Question

The value of $\frac{\sin {30}^0+\tan {45}^0-\text{cosec}{60}^0}{\sec {30}^0+\cos {60}^0+\cot {45}^0}$ is:

• A. $\frac{43-24\sqrt{3}}{11}$
• B. $\frac{27-16-24\sqrt{3}}{27-16}$
• C. $\frac{43+24\sqrt{3}}{11}$
• D. $\text{both a and b}$

Answer 1

The correct option is A $\frac{43-24\sqrt{3}}{11}$

$\frac{\sin 30°+\tan 45°-\text{cosec}60°}{\sec 30°+\cos 60°+\cot 45°}$

$=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}$

$=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}$

$=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}\times \frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}-\frac{2}{\sqrt{3}}}$ $[$On Rationalising the denominator$]$

$=\frac{\frac{9}{4}+\frac{4}{3}-2\times \frac{3}{2}\times \frac{2}{\sqrt{3}}}{\frac{11}{12}}$

$=\frac{\frac{43}{12}-2\sqrt{3}}{\frac{11}{12}}$

$=\frac{43-24\sqrt{3}}{11}$