The value of the expression $\dfrac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$ is:
(a) $\dfrac{43 - 24\sqrt{3}}{11}$
(b) $\dfrac{43 + 24\sqrt{3}}{11}$
(c) $\dfrac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$
(d) $1$
Step-by-Step Solution
Key Concept: Substitute standard values: $\sin 30^\circ=1/2, \tan 45^\circ=1, \csc 60^\circ=2/\sqrt{3}$, etc.
Numerator $= 1/2 + 1 - 2/\sqrt{3} = 3/2 - 2/\sqrt{3} = \dfrac{3\sqrt{3} - 4}{2\sqrt{3}}$. [0.5 Mark]
Denominator $= 2/\sqrt{3} + 1/2 + 1 = 3/2 + 2/\sqrt{3} = \dfrac{3\sqrt{3} + 4}{2\sqrt{3}}$. Ratio $= \dfrac{3\sqrt{3} - 4}{3\sqrt{3} + 4} = \dfrac{(3\sqrt{3}-4)^2}{27 - 16} = \dfrac{27 + 16 - 24\sqrt{3}}{11} = \dfrac{43 - 24\sqrt{3}}{11}$. [0.5 Mark]
---
🎯 Official CBSE Marking Scheme:
Substituting values: 0.5 Mark
Rationalising denominator to get $(43 - 24\sqrt{3})/11$: 0.5 Mark
Correct Answer: $\dfrac{43 - 24\sqrt{3}}{11}$