Inverse Trigonometric Functions

Let $y = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$. This implies $\sin y = \frac{1}{\sqrt{2}}$.

We know that the principal value branch of $\sin^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Since $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the principal value is $\mathbf{\frac{\pi}{4}}$.

Let $y = \cos^{-1}\left(\frac{1}{2}\right)$. This implies $\cos y = \frac{1}{2}$.

We know that the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.

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

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

We know that the principal value branch of $\tan^{-1} x$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Since $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ and $\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the principal value is $\mathbf{\frac{\pi}{3}}$.

Let $y = \text{cosec}^{-1}(2)$. This implies $\text{cosec } y = 2 \Rightarrow \sin y = \frac{1}{2}$.

We know that the principal value branch of $\text{cosec}^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$.

Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$, the principal value is $\mathbf{\frac{\pi}{6}}$.

Let $y = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$. This implies $\sec y = \frac{2}{\sqrt{3}} \Rightarrow \cos y = \frac{\sqrt{3}}{2}$.

We know that the principal value branch of $\sec^{-1} x$ is $[0, \pi] - \{\frac{\pi}{2}\}$.

Since $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ and $\frac{\pi}{6} \in [0, \pi] - \{\frac{\pi}{2}\}$, the principal value is $\mathbf{\frac{\pi}{6}}$.

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

We know that the principal value branch of $\cot^{-1} x$ is $(0, \pi)$.

Since $\cot\left(\frac{\pi}{6}\right) = \sqrt{3}$ and $\frac{\pi}{6} \in (0, \pi)$, the principal value is $\mathbf{\frac{\pi}{6}}$.

Using the property $\sin^{-1}(-x) = -\sin^{-1}x$ for $x \in [-1, 1]$:

$\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$.

Since $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$, we get $-\sin^{-1}\left(\sin\frac{\pi}{3}\right) = \mathbf{-\frac{\pi}{3}}$.

Note: $-\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, which is valid.

Using the property $\cos^{-1}(-x) = \pi - \cos^{-1}x$ for $x \in [-1, 1]$:

$\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \pi - \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$.

Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, we have $\pi - \frac{\pi}{4} = \mathbf{\frac{3\pi}{4}}$.

Note: $\frac{3\pi}{4} \in [0, \pi]$, which is valid.

Using the property $\tan^{-1}(-x) = -\tan^{-1}x$ for $x \in \mathbb{R}$:

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

Since $\tan\left(\frac{\pi}{4}\right) = 1$, we get $-\frac{\pi}{4}$.

Note: $-\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, which is valid. Hence, the value is $\mathbf{-\frac{\pi}{4}}$.

Using the property $\text{cosec}^{-1}(-x) = -\text{cosec}^{-1}x$ for $|x| \ge 1$:

$\text{cosec}^{-1}(-\sqrt{2}) = -\text{cosec}^{-1}(\sqrt{2})$.

Let $y = \text{cosec}^{-1}(\sqrt{2}) \Rightarrow \text{cosec } y = \sqrt{2} \Rightarrow \sin y = \frac{1}{\sqrt{2}} \Rightarrow y = \frac{\pi}{4}$.

Thus, the value is $-\frac{\pi}{4}$. Since $-\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$, it is the principal value: $\mathbf{-\frac{\pi}{4}}$.

We use the fundamental identity $\sin(\sin^{-1} x) = x$ for $x \in [-1, 1]$.

Since $0.5 \in [-1, 1]$, the identity directly applies.

Thus, $\sin(\sin^{-1} 0.5) = \mathbf{0.5}$ (or $\mathbf{\frac{1}{2}}$).

We use the identity $\cos(\cos^{-1} x) = x$ for $x \in [-1, 1]$.

Since $\frac{\sqrt{3}}{2} \approx 0.866 \in [-1, 1]$, the identity applies.

Thus, $\cos\left(\cos^{-1} \frac{\sqrt{3}}{2}\right) = \mathbf{\frac{\sqrt{3}}{2}}$.

