Chapter 2: Inverse Trigonometric Functions

Let $$\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = y$$. Then $$\sin y = \dfrac{1}{\sqrt{2}}$$.

The principal value branch of $$\sin^{-1}$$ is $$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$, so we need the angle $$y$$ in this interval whose sine is $$\dfrac{1}{\sqrt{2}}$$.

We know that $$\sin\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$$, and $$\dfrac{\pi}{4} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$.

Hence the principal value is $$\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$$.

Let $$\cot^{-1}\left(-\dfrac{1}{\sqrt{3}}\right) = y$$. Then $$\cot y = -\dfrac{1}{\sqrt{3}}$$.

The principal value branch of $$\cot^{-1}$$ is $$(0, \pi)$$, so we need the angle $$y$$ in this interval with $$\cot y = -\dfrac{1}{\sqrt{3}}$$.

Since $$\cot\dfrac{\pi}{3} = \dfrac{1}{\sqrt{3}}$$ and the required value is negative, $$y$$ lies in $$\left(\dfrac{\pi}{2}, \pi\right)$$.

Using $$\cot(\pi - \theta) = -\cot\theta$$, we get $$\cot\left(\pi - \dfrac{\pi}{3}\right) = -\cot\dfrac{\pi}{3} = -\dfrac{1}{\sqrt{3}}$$, i.e. $$\cot\dfrac{2\pi}{3} = -\dfrac{1}{\sqrt{3}}$$.

Since $$\dfrac{2\pi}{3} \in (0, \pi)$$, the principal value is $$\cot^{-1}\left(-\dfrac{1}{\sqrt{3}}\right) = \dfrac{2\pi}{3}$$.

Let $$\sin^{-1}\left(-\dfrac{1}{2}\right) = y$$. Then $$\sin y = -\dfrac{1}{2}$$.

The principal value branch of $$\sin^{-1}$$ is $$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$.

We know $$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$$, so $$\sin\left(-\dfrac{\pi}{6}\right) = -\sin\dfrac{\pi}{6} = -\dfrac{1}{2}$$.

Since $$-\dfrac{\pi}{6} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$, the principal value is $$\sin^{-1}\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$$.

Let $$\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = y$$. Then $$\cos y = \dfrac{\sqrt{3}}{2}$$.

The principal value branch of $$\cos^{-1}$$ is $$[0, \pi]$$.

We know $$\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$$, and $$\dfrac{\pi}{6} \in [0, \pi]$$.

Hence the principal value is $$\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{6}$$.

Let $$\operatorname{cosec}^{-1}(2) = y$$. Then $$\operatorname{cosec} y = 2$$, i.e. $$\sin y = \dfrac{1}{2}$$.

The principal value branch of $$\operatorname{cosec}^{-1}$$ is $$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}$$.

We know $$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$$, so $$\operatorname{cosec}\dfrac{\pi}{6} = 2$$, and $$\dfrac{\pi}{6}$$ lies in the principal branch.

Hence the principal value is $$\operatorname{cosec}^{-1}(2) = \dfrac{\pi}{6}$$.

Let $$\tan^{-1}(-\sqrt{3}) = y$$. Then $$\tan y = -\sqrt{3}$$.

The principal value branch of $$\tan^{-1}$$ is $$\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$.

We know $$\tan\dfrac{\pi}{3} = \sqrt{3}$$, so $$\tan\left(-\dfrac{\pi}{3}\right) = -\tan\dfrac{\pi}{3} = -\sqrt{3}$$.

Since $$-\dfrac{\pi}{3} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$, the principal value is $$\tan^{-1}(-\sqrt{3}) = -\dfrac{\pi}{3}$$.

Let $$\cos^{-1}\left(-\dfrac{1}{2}\right) = y$$. Then $$\cos y = -\dfrac{1}{2}$$.

The principal value branch of $$\cos^{-1}$$ is $$[0, \pi]$$.

We know $$\cos\dfrac{\pi}{3} = \dfrac{1}{2}$$. Using $$\cos(\pi - \theta) = -\cos\theta$$, $$\cos\left(\pi - \dfrac{\pi}{3}\right) = -\dfrac{1}{2}$$, i.e. $$\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$$.

Since $$\dfrac{2\pi}{3} \in [0, \pi]$$, the principal value is $$\cos^{-1}\left(-\dfrac{1}{2}\right) = \dfrac{2\pi}{3}$$.

Let $$\tan^{-1}(-1) = y$$. Then $$\tan y = -1$$.

The principal value branch of $$\tan^{-1}$$ is $$\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$.

We know $$\tan\dfrac{\pi}{4} = 1$$, so $$\tan\left(-\dfrac{\pi}{4}\right) = -\tan\dfrac{\pi}{4} = -1$$.

Since $$-\dfrac{\pi}{4} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$, the principal value is $$\tan^{-1}(-1) = -\dfrac{\pi}{4}$$.

Let $$\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right) = y$$. Then $$\sec y = \dfrac{2}{\sqrt{3}}$$, i.e. $$\cos y = \dfrac{\sqrt{3}}{2}$$.

The principal value branch of $$\sec^{-1}$$ is $$[0, \pi] - \left\{\dfrac{\pi}{2}\right\}$$.

We know $$\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$$, so $$\sec\dfrac{\pi}{6} = \dfrac{2}{\sqrt{3}}$$, and $$\dfrac{\pi}{6}$$ lies in the principal branch.

Hence the principal value is $$\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right) = \dfrac{\pi}{6}$$.

Let $$\cot^{-1}(\sqrt{3}) = y$$. Then $$\cot y = \sqrt{3}$$.

The principal value branch of $$\cot^{-1}$$ is $$(0, \pi)$$.

We know $$\cot\dfrac{\pi}{6} = \sqrt{3}$$, and $$\dfrac{\pi}{6} \in (0, \pi)$$.

Hence the principal value is $$\cot^{-1}(\sqrt{3}) = \dfrac{\pi}{6}$$.

