We are asked to evaluate the following expression: \[ \frac{\cos^{-1}(0) + \sin^{-1}\left( \frac{\sqrt{3}}{2} \right) + \cos^{-1}\left( \frac{1}{2} \right)}{\sin^{-1}(1) + \cos^{-1}\left( \frac{\sqrt{3}}{2} \right) + \sin^{-1}\left( \frac{1}{\sqrt{2}} \right)} \] Step 1: Simplifying the Numerator
- \( \cos^{-1}(0) \): We know that \( \cos(\frac{\pi}{2}) = 0 \), so: \[ \cos^{-1}(0) = \frac{\pi}{2} \] - \( \sin^{-1}\left( \frac{\sqrt{3}}{2} \right) \): We know that \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \), so: \[ \sin^{-1}\left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} \] - \( \cos^{-1}\left( \frac{1}{2} \right) \): We know that \( \cos(\frac{\pi}{3}) = \frac{1}{2} \), so: \[ \cos^{-1}\left( \frac{1}{2} \right) = \frac{\pi}{3} \] Thus, the numerator becomes: \[ \frac{\pi}{2} + \frac{\pi}{3} + \frac{\pi}{3} = \frac{3\pi}{6} + \frac{2\pi}{6} + \frac{2\pi}{6} = \frac{7\pi}{6} \] Step 2: Simplifying the Denominator
- \( \sin^{-1}(1) \): We know that \( \sin(\frac{\pi}{2}) = 1 \), so: \[ \sin^{-1}(1) = \frac{\pi}{2} \] - \( \cos^{-1}\left( \frac{\sqrt{3}}{2} \right) \): We know that \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \), so: \[ \cos^{-1}\left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{6} \] - \( \sin^{-1}\left( \frac{1}{\sqrt{2}} \right) \): We know that \( \sin\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \), so: \[ \sin^{-1}\left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} \] Thus, the denominator becomes: \[ \frac{\pi}{2} + \frac{\pi}{6} + \frac{\pi}{4} \] Finding a common denominator for the terms: \[ \frac{6\pi}{12} + \frac{2\pi}{12} + \frac{3\pi}{12} = \frac{11\pi}{12} \] Step 3: Final Calculation
Now, we calculate the overall expression: \[ \frac{\frac{7\pi}{6}}{\frac{11\pi}{12}} = \frac{7\pi}{6} \times \frac{12}{11\pi} = \frac{7 \times 12}{6 \times 11} = \frac{84}{66} = \frac{14}{11} \] Thus, the correct answer is option (D), \( \frac{14}{11} \).
Thus, the correct answer is option (D), \( \frac{14}{11} \).
Let \(a=i+j+2k\) and \(b=i-2j+3k\) be two vectors. Then the unit vector in the direction of \(a-b\) is
Kepler's second law (law of areas) of planetary motion leads to law of conservation of