We use the property $\sin^{-1}(\sin \theta) = \theta$ for $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Since $\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the property applies directly.

Thus, $\sin^{-1}\left(\sin \frac{\pi}{4}\right) = \mathbf{\frac{\pi}{4}}$.

We use the property $\cos^{-1}(\cos \theta) = \theta$ for $\theta \in [0, \pi]$.

Since $\frac{2\pi}{3} \in [0, \pi]$, the property applies directly.

Thus, $\cos^{-1}\left(\cos \frac{2\pi}{3}\right) = \mathbf{\frac{2\pi}{3}}$.

Note that the property $\sin^{-1}(\sin \theta) = \theta$ only holds if $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Here, $\theta = \frac{2\pi}{3} \approx 120^\circ$, which is outside the range.

We rewrite $\sin\left(\frac{2\pi}{3}\right)$ using $\sin(\pi - x) = \sin x$:

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

Now, $\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore:

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

The property $\cos^{-1}(\cos \theta) = \theta$ holds only for $\theta \in [0, \pi]$.

Here, $\theta = \frac{7\pi}{6} \approx 210^\circ \notin [0, \pi]$.

We rewrite $\cos\left(\frac{7\pi}{6}\right)$ using $\cos(2\pi - x) = \cos x$:

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

Since $\frac{5\pi}{6} \in [0, \pi]$, we have:

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

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

Here, $\frac{3\pi}{4} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. We rewrite the tangent term:

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

Since $-\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we apply the identity:

$\tan^{-1}\left(\tan \frac{3\pi}{4}\right) = \tan^{-1}\left(\tan\left(-\frac{\pi}{4}\right)\right) = \mathbf{-\frac{\pi}{4}}$.

The reciprocal property states: $\text{cosec}^{-1}(x) = \sin^{-1}\left(\frac{1}{x}\right)$ for $|x| \ge 1$.

Substituting $x = 2$ (which satisfies $|2| \ge 1$):

$\text{cosec}^{-1}(2) = \sin^{-1}\left(\frac{1}{2}\right)$.

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

By reciprocal property, $\sec^{-1}(-2) = \cos^{-1}\left(-\frac{1}{2}\right)$.

Now, using the negative argument formula $\cos^{-1}(-u) = \pi - \cos^{-1}u$:

$\cos^{-1}\left(-\frac{1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right)$.

Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$, we have $\pi - \frac{\pi}{3} = \mathbf{\frac{2\pi}{3}}$.

First evaluate the inner term: $\sin^{-1}\left(-\frac{1}{2}\right)$.

Since $\sin^{-1}(-u) = -\sin^{-1}u$, we get $-\sin^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6}$.

Substitute this back into the expression:

$\sin\left(\frac{\pi}{3} - \left(-\frac{\pi}{6}\right)\right) = \sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2}\right)$.

We know $\sin\left(\frac{\pi}{2}\right) = \mathbf{1}$.

We know that the inverse sine function is an odd function, meaning $\sin^{-1}(-x) = -\sin^{-1}(x)$ for all $x \in [-1, 1]$.

Substituting this relation in:

$\sin^{-1}(-x) + \sin^{-1}(x) = -\sin^{-1}(x) + \sin^{-1}(x) = \mathbf{0}$.

Using the property $\cos^{-1}(-u) = \pi - \cos^{-1}u$ for $u \in [-1, 1]$:

$\cos^{-1}\left(-\frac{1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right)$.

Substitute this into the original expression:

$\left[\pi - \cos^{-1}\left(\frac{1}{2}\right)\right] + \cos^{-1}\left(\frac{1}{2}\right) = \mathbf{\pi}$.

We use the complementary angle identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.

Since $\frac{1}{5} \in [-1, 1]$, the identity directly applies.

Thus, $\sin^{-1}\left(\frac{1}{5}\right) + \cos^{-1}\left(\frac{1}{5}\right) = \mathbf{\frac{\pi}{2}}$.