Let $$\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = y$$. Then $$\cos y = -\dfrac{1}{\sqrt{2}}$$.

The principal value branch of $$\cos^{-1}$$ is $$[0, \pi]$$.

We know $$\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$$. Using $$\cos(\pi - \theta) = -\cos\theta$$, $$\cos\left(\pi - \dfrac{\pi}{4}\right) = -\dfrac{1}{\sqrt{2}}$$, i.e. $$\cos\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt{2}}$$.

Since $$\dfrac{3\pi}{4} \in [0, \pi]$$, the principal value is $$\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = \dfrac{3\pi}{4}$$.

Let $$\operatorname{cosec}^{-1}(-\sqrt{2}) = y$$. Then $$\operatorname{cosec} y = -\sqrt{2}$$, i.e. $$\sin y = -\dfrac{1}{\sqrt{2}}$$.

The principal value branch of $$\operatorname{cosec}^{-1}$$ is $$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}$$.

We know $$\sin\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$$, so $$\sin\left(-\dfrac{\pi}{4}\right) = -\dfrac{1}{\sqrt{2}}$$, giving $$\operatorname{cosec}\left(-\dfrac{\pi}{4}\right) = -\sqrt{2}$$.

Since $$-\dfrac{\pi}{4}$$ lies in the principal branch, the principal value is $$\operatorname{cosec}^{-1}(-\sqrt{2}) = -\dfrac{\pi}{4}$$.

We evaluate each inverse trigonometric value using its principal value branch.

$$\tan^{-1}(1)$$: since $$\tan\dfrac{\pi}{4} = 1$$ and $$\dfrac{\pi}{4} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$, $$\tan^{-1}(1) = \dfrac{\pi}{4}$$.

$$\cos^{-1}\left(-\dfrac{1}{2}\right)$$: since $$\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$$ and $$\dfrac{2\pi}{3} \in [0, \pi]$$, $$\cos^{-1}\left(-\dfrac{1}{2}\right) = \dfrac{2\pi}{3}$$.

$$\sin^{-1}\left(-\dfrac{1}{2}\right)$$: since $$\sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2}$$ and $$-\dfrac{\pi}{6} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$, $$\sin^{-1}\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$$.

Adding the three values:

$$\tan^{-1}(1) + \cos^{-1}\left(-\dfrac{1}{2}\right) + \sin^{-1}\left(-\dfrac{1}{2}\right) = \dfrac{\pi}{4} + \dfrac{2\pi}{3} - \dfrac{\pi}{6}$$

Taking the LCM $$12$$: $$\dfrac{3\pi}{12} + \dfrac{8\pi}{12} - \dfrac{2\pi}{12} = \dfrac{9\pi}{12} = \dfrac{3\pi}{4}$$.

We evaluate each value using its principal value branch.

$$\cos^{-1}\left(\dfrac{1}{2}\right)$$: since $$\cos\dfrac{\pi}{3} = \dfrac{1}{2}$$ and $$\dfrac{\pi}{3} \in [0, \pi]$$, $$\cos^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3}$$.

$$\sin^{-1}\left(\dfrac{1}{2}\right)$$: since $$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$$ and $$\dfrac{\pi}{6} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$, $$\sin^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{6}$$.

Therefore:

$$\cos^{-1}\left(\dfrac{1}{2}\right) + 2\sin^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3} + 2 \times \dfrac{\pi}{6} = \dfrac{\pi}{3} + \dfrac{\pi}{3} = \dfrac{2\pi}{3}$$

If $$\sin^{-1} x = y$$, then $$y$$ is the value of the inverse sine function, which is defined to lie in its principal value branch.

The principal value branch of $$\sin^{-1}$$ is the closed interval $$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$. It is a closed interval because $$\sin^{-1}$$ is defined for all $$x \in [-1, 1]$$, and it attains the endpoint values $$y = \pm\dfrac{\pi}{2}$$ at $$x = \pm 1$$.

Hence $$-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$$, which is option (B). Options (C) and (D) wrongly use open intervals, and option (A) gives the range of $$\cos^{-1}$$, not $$\sin^{-1}$$.

$$\tan^{-1}\sqrt{3}$$: since $$\tan\dfrac{\pi}{3} = \sqrt{3}$$ and $$\dfrac{\pi}{3} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$, $$\tan^{-1}\sqrt{3} = \dfrac{\pi}{3}$$.

$$\sec^{-1}(-2)$$: we need $$\sec y = -2$$, i.e. $$\cos y = -\dfrac{1}{2}$$, with $$y \in [0, \pi] - \left\{\dfrac{\pi}{2}\right\}$$. Since $$\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$$, we get $$\sec^{-1}(-2) = \dfrac{2\pi}{3}$$.

Therefore:

$$\tan^{-1}\sqrt{3} - \sec^{-1}(-2) = \dfrac{\pi}{3} - \dfrac{2\pi}{3} = -\dfrac{\pi}{3}$$

This is option (B).

Write the numerator and denominator in terms of the half angle $$\dfrac{x}{2}$$.

Numerator: $$\cos x = \cos^2\dfrac{x}{2} - \sin^2\dfrac{x}{2} = \left(\cos\dfrac{x}{2} - \sin\dfrac{x}{2}\right)\left(\cos\dfrac{x}{2} + \sin\dfrac{x}{2}\right)$$.

Denominator: $$1 - \sin x = \cos^2\dfrac{x}{2} + \sin^2\dfrac{x}{2} - 2\sin\dfrac{x}{2}\cos\dfrac{x}{2} = \left(\cos\dfrac{x}{2} - \sin\dfrac{x}{2}\right)^2$$.

