Practice Exercise 2.2

We know that \( \tan^{-1} \frac{2x}{1 - x^2} = 2 \tan^{-1} x \).
So, LHS = \( \tan^{-1} x + 2 \tan^{-1} x = 3 \tan^{-1} x \).

Let \( x = \tan \theta \).
RHS = \( \tan^{-1} \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \)
= \( \tan^{-1} (\tan 3\theta) \) = \( 3\theta \).

Since \( \theta = \tan^{-1} x \), RHS = \( 3 \tan^{-1} x \).
Thus, LHS = RHS.

Let \( x = \sin \theta \). Then \( \theta = \sin^{-1} x \).
LHS = \( \cos^{-1}(1 - 2 \sin^2 \theta) \)

We know \( \cos 2\theta = 1 - 2 \sin^2 \theta \).
So, LHS = \( \cos^{-1}(\cos 2\theta) \).

LHS = \( 2\theta = 2 \sin^{-1} x = \text{RHS} \).

Use \( \cos x = \sin(\frac{\pi}{2} - x) \) and \( \sin x = \cos(\frac{\pi}{2} - x) \).
Exp = \( \tan^{-1} \frac{\sin(\frac{\pi}{2} - x)}{1 + \cos(\frac{\pi}{2} - x)} \)

Numerator: \( 2 \sin(\frac{\pi}{4} - \frac{x}{2}) \cos(\frac{\pi}{4} - \frac{x}{2}) \)
Denominator: \( 2 \cos^2(\frac{\pi}{4} - \frac{x}{2}) \)
Result = \( \tan^{-1} \tan(\frac{\pi}{4} - \frac{x}{2}) \).

\( 1 - \cos x = 2 \sin^2 \frac{x}{2} \)
\( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \)

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

Divide numerator and denominator by \( b \cos x \):
\( \tan^{-1} \left( \frac{\frac{a}{b} - \tan x}{1 + \frac{a}{b} \tan x} \right) \)

This matches formula \( \tan^{-1} x - \tan^{-1} y \).
Here \( x = a/b \) and \( y = \tan x \).
Result = \( \tan^{-1} \frac{a}{b} - \tan^{-1}(\tan x) \).

Put \( x = a \tan \theta \). Then \( \sqrt{x^2 + a^2} = a \sec \theta \).

\( \sin^{-1} \left( \frac{a \tan \theta}{a \sec \theta} \right) = \sin^{-1} (\sin \theta) = \theta \).
Since \( \theta = \tan^{-1} \frac{x}{a} \).

Put \( x = \tan \theta \).
Expression = \( \tan^{-1} \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) \).

This is the formula for \( \tan 3\theta \).
So, \( \tan^{-1}( \tan 3\theta ) = 3\theta \).

\( \cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{6} \).

\( 4 \times (\frac{\pi}{6}) = \frac{2\pi}{3} \). Then \( \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} \).

\( \tan^{-1}[ 2 \times \frac{\sqrt{3}}{2} ] = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \).

Note that \( \cos^{-1} \frac{4}{5} \) corresponds to the same triangle as \( \sin^{-1} \frac{3}{5} \).
So, \( \cos^{-1} \frac{4}{5} = \sin^{-1} \frac{3}{5} \).
Expression = \( \cos( \sin^{-1} \frac{3}{5} + \sin^{-1} \frac{3}{5} ) = \cos( 2 \sin^{-1} \frac{3}{5} ) \).

Let \( \theta = \sin^{-1} \frac{3}{5} \). We need \( \cos 2\theta \).
\( \cos 2\theta = 1 - 2 \sin^2 \theta = 1 - 2(3/5)^2 \)
\( = 1 - 18/25 = 7/25 \).

\( [0, \pi] \). \( \frac{5\pi}{4} \) is outside.

\( \cos(5\pi/4) = \cos(2\pi - 3\pi/4) = \cos(3\pi/4) \).
Alternatively \( \cos(5\pi/4) = \cos(\pi + \pi/4) = -\cos(\pi/4) \).
\( \cos^{-1}(-\frac{1}{\sqrt{2}}) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).

\( [-\pi/2, \pi/2] \).

\( \sin(7\pi/6) = \sin(\pi + \pi/6) = -\sin(\pi/6) \).
\( \sin^{-1}(-1/2) = -\pi/6 \).

Let \( A = \tan^{-1}(3/4) \) so \( \tan A = 3/4 \), \( \sin A=3/5, \cos A=4/5 \).
Let \( B = \sin^{-1}(3/5) \) so \( \sin B = 3/5, \cos B=4/5 \).

\( \cos(A+B) = \cos A \cos B - \sin A \sin B \)
\( = (4/5)(4/5) - (3/5)(3/5) = 16/25 - 9/25 \).

\( (-\pi/2, \pi/2) \). \( 5\pi/6 \) is outside.

\( \tan(5\pi/6) = \tan(\pi - \pi/6) = -\tan(\pi/6) \).
Inverse is \( -\pi/6 \).

\( \cos^{-1}(-\frac{\sqrt{3}}{2}) = \frac{5\pi}{6} \).

\( \pi - \frac{5\pi}{6} = \frac{\pi}{6} \).
\( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \).

We know \( \sin^{-1} x + \cos^{-1} x = \pi/2 \) for all \( x \in [-1, 1] \).
Here \( x = -1/2 \), which is valid.