We use the complementary angle identity: $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ for all $x \in \mathbb{R}$.

Since $\frac{2}{3} \in \mathbb{R}$, the identity directly holds.

Thus, $\tan^{-1}\left(\frac{2}{3}\right) + \cot^{-1}\left(\frac{2}{3}\right) = \mathbf{\frac{\pi}{2}}$.

Identify the inner sum: $\sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2}$.

By complementary sum identity, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$. Since $\frac{1}{2} \in [-1, 1]$, the sum is $\frac{\pi}{2}$.

Substitute this back: $\sin\left(\frac{\pi}{2}\right) = \mathbf{1}$.

We use the complementary angle identity: $\sec^{-1}x + \text{cosec}^{-1}x = \frac{\pi}{2}$ for $|x| \ge 1$.

Substituting this directly into the expression:

$\cos(\sec^{-1}x + \text{cosec}^{-1}x) = \cos\left(\frac{\pi}{2}\right)$.

Since $\cos\left(\frac{\pi}{2}\right) = \mathbf{0}$.

We use the identity $\cos^{-1}(-x) = \pi - \cos^{-1}x$ for $x \in [-1, 1]$.

Substituting this relation in:

$\cos^{-1}(-x) + \cos^{-1}(x) = \left[\pi - \cos^{-1}x\right] + \cos^{-1}x = \mathbf{\pi}$.

Using the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$:

$\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$.

Given $\sin^{-1}x = \frac{\pi}{5}$, substitute to find $\cos^{-1}x$:

$\cos^{-1}x = \frac{\pi}{2} - \frac{\pi}{5} = \frac{5\pi - 2\pi}{10} = \mathbf{\frac{3\pi}{10}}$.

Using the complementary sum identity $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$:

$\cot^{-1}x = \frac{\pi}{2} - \tan^{-1}x$.

Given $\tan^{-1}x = \frac{\pi}{4}$, substitute to get:

$\cot^{-1}x = \frac{\pi}{2} - \frac{\pi}{4} = \mathbf{\frac{\pi}{4}}$.

Using the property $\tan^{-1}a + \cot^{-1}a = \frac{\pi}{2}$ for all $a \in \mathbb{R}$:

The expression becomes $\cot\left(\frac{\pi}{2}\right)$.

Since $\cot\left(\frac{\pi}{2}\right) = \frac{\cos(\pi/2)}{\sin(\pi/2)} = \frac{0}{1} = \mathbf{0}$.

We use the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ which holds when $xy < 1$.

Here, $x = \frac{1}{2}$ and $y = \frac{1}{3}$. The product $xy = \frac{1}{6} < 1$, so the formula is valid.

Substitute values:

$\tan^{-1}\left(\frac{1/2 + 1/3}{1 - (1/2)(1/3)}\right) = \tan^{-1}\left(\frac{5/6}{1 - 1/6}\right) = \tan^{-1}\left(\frac{5/6}{5/6}\right) = \tan^{-1}(1)$.

Since $\tan\left(\frac{\pi}{4}\right) = 1$, the answer is $\mathbf{\frac{\pi}{4}}$.

Warning: $x = 2 > 0$, $y = 3 > 0$ and $xy = 6 > 1$.

When $xy > 1$ with positive $x, y$, the standard formula changes to:

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

Substitute the values:

$\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}\left(\frac{2+3}{1-6}\right) = \pi + \tan^{-1}\left(\frac{5}{-5}\right)$

$= \pi + \tan^{-1}(-1) = \pi - \tan^{-1}(1) = \pi - \frac{\pi}{4} = \mathbf{\frac{3\pi}{4}}$.

Using the identity $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ since $xy = \frac{14}{264} < 1$:

$\tan^{-1}\left(\frac{2/11 + 7/24}{1 - (2/11)(7/24)}\right) = \tan^{-1}\left(\frac{(48+77)/264}{1 - 14/264}\right)$

$= \tan^{-1}\left(\frac{125/264}{250/264}\right) = \tan^{-1}\left(\frac{125}{250}\right) = \mathbf{\tan^{-1}\left(\frac{1}{2}\right)}$.