Therefore:

$$\dfrac{\cos x}{1 - \sin x} = \dfrac{\left(\cos\dfrac{x}{2} - \sin\dfrac{x}{2}\right)\left(\cos\dfrac{x}{2} + \sin\dfrac{x}{2}\right)}{\left(\cos\dfrac{x}{2} - \sin\dfrac{x}{2}\right)^2} = \dfrac{\cos\dfrac{x}{2} + \sin\dfrac{x}{2}}{\cos\dfrac{x}{2} - \sin\dfrac{x}{2}}$$

Dividing numerator and denominator by $$\cos\dfrac{x}{2}$$:

$$\dfrac{\cos x}{1 - \sin x} = \dfrac{1 + \tan\dfrac{x}{2}}{1 - \tan\dfrac{x}{2}} = \tan\left(\dfrac{\pi}{4} + \dfrac{x}{2}\right)$$

Hence $$\tan^{-1}\left(\dfrac{\cos x}{1 - \sin x}\right) = \tan^{-1}\left[\tan\left(\dfrac{\pi}{4} + \dfrac{x}{2}\right)\right]$$.

For $$-\dfrac{3\pi}{2} < x < \dfrac{\pi}{2}$$ we have $$-\dfrac{\pi}{2} < \dfrac{\pi}{4} + \dfrac{x}{2} < \dfrac{\pi}{2}$$, so the angle lies in the principal branch of $$\tan^{-1}$$.

Therefore $$\tan^{-1}\left(\dfrac{\cos x}{1 - \sin x}\right) = \dfrac{\pi}{4} + \dfrac{x}{2}$$.

Put $$x = \sec\theta$$, so that $$\theta = \sec^{-1} x$$. Since $$x > 1$$, the angle $$\theta \in \left(0, \dfrac{\pi}{2}\right)$$.

Then $$\sqrt{x^2 - 1} = \sqrt{\sec^2\theta - 1} = \sqrt{\tan^2\theta} = \tan\theta$$ (positive, since $$\theta$$ is acute).

Therefore:

$$\cot^{-1}\left(\dfrac{1}{\sqrt{x^2 - 1}}\right) = \cot^{-1}\left(\dfrac{1}{\tan\theta}\right) = \cot^{-1}(\cot\theta)$$

Since $$\theta \in \left(0, \dfrac{\pi}{2}\right) \subset (0, \pi)$$, the principal value branch of $$\cot^{-1}$$, we get $$\cot^{-1}(\cot\theta) = \theta$$.

Hence $$\cot^{-1}\left(\dfrac{1}{\sqrt{x^2 - 1}}\right) = \theta = \sec^{-1} x$$.

Put $$x = \sin\theta$$, so that $$\theta = \sin^{-1} x$$. Since $$x \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$$, the angle satisfies $$-\dfrac{\pi}{6} \le \theta \le \dfrac{\pi}{6}$$.

Using the triple-angle identity $$\sin 3\theta = 3\sin\theta - 4\sin^3\theta$$:

$$3x - 4x^3 = 3\sin\theta - 4\sin^3\theta = \sin 3\theta$$

Therefore $$\sin^{-1}(3x - 4x^3) = \sin^{-1}(\sin 3\theta)$$.

Since $$-\dfrac{\pi}{6} \le \theta \le \dfrac{\pi}{6}$$, we have $$-\dfrac{\pi}{2} \le 3\theta \le \dfrac{\pi}{2}$$, which is the principal value branch of $$\sin^{-1}$$. Hence $$\sin^{-1}(\sin 3\theta) = 3\theta$$.

Therefore $$\sin^{-1}(3x - 4x^3) = 3\theta = 3\sin^{-1} x$$, which proves $$3\sin^{-1} x = \sin^{-1}(3x - 4x^3)$$.

Put $$x = \cos\theta$$, so that $$\theta = \cos^{-1} x$$. Since $$x \in \left[\dfrac{1}{2}, 1\right]$$, the angle satisfies $$0 \le \theta \le \dfrac{\pi}{3}$$.

Using the triple-angle identity $$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$$:

$$4x^3 - 3x = 4\cos^3\theta - 3\cos\theta = \cos 3\theta$$

Therefore $$\cos^{-1}(4x^3 - 3x) = \cos^{-1}(\cos 3\theta)$$.

Since $$0 \le \theta \le \dfrac{\pi}{3}$$, we have $$0 \le 3\theta \le \pi$$, which is the principal value branch of $$\cos^{-1}$$. Hence $$\cos^{-1}(\cos 3\theta) = 3\theta$$.

Therefore $$\cos^{-1}(4x^3 - 3x) = 3\theta = 3\cos^{-1} x$$, which proves $$3\cos^{-1} x = \cos^{-1}(4x^3 - 3x)$$.

Put $$x = \tan\theta$$, so that $$\theta = \tan^{-1} x$$. Then $$\sqrt{1 + x^2} = \sqrt{1 + \tan^2\theta} = \sec\theta$$.

Substitute into the expression:

$$\dfrac{\sqrt{1 + x^2} - 1}{x} = \dfrac{\sec\theta - 1}{\tan\theta} = \dfrac{\dfrac{1}{\cos\theta} - 1}{\dfrac{\sin\theta}{\cos\theta}} = \dfrac{1 - \cos\theta}{\sin\theta}$$

Using $$1 - \cos\theta = 2\sin^2\dfrac{\theta}{2}$$ and $$\sin\theta = 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}$$:

$$\dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{2\sin^2\dfrac{\theta}{2}}{2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}} = \tan\dfrac{\theta}{2}$$

Therefore:

$$\tan^{-1}\dfrac{\sqrt{1 + x^2} - 1}{x} = \tan^{-1}\left(\tan\dfrac{\theta}{2}\right) = \dfrac{\theta}{2} = \dfrac{1}{2}\tan^{-1} x$$

Use the half-angle identities $$1 - \cos x = 2\sin^2\dfrac{x}{2}$$ and $$1 + \cos x = 2\cos^2\dfrac{x}{2}$$.

$$\dfrac{1 - \cos x}{1 + \cos x} = \dfrac{2\sin^2\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}} = \tan^2\dfrac{x}{2}$$

Hence:

$$\sqrt{\dfrac{1 - \cos x}{1 + \cos x}} = \sqrt{\tan^2\dfrac{x}{2}} = \left|\tan\dfrac{x}{2}\right|$$

For $$0 < x < \pi$$ we have $$0 < \dfrac{x}{2} < \dfrac{\pi}{2}$$, so $$\tan\dfrac{x}{2} > 0$$ and the modulus can be dropped.

