1. For \(\sin^{-1}(-\sqrt{3}/2)\):
   • This is the angle whose sine is \(-\sqrt{3}/2\)
   • We know that \(\sin(5\pi/6) = -\sqrt{3}/2\)
   • Therefore, \(\sin^{-1}(-\sqrt{3}/2) = -\pi/3\)
2. For \(\tan^{-1}(1) - \tan^{-1}(-1)\):
   • \(\tan^{-1}(1) = \pi/4\)
   • \(\tan^{-1}(-1) = -\pi/4\)
   • Therefore, \(\tan^{-1}(1) - \tan^{-1}(-1) = \pi/2\)
3. For \(\cot^{-1}(1/\sqrt{3}) - \cot^{-1}(-1/\sqrt{3})\):
   • \(\cot^{-1}(1/\sqrt{3}) = \pi/6\)
   • \(\cot^{-1}(-1/\sqrt{3}) = 5\pi/6\)
   • Therefore, the difference is \(-\pi/3\)
4. For \(\cosh^{-1}\sqrt{2}\):
   • By definition, \(\cosh^{-1}x = \ln(x + \sqrt{x^2-1})\)
   • Substituting \(x = \sqrt{2}\)
   • Result is \(\ln(1 + \sqrt{2})\)
5. For \(e^{[-\coth^{-1}(25/7)]}\):
   • Recall that \(\coth^{-1}x = \frac{1}{2}\ln(\frac{x+1}{x-1})\)
   • Substituting x = 25/7: \[\coth^{-1}(\frac{25}{7}) = \frac{1}{2}\ln(\frac{32/7}{18/7}) = \frac{1}{2}\ln(\frac{32}{18})\]
   • Therefore: \[e^{[-\coth^{-1}(25/7)]} = e^{-\frac{1}{2}\ln(\frac{32}{18})} = (\frac{32}{18})^{-1/2} = \frac{3}{4}\]

Answers:

• a) \(-\pi/3\)
• b) \(\pi/2\)
• c) \(-\pi/3\)
• d) \(\ln(1 + \sqrt{2})\)
• e) \(\frac{3}{4}\)

1. Let's start by noting that \(\cot^{-1}x = \tan^{-1}(\frac{1}{x})\)
2. Therefore, we need to prove: \[\tan^{-1}x + \tan^{-1}(\frac{1}{x}) = \frac{\pi}{2}\]
3. Using the tangent addition formula: \[\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\]
4. Let \(A = \tan^{-1}x\) and \(B = \tan^{-1}(\frac{1}{x})\)
5. Then \(\tan(A + B) = \frac{x + \frac{1}{x}}{1 - x \cdot \frac{1}{x}} = \infty\)
6. Therefore, \(A + B = \frac{\pi}{2}\)

The graph above visually confirms that this sum remains constant at π/2 for all valid x.

1. Recall the definitions: \[\sinh x = \frac{e^x - e^{-x}}{2}\] \[\cosh x = \frac{e^x + e^{-x}}{2}\]
2. For the first identity: \[\sinh(x \pm y) = \frac{e^{x\pm y} - e^{-(x\pm y)}}{2}\] \[= \frac{e^x e^{\pm y} - e^{-x} e^{\mp y}}{2}\]
3. Expand this using the definitions above to get: \[\sinh x \cosh y \pm \cosh x \sinh y\]
4. Similarly for the second identity, starting with: \[\cosh(x \pm y) = \frac{e^{x\pm y} + e^{-(x\pm y)}}{2}\]
5. This expands to: \[\cosh x \cosh y \pm \sinh x \sinh y\]

The plot above shows how these functions behave and their relationships to each other.

1. For the triple angle formula: \[\cos(3x) = 4\cos^3(x) - 3\cos(x)\]
2. Start with the double angle formula: \[\cos(2x) = 2\cos^2(x) - 1\]
3. Then use the angle addition formula: \[\cos(3x) = \cos(2x + x)\]
4. Substitute the double angle formula into this to get the triple angle formula
5. For the hyperbolic formula: \[\cosh(2x) = 2\cosh^2(x) - 1\]
6. Use the same process to derive the hyperbolic triple angle formula

1. For the inverse hyperbolic functions: \[\text{arsinh}(x) = \ln(x + \sqrt{x^2 + 1})\] \[\text{arcosh}(x) = \ln(x + \sqrt{x^2 - 1})\] \[\text{artanh}(x) = \frac{1}{2}\ln(\frac{1+x}{1-x})\]
2. These are derived by inverting the hyperbolic functions
3. For the first identity: \[\text{arsinh}(x) = \ln(x + \sqrt{x^2 + 1})\]
4. Start with \(y = \text{arsinh}(x)\) and use the definition of sinh
5. Then solve for x to get the inverse function

1. For the domain-restricted functions: \[\ln(\sin x), \frac{1}{\sqrt{|\cos x|}}, \ln|\ln x|\]
2. These functions have specific domains where they are defined
3. For the first function: \[\ln(\sin x)\]
4. Since the natural logarithm is only defined for positive values, the domain is: \[0 < \sin x \leq 1\]
5. Similarly, analyze the other two functions to find their domains