We use the double-angle formula: $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ for $|x| < 1$.

Since $x = \frac{1}{3} < 1$, we substitute directly:

$2\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{2(1/3)}{1 - (1/3)^2}\right) = \tan^{-1}\left(\frac{2/3}{1 - 1/9}\right)$

$= \tan^{-1}\left(\frac{2/3}{8/9}\right) = \tan^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right) = \mathbf{\tan^{-1}\left(\frac{3}{4}\right)}$.

We use the double-angle conversion formula: $2\tan^{-1}x = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ for $|x| \le 1$.

Here, $x = \frac{1}{2} \le 1$. Substituting:

$2\tan^{-1}\left(\frac{1}{2}\right) = \sin^{-1}\left(\frac{2(1/2)}{1 + (1/2)^2}\right) = \sin^{-1}\left(\frac{1}{1 + 1/4}\right)$

$= \sin^{-1}\left(\frac{1}{5/4}\right) = \mathbf{\sin^{-1}\left(\frac{4}{5}\right)}$.

We use the double-angle conversion formula: $2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ for $x \ge 0$.

Here, $x = \frac{1}{3} \ge 0$. Substituting:

$2\tan^{-1}\left(\frac{1}{3}\right) = \cos^{-1}\left(\frac{1 - (1/3)^2}{1 + (1/3)^2}\right) = \cos^{-1}\left(\frac{1 - 1/9}{1 + 1/9}\right)$

$= \cos^{-1}\left(\frac{8/9}{10/9}\right) = \mathbf{\cos^{-1}\left(\frac{4}{5}\right)}$.

Let $\theta = \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9}$. Since $xy = \frac{2}{36} = \frac{1}{18} < 1$, we simplify $\theta$:

$\theta = \tan^{-1}\left(\frac{1/4 + 2/9}{1 - (1/4)(2/9)}\right) = \tan^{-1}\left(\frac{(9+8)/36}{1 - 2/36}\right)$

$= \tan^{-1}\left(\frac{17/36}{34/36}\right) = \tan^{-1}\left(\frac{17}{34}\right) = \tan^{-1}\left(\frac{1}{2}\right)$.

Substitute this back: $\tan(\theta) = \tan\left(\tan^{-1}\frac{1}{2}\right) = \mathbf{\frac{1}{2}}$.

Let's simplify the second term: $\tan^{-1}\left(\frac{x-y}{x+y}\right)$.

Divide the numerator and denominator by $y$:

$\tan^{-1}\left(\frac{x/y - 1}{x/y + 1}\right) = \tan^{-1}\left(\frac{x/y - 1}{1 + (x/y)(1)}\right)$.

Recall the identity $\tan^{-1}\left(\frac{u-v}{1+uv}\right) = \tan^{-1}u - \tan^{-1}v$. Let $u = x/y$ and $v = 1$:

$\tan^{-1}\left(\frac{x/y - 1}{1 + (x/y)(1)}\right) = \tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}(1) = \tan^{-1}\left(\frac{x}{y}\right) - \frac{\pi}{4}$.

Substitute this back into the original expression:

$\tan^{-1}\left(\frac{x}{y}\right) - \left[\tan^{-1}\left(\frac{x}{y}\right) - \frac{\pi}{4}\right] = \mathbf{\frac{\pi}{4}}$.

We first group the last two terms: $\tan^{-1}(2) + \tan^{-1}(3)$.

Using the result from Q32 (since $2 \times 3 = 6 > 1$), we have:

$\tan^{-1}(2) + \tan^{-1}(3) = \frac{3\pi}{4}$.

Now, add $\tan^{-1}(1) = \frac{\pi}{4}$:

$\tan^{-1}(1) + \left[\tan^{-1}(2) + \tan^{-1}(3)\right] = \frac{\pi}{4} + \frac{3\pi}{4} = \frac{4\pi}{4} = \mathbf{\pi}$.