Therefore:

$$\tan^{-1}\left(\sqrt{\dfrac{1 - \cos x}{1 + \cos x}}\right) = \tan^{-1}\left(\tan\dfrac{x}{2}\right) = \dfrac{x}{2}$$

Divide the numerator and denominator of the fraction by $$\cos x$$:

$$\dfrac{\cos x - \sin x}{\cos x + \sin x} = \dfrac{1 - \tan x}{1 + \tan x}$$

Using $$\tan\dfrac{\pi}{4} = 1$$ and the formula $$\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A\tan B}$$:

$$\dfrac{1 - \tan x}{1 + \tan x} = \dfrac{\tan\dfrac{\pi}{4} - \tan x}{1 + \tan\dfrac{\pi}{4}\tan x} = \tan\left(\dfrac{\pi}{4} - x\right)$$

Therefore:

$$\tan^{-1}\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right) = \tan^{-1}\left[\tan\left(\dfrac{\pi}{4} - x\right)\right]$$

For $$-\dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$$ we have $$-\dfrac{\pi}{2} < \dfrac{\pi}{4} - x < \dfrac{\pi}{2}$$, which is the principal branch of $$\tan^{-1}$$.

Hence the expression equals $$\dfrac{\pi}{4} - x$$.

Put $$x = a\sin\theta$$, so that $$\theta = \sin^{-1}\dfrac{x}{a}$$. Since $$|x| < a$$, the angle $$\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$, where $$\cos\theta > 0$$.

Then:

$$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\sqrt{\cos^2\theta} = a\cos\theta$$

Therefore:

$$\dfrac{x}{\sqrt{a^2 - x^2}} = \dfrac{a\sin\theta}{a\cos\theta} = \tan\theta$$

Hence:

$$\tan^{-1}\dfrac{x}{\sqrt{a^2 - x^2}} = \tan^{-1}(\tan\theta) = \theta = \sin^{-1}\dfrac{x}{a}$$

Divide the numerator and denominator of the fraction by $$a^3$$:

$$\dfrac{3a^2 x - x^3}{a^3 - 3ax^2} = \dfrac{3\dfrac{x}{a} - \dfrac{x^3}{a^3}}{1 - 3\dfrac{x^2}{a^2}} = \dfrac{3\left(\dfrac{x}{a}\right) - \left(\dfrac{x}{a}\right)^3}{1 - 3\left(\dfrac{x}{a}\right)^2}$$

Put $$\dfrac{x}{a} = \tan\theta$$, so that $$\theta = \tan^{-1}\dfrac{x}{a}$$. Using the triple-angle identity $$\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$$:

$$\dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan 3\theta$$

Therefore:

$$\tan^{-1}\left(\dfrac{3a^2 x - x^3}{a^3 - 3ax^2}\right) = \tan^{-1}(\tan 3\theta)$$

For $$-\dfrac{a}{\sqrt{3}} < x < \dfrac{a}{\sqrt{3}}$$ we have $$-\dfrac{1}{\sqrt{3}} < \dfrac{x}{a} < \dfrac{1}{\sqrt{3}}$$, so $$-\dfrac{\pi}{6} < \theta < \dfrac{\pi}{6}$$ and hence $$-\dfrac{\pi}{2} < 3\theta < \dfrac{\pi}{2}$$, the principal branch of $$\tan^{-1}$$.

Therefore the expression equals $$3\theta = 3\tan^{-1}\dfrac{x}{a}$$.

Work from the innermost function outward.

$$\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$$, since $$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$$ and $$\dfrac{\pi}{6}$$ lies in the principal branch.

So $$2\sin^{-1}\dfrac{1}{2} = 2 \times \dfrac{\pi}{6} = \dfrac{\pi}{3}$$.

Then $$\cos\left(2\sin^{-1}\dfrac{1}{2}\right) = \cos\dfrac{\pi}{3} = \dfrac{1}{2}$$.

So $$2\cos\left(2\sin^{-1}\dfrac{1}{2}\right) = 2 \times \dfrac{1}{2} = 1$$.

Finally $$\tan^{-1}\left[2\cos\left(2\sin^{-1}\dfrac{1}{2}\right)\right] = \tan^{-1}(1) = \dfrac{\pi}{4}$$.

Use the standard identities, valid for the given ranges of $$x$$ and $$y$$:

$$\sin^{-1}\dfrac{2x}{1 + x^2} = 2\tan^{-1} x \qquad\text{and}\qquad \cos^{-1}\dfrac{1 - y^2}{1 + y^2} = 2\tan^{-1} y$$

Substituting:

$$\dfrac{1}{2}\left[\sin^{-1}\dfrac{2x}{1 + x^2} + \cos^{-1}\dfrac{1 - y^2}{1 + y^2}\right] = \dfrac{1}{2}\left[2\tan^{-1} x + 2\tan^{-1} y\right] = \tan^{-1} x + \tan^{-1} y$$

So we need $$\tan\left(\tan^{-1} x + \tan^{-1} y\right)$$. Using $$\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$$ with $$A = \tan^{-1} x$$, $$B = \tan^{-1} y$$ (and $$xy < 1$$):

$$\tan\left(\tan^{-1} x + \tan^{-1} y\right) = \dfrac{x + y}{1 - xy}$$

The angle $$\dfrac{2\pi}{3}$$ does not lie in the principal value branch $$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$ of $$\sin^{-1}$$, so we cannot write the answer as $$\dfrac{2\pi}{3}$$ directly.

