The value of $\dfrac{\sin30^0+\tan45^0-\text{cosec}60^0}{\sec30^0+\cos60^0+\cot45^0}$ is:
- $\dfrac{43-24\sqrt3}{11}$
- $\dfrac{27-16-24\sqrt3}{27-16}$
- $\dfrac{43+24\sqrt3}{11}$
- $\text {both a and b}$
Substituting values: sin30=1/2, tan45=1, cosec60=2/sqrt(3), sec30=2/sqrt(3), cos60=1/2, cot45=1. The numerator is (3/2 - 2/sqrt(3)) and the denominator is (3/2 + 2/sqrt(3)). Simplifying gives (3sqrt(3)-4)/(3sqrt(3)+4). Rationalizing results in (43-24sqrt(3))/11.
Substitute the standard trigonometric values sin(30 degrees) = 1/2, tan(45 degrees) = 1, cosec(60 degrees) = 2 / sqrt(3), sec(30 degrees) = 2 / sqrt(3), cos(60 degrees) = 1/2, and cot(45 degrees) = 1 into the expression. The numerator becomes 1/2 + 1 - 2 / sqrt(3) = (3 - 4 * sqrt(3)) / (2 * sqrt(3)). The denominator becomes 2 / sqrt(3) + 1/2 + 1 = (3 + 4 * sqrt(3)) / (2 * sqrt(3)). Dividing the numerator by the denominator gives ((3 - 4 * sqrt(3)) / (2 * sqrt(3))) * ((2 * sqrt(3)) / (3 + 4 * sqrt(3))), which simplifies to (3 - 4 * sqrt(3)) / (3 + 4 * sqrt(3)). Rationalizing this by multiplying by (3 - 4 * sqrt(3)) / (3 - 4 * sqrt(3)) results in (43 - 24 * sqrt(3)) / 11.