Since $xy < 1$, we apply the addition identity:

$\tan^{-1}\left(\frac{x+y}{1-xy}\right) = \frac{\pi}{4}$.

Take the tangent of both sides:

$\frac{x+y}{1-xy} = \tan\left(\frac{\pi}{4}\right) \Rightarrow \frac{x+y}{1-xy} = 1$.

Cross-multiplying, we obtain: $\mathbf{x + y = 1 - xy}$ (or $\mathbf{x + y + xy = 1}$).

Use double-angle and half-angle identities:

$\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$ and $1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$.

Substitute these into the expression:

$\frac{\sin x}{1+\cos x} = \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{\sin(x/2)}{\cos(x/2)} = \tan\left(\frac{x}{2}\right)$.

The expression becomes $\tan^{-1}\left(\tan\frac{x}{2}\right)$.

Since $-\pi < x < \pi$, we have $-\frac{\pi}{2} < \frac{x}{2} < \frac{\pi}{2}$, which is the principal branch. Thus, the simplified form is $\mathbf{\frac{x}{2}}$.

Using identities, rewrite: $\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)$.

Also, $1 - \sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2$.

Substituting these:

$\frac{\cos x}{1-\sin x} = \frac{(\cos(x/2)-\sin(x/2))(\cos(x/2)+\sin(x/2))}{(\cos(x/2)-\sin(x/2))^2} = \frac{\cos(x/2)+\sin(x/2)}{\cos(x/2)-\sin(x/2)}$.

Divide numerator and denominator by $\cos\frac{x}{2}$:

$\frac{1 + \tan(x/2)}{1 - \tan(x/2)} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$.

Thus, $\tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right) = \mathbf{\frac{\pi}{4} + \frac{x}{2}}$.

Recall half-angle forms: $1-\cos x = 2\sin^2\frac{x}{2}$ and $1+\cos x = 2\cos^2\frac{x}{2}$.

Substitute to resolve the square root:

$\sqrt{\frac{1-\cos x}{1+\cos x}} = \sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}} = \sqrt{\tan^2\left(\frac{x}{2}\right)} = \left|\tan\frac{x}{2}\right|$.

Since $0 < x < \pi$, we have $0 < \frac{x}{2} < \frac{\pi}{2}$ where tangent is positive. Thus, $\left|\tan\frac{x}{2}\right| = \tan\frac{x}{2}$.

The expression simplifies to $\tan^{-1}\left(\tan\frac{x}{2}\right) = \mathbf{\frac{x}{2}}$.

Let $x = \sec\theta$, which means $\theta = \sec^{-1}x$. Since $x > 1$, we have $\theta \in (0, \frac{\pi}{2})$.

Substitute $x$ in the term:

$\frac{1}{\sqrt{x^2-1}} = \frac{1}{\sqrt{\sec^2\theta - 1}} = \frac{1}{\sqrt{\tan^2\theta}} = \frac{1}{\tan\theta} = \cot\theta$ (since $\tan\theta > 0$ for $\theta \in (0, \frac{\pi}{2})$).

The expression becomes $\cot^{-1}(\cot\theta) = \theta$.

Substituting back $\theta = \sec^{-1}x$, the simplified form is $\mathbf{\sec^{-1}x}$.

Let $x = \tan\theta$, so $\theta = \tan^{-1}x$. Since $-1 \le x \le 1$, we have $\theta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$.

Substitute $x$:

$\frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$.

The expression becomes $\sin^{-1}(\sin 2\theta)$.

Since $\theta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$, we have $2\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, which lies in the principal value branch of sine.

Thus, $\sin^{-1}(\sin 2\theta) = 2\theta = \mathbf{2\tan^{-1}x}$.

Let $x = \tan\theta$, so $\theta = \tan^{-1}x$. Since $x \ge 0$, we have $\theta \in [0, \frac{\pi}{2})$.