Using $$\sin(\pi - \theta) = \sin\theta$$:

$$\sin\dfrac{2\pi}{3} = \sin\left(\pi - \dfrac{2\pi}{3}\right) = \sin\dfrac{\pi}{3}$$

Now $$\dfrac{\pi}{3} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$, so:

$$\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right) = \sin^{-1}\left(\sin\dfrac{\pi}{3}\right) = \dfrac{\pi}{3}$$

The angle $$\dfrac{3\pi}{4}$$ does not lie in the principal value branch $$\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$ of $$\tan^{-1}$$.

Since the tangent function has period $$\pi$$:

$$\tan\dfrac{3\pi}{4} = \tan\left(\dfrac{3\pi}{4} - \pi\right) = \tan\left(-\dfrac{\pi}{4}\right)$$

Now $$-\dfrac{\pi}{4} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$, so:

$$\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right) = \tan^{-1}\left[\tan\left(-\dfrac{\pi}{4}\right)\right] = -\dfrac{\pi}{4}$$

Let $$\sin^{-1}\dfrac{3}{5} = \alpha$$, so $$\sin\alpha = \dfrac{3}{5}$$ with $$\alpha$$ acute. Then $$\cos\alpha = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$$, and $$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{3}{4}$$.

Let $$\cot^{-1}\dfrac{3}{2} = \beta$$, so $$\cot\beta = \dfrac{3}{2}$$, giving $$\tan\beta = \dfrac{2}{3}$$.

We need $$\tan(\alpha + \beta)$$. Using $$\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$$:

$$\tan(\alpha + \beta) = \dfrac{\dfrac{3}{4} + \dfrac{2}{3}}{1 - \dfrac{3}{4} \cdot \dfrac{2}{3}} = \dfrac{\dfrac{9 + 8}{12}}{1 - \dfrac{1}{2}} = \dfrac{\dfrac{17}{12}}{\dfrac{1}{2}} = \dfrac{17}{12} \times 2 = \dfrac{17}{6}$$

Therefore $$\tan\left(\sin^{-1}\dfrac{3}{5} + \cot^{-1}\dfrac{3}{2}\right) = \dfrac{17}{6}$$.

The angle $$\dfrac{7\pi}{6}$$ does not lie in the principal value branch $$[0, \pi]$$ of $$\cos^{-1}$$, so option (A) is wrong.

Using $$\cos(2\pi - \theta) = \cos\theta$$:

$$\cos\dfrac{7\pi}{6} = \cos\left(2\pi - \dfrac{7\pi}{6}\right) = \cos\dfrac{5\pi}{6}$$

Now $$\dfrac{5\pi}{6} \in [0, \pi]$$, so:

$$\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right) = \cos^{-1}\left(\cos\dfrac{5\pi}{6}\right) = \dfrac{5\pi}{6}$$

This is option (B).

First find $$\sin^{-1}\left(-\dfrac{1}{2}\right)$$. Since $$\sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2}$$ and $$-\dfrac{\pi}{6}$$ lies in the principal branch, $$\sin^{-1}\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$$.

Therefore:

$$\dfrac{\pi}{3} - \sin^{-1}\left(-\dfrac{1}{2}\right) = \dfrac{\pi}{3} - \left(-\dfrac{\pi}{6}\right) = \dfrac{\pi}{3} + \dfrac{\pi}{6} = \dfrac{\pi}{2}$$

Hence:

$$\sin\left(\dfrac{\pi}{3} - \sin^{-1}\left(-\dfrac{1}{2}\right)\right) = \sin\dfrac{\pi}{2} = 1$$

This is option (D).

$$\tan^{-1}\sqrt{3}$$: since $$\tan\dfrac{\pi}{3} = \sqrt{3}$$ and $$\dfrac{\pi}{3} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$, we get $$\tan^{-1}\sqrt{3} = \dfrac{\pi}{3}$$.

$$\cot^{-1}(-\sqrt{3})$$: we need $$\cot y = -\sqrt{3}$$ with $$y \in (0, \pi)$$. Using $$\cot(\pi - \theta) = -\cot\theta$$ and $$\cot\dfrac{\pi}{6} = \sqrt{3}$$, we get $$\cot\dfrac{5\pi}{6} = -\sqrt{3}$$, so $$\cot^{-1}(-\sqrt{3}) = \dfrac{5\pi}{6}$$.

Therefore:

$$\tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3}) = \dfrac{\pi}{3} - \dfrac{5\pi}{6} = \dfrac{2\pi - 5\pi}{6} = -\dfrac{3\pi}{6} = -\dfrac{\pi}{2}$$

This is option (B).

The angle $$\dfrac{3\pi}{5}$$ does not lie in the principal value branch $$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$ of $$\sin^{-1}$$, because $$\dfrac{3\pi}{5} > \dfrac{\pi}{2}$$.

Using $$\sin(\pi - \theta) = \sin\theta$$:

$$\sin\dfrac{3\pi}{5} = \sin\left(\pi - \dfrac{3\pi}{5}\right) = \sin\dfrac{2\pi}{5}$$

Now $$\dfrac{2\pi}{5} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$, so:

$$\sin^{-1}\left(\sin\dfrac{3\pi}{5}\right) = \sin^{-1}\left(\sin\dfrac{2\pi}{5}\right) = \dfrac{2\pi}{5}$$

The angle $$\dfrac{13\pi}{6}$$ does not lie in the principal value branch $$[0, \pi]$$ of $$\cos^{-1}$$.

Since the cosine function has period $$2\pi$$:

$$\cos\dfrac{13\pi}{6} = \cos\left(2\pi + \dfrac{\pi}{6}\right) = \cos\dfrac{\pi}{6}$$

Now $$\dfrac{\pi}{6} \in [0, \pi]$$, so:

$$\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right) = \cos^{-1}\left(\cos\dfrac{\pi}{6}\right) = \dfrac{\pi}{6}$$

The angle $$\dfrac{7\pi}{6}$$ does not lie in the principal value branch $$\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$ of $$\tan^{-1}$$.

