain, \frac{\pi}{4}-x stays within \left(-\frac{\pi}{2},\frac{\pi}{2}\right), the principal range of \tan^{-1}.

Let x=a\sin\theta, \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right), so \sqrt{a^2-x^2}=a\cos\theta (positive throughout).

The expression reduces to \tan\theta, so \tan^{-1}[\tan\theta]=\theta.

Let x=a\tan\theta, \theta\in\left(-\frac{\pi}{6},\frac{\pi}{6}\right), matching the given bound on x.

Substituting throughout, the numerator becomes a^3(3\tan\theta-\tan^3\theta) and the denominator a^3(1-3\tan^2\theta) — exactly the tangent triple-angle ratio, giving \tan3\theta.

Since \theta\in\left(-\frac{\pi}{6},\frac{\pi}{6}\right), 3\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right), the principal range of \tan^{-1}.

So \tan^{-1}[\tan3\theta]=3\theta directly.

\sin^{-1}\frac12=\frac{\pi}{6}, so 2\sin^{-1}\frac12=\frac{\pi}{3}. Then \cos\frac{\pi}{3}=\frac12, so 2\cos\frac{\pi}{3}=1, and \tan^{-1}(1)=\frac{\pi}{4}.

Let x=\tan\alpha, \alpha\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right). Then \frac{2x}{1+x^2}=\sin2\alpha.

Since 2\alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right), \sin^{-1}\frac{2x}{1+x^2}=2\alpha=2\tan^{-1}x.

Let y=\tan\beta, \beta\in\left(0,\frac{\pi}{2}\right). Then \frac{1-y^2}{1+y^2}=\cos2\beta.

Since 2\beta\in(0,\pi), \cos^{-1}\frac{1-y^2}{1+y^2}=2\beta=2\tan^{-1}y.

Substituting, the expression becomes \tan\left[\tan^{-1}x+\tan^{-1}y\right], which by the tangent addition formula equals \dfrac{x+y}{1-xy} — valid here since xy \lt 1 is given.