Substitute $x$:

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

The expression becomes $\cos^{-1}(\cos 2\theta)$.

Since $\theta \in [0, \frac{\pi}{2})$, we have $2\theta \in [0, \pi)$, which lies in the principal value branch of cosine.

Thus, $\cos^{-1}(\cos 2\theta) = 2\theta = \mathbf{2\tan^{-1}x}$.

Let $x = a\tan\theta$, which means $\tan\theta = \frac{x}{a} \Rightarrow \theta = \tan^{-1}\left(\frac{x}{a}\right)$.

Since $-\frac{a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$, we have $-\frac{1}{\sqrt{3}} < \tan\theta < \frac{1}{\sqrt{3}} \Rightarrow \theta \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$.

Substitute $x$:

$\frac{3a^2(a\tan\theta) - a^3\tan^3\theta}{a^3 - 3a(a^2\tan^2\theta)} = \frac{a^3(3\tan\theta - \tan^3\theta)}{a^3(1 - 3\tan^2\theta)} = \tan 3\theta$.

The expression is $\tan^{-1}(\tan 3\theta)$. Since $\theta \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$, we have $3\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (principal range).

Thus, $\tan^{-1}(\tan 3\theta) = 3\theta = \mathbf{3\tan^{-1}\left(\frac{x}{a}\right)}$.

Let $x = \sin\theta$, so $\theta = \sin^{-1}x$. Since $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$, we have $\theta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$.

Substitute $x$ in the expression:

$2x\sqrt{1-x^2} = 2\sin\theta\sqrt{1-\sin^2\theta} = 2\sin\theta\cos\theta$ (since $\cos\theta \ge 0$ for $\theta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$)

$= \sin 2\theta$.

So the expression becomes $\sin^{-1}(\sin 2\theta)$.

Since $\theta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$, $2\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, which fits the principal range of $\sin^{-1}$.

Thus, $\sin^{-1}(\sin 2\theta) = 2\theta = \mathbf{2\sin^{-1}x}$.

Let $x = \cos\theta$, so $\theta = \cos^{-1}x$. Since $\frac{1}{2} \le x \le 1$, we have $\theta \in [0, \frac{\pi}{3}]$.

Substitute $x$ in the term:

$4x^3 - 3x = 4\cos^3\theta - 3\cos\theta = \cos 3\theta$ (standard triple angle identity).

The expression becomes $\cos^{-1}(\cos 3\theta)$.

Since $\theta \in [0, \frac{\pi}{3}]$, we have $3\theta \in [0, \pi]$, which lies in the principal value branch of cosine ($[0, \pi]$).

Thus, $\cos^{-1}(\cos 3\theta) = 3\theta = \mathbf{3\cos^{-1}x}$.

Let $x = \tan\theta$, so $\theta = \tan^{-1}x$ ($\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ and $\theta \neq 0$).

Substitute $x$ in the term:

$\frac{\sqrt{1+x^2}-1}{x} = \frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta} = \frac{\sec\theta-1}{\tan\theta}$ (since $\sec\theta > 0$ for $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$)

$= \frac{1/\cos\theta - 1}{\sin\theta/\cos\theta} = \frac{1-\cos\theta}{\sin\theta} = \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \tan\left(\frac{\theta}{2}\right)$.

The expression simplifies to $\tan^{-1}\left(\tan\frac{\theta}{2}\right)$.

Since $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, we have $\frac{\theta}{2} \in (-\frac{\pi}{4}, \frac{\pi}{4})$, which lies in the principal branch.

Thus, the simplified form is $\frac{\theta}{2} = \mathbf{\frac{1}{2}\tan^{-1}x}$.

Taking sine on both sides of the equation:

$\sin\left(\sin^{-1}x\right) = \sin\left(\frac{\pi}{6}\right)$.

Since $\sin\left(\sin^{-1}x\right) = x$ for $x \in [-1, 1]$:

$x = \sin\left(\frac{\pi}{6}\right) = \mathbf{\frac{1}{2}}$.