Since the tangent function has period $$\pi$$:

$$\tan\dfrac{7\pi}{6} = \tan\left(\dfrac{7\pi}{6} - \pi\right) = \tan\dfrac{\pi}{6}$$

Now $$\dfrac{\pi}{6} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$, so:

$$\tan^{-1}\left(\tan\dfrac{7\pi}{6}\right) = \tan^{-1}\left(\tan\dfrac{\pi}{6}\right) = \dfrac{\pi}{6}$$

Let $$\sin^{-1}\dfrac{3}{5} = \theta$$, so $$\sin\theta = \dfrac{3}{5}$$ with $$\theta$$ acute.

Then $$\cos\theta = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$$ and $$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3}{4}$$.

Using the double-angle formula $$\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}$$:

$$\tan 2\theta = \dfrac{2 \cdot \dfrac{3}{4}}{1 - \dfrac{9}{16}} = \dfrac{\dfrac{3}{2}}{\dfrac{7}{16}} = \dfrac{3}{2} \times \dfrac{16}{7} = \dfrac{24}{7}$$

Since $$\sin\theta = \dfrac{3}{5} < \dfrac{1}{\sqrt{2}}$$, we have $$\theta < \dfrac{\pi}{4}$$, so $$2\theta < \dfrac{\pi}{2}$$ and $$2\theta$$ lies in the principal branch of $$\tan^{-1}$$. Hence $$2\theta = \tan^{-1}\dfrac{24}{7}$$.

Therefore $$2\sin^{-1}\dfrac{3}{5} = 2\theta = \tan^{-1}\dfrac{24}{7}$$.

Let $$\sin^{-1}\dfrac{8}{17} = x$$ and $$\sin^{-1}\dfrac{3}{5} = y$$, both acute angles.

For $$x$$: $$\sin x = \dfrac{8}{17}$$, so $$\cos x = \sqrt{1 - \dfrac{64}{289}} = \dfrac{15}{17}$$ and $$\tan x = \dfrac{8}{15}$$.

For $$y$$: $$\sin y = \dfrac{3}{5}$$, so $$\cos y = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$$ and $$\tan y = \dfrac{3}{4}$$.

Using $$\tan(x + y) = \dfrac{\tan x + \tan y}{1 - \tan x\tan y}$$:

$$\tan(x + y) = \dfrac{\dfrac{8}{15} + \dfrac{3}{4}}{1 - \dfrac{8}{15} \cdot \dfrac{3}{4}} = \dfrac{\dfrac{32 + 45}{60}}{1 - \dfrac{24}{60}} = \dfrac{\dfrac{77}{60}}{\dfrac{36}{60}} = \dfrac{77}{36}$$

Since $$x$$ and $$y$$ are acute and $$\tan(x + y) > 0$$, the sum $$x + y$$ is acute and lies in the principal branch of $$\tan^{-1}$$, so $$x + y = \tan^{-1}\dfrac{77}{36}$$.

Therefore $$\sin^{-1}\dfrac{8}{17} + \sin^{-1}\dfrac{3}{5} = \tan^{-1}\dfrac{77}{36}$$.

Let $$\cos^{-1}\dfrac{4}{5} = x$$ and $$\cos^{-1}\dfrac{12}{13} = y$$, both acute angles.

For $$x$$: $$\cos x = \dfrac{4}{5}$$, so $$\sin x = \sqrt{1 - \dfrac{16}{25}} = \dfrac{3}{5}$$.

For $$y$$: $$\cos y = \dfrac{12}{13}$$, so $$\sin y = \sqrt{1 - \dfrac{144}{169}} = \dfrac{5}{13}$$.

Using $$\cos(x + y) = \cos x\cos y - \sin x\sin y$$:

$$\cos(x + y) = \dfrac{4}{5} \cdot \dfrac{12}{13} - \dfrac{3}{5} \cdot \dfrac{5}{13} = \dfrac{48}{65} - \dfrac{15}{65} = \dfrac{33}{65}$$

Since $$x, y \in \left(0, \dfrac{\pi}{2}\right)$$, the sum $$x + y \in (0, \pi)$$, the principal branch of $$\cos^{-1}$$. Hence $$x + y = \cos^{-1}\dfrac{33}{65}$$.

Therefore $$\cos^{-1}\dfrac{4}{5} + \cos^{-1}\dfrac{12}{13} = \cos^{-1}\dfrac{33}{65}$$.

Let $$\cos^{-1}\dfrac{12}{13} = x$$ and $$\sin^{-1}\dfrac{3}{5} = y$$, both acute angles.

For $$x$$: $$\cos x = \dfrac{12}{13}$$, so $$\sin x = \sqrt{1 - \dfrac{144}{169}} = \dfrac{5}{13}$$.

For $$y$$: $$\sin y = \dfrac{3}{5}$$, so $$\cos y = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$$.

Using $$\sin(x + y) = \sin x\cos y + \cos x\sin y$$:

$$\sin(x + y) = \dfrac{5}{13} \cdot \dfrac{4}{5} + \dfrac{12}{13} \cdot \dfrac{3}{5} = \dfrac{20}{65} + \dfrac{36}{65} = \dfrac{56}{65}$$

Here $$\cos x = \dfrac{12}{13}$$ is large, so $$x$$ is small; numerically $$x \approx 0.39$$ and $$y \approx 0.64$$, giving $$x + y \approx 1.03 < \dfrac{\pi}{2}$$. So $$x + y$$ lies in the principal branch of $$\sin^{-1}$$, and $$x + y = \sin^{-1}\dfrac{56}{65}$$.

Therefore $$\cos^{-1}\dfrac{12}{13} + \sin^{-1}\dfrac{3}{5} = \sin^{-1}\dfrac{56}{65}$$.

Let $$\sin^{-1}\dfrac{5}{13} = x$$ and $$\cos^{-1}\dfrac{3}{5} = y$$, both acute angles.