Since $\frac{1}{2} \in [-1, 1]$, the solution is valid and correct.

Taking tangent on both sides of the equation:

$\tan\left(\tan^{-1}x\right) = \tan\left(-\frac{\pi}{4}\right)$.

Since $\tan\left(\tan^{-1}x\right) = x$ for all $x \in \mathbb{R}$:

$x = -\tan\left(\frac{\pi}{4}\right) = \mathbf{-1}$.

Thus, the valid real solution is $\mathbf{x = -1}$.

Take tangent on both sides of the equation:

$\tan\left(\tan^{-1}(2x) + \tan^{-1}(3x)\right) = \tan\left(\frac{\pi}{4}\right) \Rightarrow \frac{2x+3x}{1-(2x)(3x)} = 1$.

$\frac{5x}{1-6x^2} = 1 \Rightarrow 5x = 1 - 6x^2 \Rightarrow 6x^2 + 5x - 1 = 0$.

Factoring the quadratic equation:

$6x^2 + 6x - x - 1 = 0 \Rightarrow 6x(x+1) - (x+1) = 0 \Rightarrow (6x-1)(x+1) = 0$.

This gives candidate roots: $x = \frac{1}{6}$ or $x = -1$.

Verification:

- If $x = -1$, LHS $= \tan^{-1}(-2) + \tan^{-1}(-3) = -(\tan^{-1}2 + \tan^{-1}3) = -\frac{3\pi}{4} \neq \frac{\pi}{4}$. Reject!

- If $x = \frac{1}{6}$, LHS $= \tan^{-1}(1/3) + \tan^{-1}(1/2) = \frac{\pi}{4}$. Valid!

Thus, the unique valid solution is $\mathbf{x = \frac{1}{6}}$.

Let $\sin^{-1}x = y \Rightarrow x = \sin y$. The equation is $\sin^{-1}(1-x) - 2y = \frac{\pi}{2}$.

$\sin^{-1}(1-x) = \frac{\pi}{2} + 2y \Rightarrow 1-x = \sin\left(\frac{\pi}{2} + 2y\right)$.

Using $\sin(\frac{\pi}{2} + \theta) = \cos\theta$, we get:

$1-x = \cos 2y \Rightarrow 1-x = 1 - 2\sin^2 y$.

Substitute $x = \sin y$:

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

Candidate solutions are $x = 0$ or $x = \frac{1}{2}$.

Verification:

- If $x = 0$, LHS $= \sin^{-1}(1) - 2\sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2} =$ RHS. (Valid)

- If $x = \frac{1}{2}$, LHS $= \sin^{-1}(1/2) - 2\sin^{-1}(1/2) = -\sin^{-1}(1/2) = -\frac{\pi}{6} \neq \frac{\pi}{2}$. (Reject)

Thus, the unique valid solution is $\mathbf{x = 0}$.

Applying the tangent sum formula:

$\tan^{-1}\left(\frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}\right) = \frac{\pi}{4} \Rightarrow \frac{\frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2) - (x-1)(x+1)}{(x-2)(x+2)}} = \tan\frac{\pi}{4} = 1$.

$\frac{(x^2+x-2) + (x^2-x-2)}{(x^2-4) - (x^2-1)} = 1 \Rightarrow \frac{2x^2-4}{-3} = 1$.

$2x^2 - 4 = -3 \Rightarrow 2x^2 = 1 \Rightarrow x^2 = \frac{1}{2}$.

Thus, $x = \pm\frac{1}{\sqrt{2}}$.

Both roots satisfy the domain limitations and equation parameters, so the solutions are $\mathbf{x = \pm\frac{1}{\sqrt{2}}}$.

Using the identity $2\tan^{-1}u = \tan^{-1}\left(\frac{2u}{1-u^2}\right)$, the LHS becomes:

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

Now equate arguments with the RHS $\tan^{-1}(2\text{cosec } x) = \tan^{-1}\left(\frac{2}{\sin x}\right)$:

$\frac{2\cos x}{\sin^2x} = \frac{2}{\sin x} \Rightarrow \frac{\cos x}{\sin x} = 1$ (for $\sin x \neq 0$).