For $$x$$: $$\sin x = \dfrac{5}{13}$$, so $$\cos x = \sqrt{1 - \dfrac{25}{169}} = \dfrac{12}{13}$$ and $$\tan x = \dfrac{5}{12}$$.

For $$y$$: $$\cos y = \dfrac{3}{5}$$, so $$\sin y = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$$ and $$\tan y = \dfrac{4}{3}$$.

Using $$\tan(x + y) = \dfrac{\tan x + \tan y}{1 - \tan x\tan y}$$:

$$\tan(x + y) = \dfrac{\dfrac{5}{12} + \dfrac{4}{3}}{1 - \dfrac{5}{12} \cdot \dfrac{4}{3}} = \dfrac{\dfrac{5 + 16}{12}}{1 - \dfrac{20}{36}} = \dfrac{\dfrac{21}{12}}{\dfrac{16}{36}} = \dfrac{21}{12} \times \dfrac{36}{16} = \dfrac{63}{16}$$

Since $$\tan(x + y) > 0$$ and $$x + y$$ is acute, it lies in the principal branch of $$\tan^{-1}$$, so $$x + y = \tan^{-1}\dfrac{63}{16}$$.

Therefore $$\tan^{-1}\dfrac{63}{16} = \sin^{-1}\dfrac{5}{13} + \cos^{-1}\dfrac{3}{5}$$.

Let $$\tan^{-1}\sqrt{x} = \theta$$, so that $$\tan\theta = \sqrt{x}$$, which gives $$\tan^2\theta = x$$.

Since $$x \in [0, 1]$$, we have $$\sqrt{x} \in [0, 1]$$, so $$\theta \in \left[0, \dfrac{\pi}{4}\right]$$.

Using the identity $$\cos 2\theta = \dfrac{1 - \tan^2\theta}{1 + \tan^2\theta}$$:

$$\cos 2\theta = \dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \dfrac{1 - x}{1 + x}$$

Since $$\theta \in \left[0, \dfrac{\pi}{4}\right]$$, we have $$2\theta \in \left[0, \dfrac{\pi}{2}\right] \subset [0, \pi]$$, the principal branch of $$\cos^{-1}$$. Therefore:

$$2\theta = \cos^{-1}\left(\dfrac{1 - x}{1 + x}\right)$$

Hence $$\theta = \dfrac{1}{2}\cos^{-1}\left(\dfrac{1 - x}{1 + x}\right)$$, which proves $$\tan^{-1}\sqrt{x} = \dfrac{1}{2}\cos^{-1}\left(\dfrac{1 - x}{1 + x}\right)$$.

Express the quantities under the roots using $$1 \pm \sin x = \sin^2\dfrac{x}{2} + \cos^2\dfrac{x}{2} \pm 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}$$:

$$1 + \sin x = \left(\cos\dfrac{x}{2} + \sin\dfrac{x}{2}\right)^2, \qquad 1 - \sin x = \left(\cos\dfrac{x}{2} - \sin\dfrac{x}{2}\right)^2$$

For $$x \in \left(0, \dfrac{\pi}{4}\right)$$, we have $$\dfrac{x}{2} \in \left(0, \dfrac{\pi}{8}\right)$$, so both $$\cos\dfrac{x}{2}$$ and $$\sin\dfrac{x}{2}$$ are positive, and $$\cos\dfrac{x}{2} > \sin\dfrac{x}{2}$$. Hence:

$$\sqrt{1 + \sin x} = \cos\dfrac{x}{2} + \sin\dfrac{x}{2}, \qquad \sqrt{1 - \sin x} = \cos\dfrac{x}{2} - \sin\dfrac{x}{2}$$

Therefore the numerator and denominator become:

$$\sqrt{1 + \sin x} + \sqrt{1 - \sin x} = 2\cos\dfrac{x}{2}$$

$$\sqrt{1 + \sin x} - \sqrt{1 - \sin x} = 2\sin\dfrac{x}{2}$$

So the fraction simplifies:

$$\dfrac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \dfrac{2\cos\dfrac{x}{2}}{2\sin\dfrac{x}{2}} = \cot\dfrac{x}{2}$$

Since $$\dfrac{x}{2} \in \left(0, \dfrac{\pi}{8}\right) \subset (0, \pi)$$, the principal branch of $$\cot^{-1}$$:

$$\cot^{-1}\left(\cot\dfrac{x}{2}\right) = \dfrac{x}{2}$$

This proves the required identity.

Following the hint, put $$x = \cos 2\theta$$, so that $$2\theta = \cos^{-1} x$$, i.e. $$\theta = \dfrac{1}{2}\cos^{-1} x$$.

Since $$-\dfrac{1}{\sqrt{2}} \le x \le 1$$, we have $$\cos^{-1} x \in \left[0, \dfrac{3\pi}{4}\right]$$, so $$\theta \in \left[0, \dfrac{3\pi}{8}\right]$$, where $$\cos\theta \ge 0$$ and $$\sin\theta \ge 0$$.

Now $$1 + x = 1 + \cos 2\theta = 2\cos^2\theta$$ and $$1 - x = 1 - \cos 2\theta = 2\sin^2\theta$$, so:

$$\sqrt{1 + x} = \sqrt{2}\cos\theta, \qquad \sqrt{1 - x} = \sqrt{2}\sin\theta$$

Substitute into the fraction:

$$\dfrac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} = \dfrac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta} = \dfrac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}$$

Dividing numerator and denominator by $$\cos\theta$$:

$$\dfrac{1 - \tan\theta}{1 + \tan\theta} = \tan\left(\dfrac{\pi}{4} - \theta\right)$$

Therefore:

$$\tan^{-1}\left(\dfrac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}\right) = \tan^{-1}\left[\tan\left(\dfrac{\pi}{4} - \theta\right)\right]$$

Since $$\theta \in \left[0, \dfrac{3\pi}{8}\right]$$, the angle $$\dfrac{\pi}{4} - \theta \in \left[-\dfrac{\pi}{8}, \dfrac{\pi}{4}\right]$$, which lies in the principal branch of $$\tan^{-1}$$. So the expression equals $$\dfrac{\pi}{4} - \theta$$.

Finally, replacing $$\theta = \dfrac{1}{2}\cos^{-1} x$$:

$$\tan^{-1}\left(\dfrac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}\right) = \dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1} x$$

We solve $$2\tan^{-1}(\cos x) = \tan^{-1}(2\,\operatorname{cosec} x)$$, with $$\sin x \ne 0$$ so that $$\operatorname{cosec} x$$ is defined.

Apply the identity $$2\tan^{-1}a = \tan^{-1}\!\dfrac{2a}{1-a^{2}}$$ with $$a = \cos x$$:

$$2\tan^{-1}(\cos x) = \tan^{-1}\!\dfrac{2\cos x}{1 - \cos^{2} x} = \tan^{-1}\!\dfrac{2\cos x}{\sin^{2} x}.$$

The equation becomes

$$\tan^{-1}\!\dfrac{2\cos x}{\sin^{2} x} = \tan^{-1}\!\dfrac{2}{\sin x}.$$

Take $$\tan$$ of both sides. Because $$\tan$$ has period $$\pi$$, equality of $$\tan^{-1}$$ values implies equality of arguments only up to integer multiples of $$\pi$$ in $$x$$ — we'll capture that in the general solution at the end.

$$\dfrac{2\cos x}{\sin^{2} x} = \dfrac{2}{\sin x}.$$

Multiply both sides by $$\sin^{2} x$$ (allowed since $$\sin x \ne 0$$):

$$2\cos x = 2\sin x \;\Rightarrow\; \sin x = \cos x \;\Rightarrow\; \tan x = 1.$$

Since $$\tan$$ has period $$\pi$$, the general solution is

$$x = \dfrac{\pi}{4} + n\pi, \qquad n \in \mathbb{Z}, \quad \sin x \ne 0.$$

A quick check at $$x = \pi/4$$: both $$\cos x$$ and $$\sin x$$ equal $$1/\sqrt{2}$$ > 0, and substitution gives $$2\tan^{-1}(1/\sqrt{2}) = \tan^{-1}(2\sqrt{2})$$ — which holds. The principal solution in $$[0, 2\pi)$$ is $$x = \pi/4$$.

We solve $$\tan^{-1}\dfrac{1 - x}{1 + x} = \dfrac{1}{2}\tan^{-1} x$$ for $$x > 0$$.

Rewrite the left side. Since $$\dfrac{1 - x}{1 + x} = \dfrac{1 - x}{1 + 1 \cdot x}$$, use the identity $$\tan^{-1} u - \tan^{-1} v = \tan^{-1}\dfrac{u - v}{1 + uv}$$ with $$u = 1$$, $$v = x$$:

$$\tan^{-1}\dfrac{1 - x}{1 + x} = \tan^{-1}(1) - \tan^{-1} x = \dfrac{\pi}{4} - \tan^{-1} x$$

The equation becomes:

$$\dfrac{\pi}{4} - \tan^{-1} x = \dfrac{1}{2}\tan^{-1} x$$

Bringing the $$\tan^{-1} x$$ terms together:

$$\dfrac{\pi}{4} = \tan^{-1} x + \dfrac{1}{2}\tan^{-1} x = \dfrac{3}{2}\tan^{-1} x$$

So $$\tan^{-1} x = \dfrac{\pi}{6}$$, which gives:

$$x = \tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}$$

Since $$\dfrac{1}{\sqrt{3}} > 0$$, it satisfies the condition $$x > 0$$.

Let $$\tan^{-1} x = \theta$$, so that $$\tan\theta = x$$ with $$\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$. We need $$\sin\theta$$.

From $$\tan\theta = \dfrac{x}{1}$$, construct a right triangle with side opposite to $$\theta$$ equal to $$x$$ and adjacent side equal to $$1$$, so the hypotenuse is $$\sqrt{1 + x^2}$$.

Therefore:

$$\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{x}{\sqrt{1 + x^2}}$$

Hence $$\sin(\tan^{-1} x) = \dfrac{x}{\sqrt{1 + x^2}}$$, which is option (D).

We solve $$\sin^{-1}(1 - x) - 2\sin^{-1} x = \dfrac{\pi}{2}$$.

Rearrange: $$\sin^{-1}(1 - x) = \dfrac{\pi}{2} + 2\sin^{-1} x$$.

Take the sine of both sides:

$$1 - x = \sin\left(\dfrac{\pi}{2} + 2\sin^{-1} x\right) = \cos\left(2\sin^{-1} x\right)$$

Let $$\sin^{-1} x = \phi$$, so $$\sin\phi = x$$. Using $$\cos 2\phi = 1 - 2\sin^2\phi$$:

$$\cos\left(2\sin^{-1} x\right) = 1 - 2x^2$$

So the equation becomes:

$$1 - x = 1 - 2x^2 \;\Rightarrow\; 2x^2 - x = 0 \;\Rightarrow\; x(2x - 1) = 0$$

This gives $$x = 0$$ or $$x = \dfrac{1}{2}$$.

Check $$x = \dfrac{1}{2}$$: $$\sin^{-1}\left(1 - \dfrac{1}{2}\right) - 2\sin^{-1}\dfrac{1}{2} = \sin^{-1}\dfrac{1}{2} - 2\sin^{-1}\dfrac{1}{2} = -\sin^{-1}\dfrac{1}{2} = -\dfrac{\pi}{6} \ne \dfrac{\pi}{2}$$. So $$x = \dfrac{1}{2}$$ is rejected.

Check $$x = 0$$: $$\sin^{-1}(1) - 2\sin^{-1}(0) = \dfrac{\pi}{2} - 0 = \dfrac{\pi}{2}$$. This satisfies the equation.

Hence $$x = 0$$, which is option (C).