$\cot x = 1 \Rightarrow \tan x = 1$.

For principal domain value, we get $\mathbf{x = \frac{\pi}{4}}$.

Let $\theta = \tan^{-1}x \Rightarrow \tan\theta = x$. Then, $\cos\theta = \frac{1}{\sec\theta} = \frac{1}{\sqrt{1+x^2}}$. So LHS $= \frac{1}{\sqrt{1+x^2}}$.

Let $\phi = \cot^{-1}\frac{3}{4} \Rightarrow \cot\phi = \frac{3}{4}$. Using triangle properties where adjacent $= 3$, opposite $= 4$, the hypotenuse is $5$.

Thus, $\sin\phi = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}$. So RHS $= \frac{4}{5}$.

Equate LHS and RHS:

$\frac{1}{\sqrt{1+x^2}} = \frac{4}{5} \Rightarrow \sqrt{1+x^2} = \frac{5}{4}$.

Square both sides:

$1+x^2 = \frac{25}{16} \Rightarrow x^2 = \frac{9}{16} \Rightarrow \mathbf{x = \pm\frac{3}{4}}$.

Recall the difference identity: $\tan^{-1}\left(\frac{1-x}{1+x}\right) = \tan^{-1}(1) - \tan^{-1}x = \frac{\pi}{4} - \tan^{-1}x$.

Substitute this back into the original equation:

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

$\frac{\pi}{4} = \tan^{-1}x + \frac{1}{2}\tan^{-1}x \Rightarrow \frac{\pi}{4} = \frac{3}{2}\tan^{-1}x$.

Multiply by $\frac{2}{3}$:

$\tan^{-1}x = \frac{\pi}{4} \times \frac{2}{3} = \frac{\pi}{6}$.

Taking tangent on both sides: $x = \tan\left(\frac{\pi}{6}\right) = \mathbf{\frac{1}{\sqrt{3}}}$. Since $\frac{1}{\sqrt{3}} > 0$, the solution is valid.

We use the standard complementary identity: $\sin^{-1}u + \cos^{-1}u = \frac{\pi}{2}$ for $u \in [-1, 1]$.

Comparing the given equation $\sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}x = \frac{\pi}{2}$ directly to this identity, we match arguments:

$u = \frac{5}{13} \Rightarrow \mathbf{x = \frac{5}{13}}$.

Since $\frac{5}{13} \approx 0.38 \in [-1, 1]$, the solution is mathematically correct and valid.

Rewrite the equation as: $\sin^{-1}(2x) = \frac{\pi}{3} - \sin^{-1}x$.

Take the sine of both sides:

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

Apply the subtraction formula $\sin(A-B) = \sin A\cos B - \cos A\sin B$, letting $B = \sin^{-1}x \Rightarrow \sin B = x, \cos B = \sqrt{1-x^2}$:

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

$2x = \frac{\sqrt{3}}{2}\sqrt{1-x^2} - \frac{1}{2}x$.

Multiply by 2 to clear fractions:

$4x = \sqrt{3}\sqrt{1-x^2} - x \Rightarrow 5x = \sqrt{3}\sqrt{1-x^2}$.

Square both sides (requiring $x \ge 0$ since $5x \ge 0$):

$25x^2 = 3(1-x^2) \Rightarrow 25x^2 = 3 - 3x^2 \Rightarrow 28x^2 = 3$.

$x^2 = \frac{3}{28} \Rightarrow x = \pm\sqrt{\frac{3}{28}} = \pm\frac{1}{2}\sqrt{\frac{3}{7}}$.

Since we require $x \ge 0$, the unique valid solution is $\mathbf{x = \frac{1}{2}\sqrt{\frac{3}{7}}}$ (or $\sqrt{\frac{3}{28}}$).