Q1(iv):

Evaluate the following : (iv) $\frac{\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$

Given: The trigonometric expression $\frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$.

To Find: The numerical value of the given expression.

Step 1: Identify the values of the trigonometric ratios.
Using the standard trigonometric table for specific angles, we have:
$\sin 30^\circ = \frac{1}{2}$
$\tan 45^\circ = 1$
$\text{cosec } 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$
$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$
$\cos 60^\circ = \frac{1}{2}$
$\cot 45^\circ = \frac{1}{\tan 45^\circ} = \frac{1}{1} = 1$

Step 2: Substitute the values into the expression.
Substituting the values identified in Step 1 into the given expression:
$\text{Expression} = \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1}$

Step 3: Simplify the numerator and the denominator.
For the numerator: $\frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}}$
Finding a common denominator ($2\sqrt{3}$):
$\frac{3\sqrt{3} - 4}{2\sqrt{3}}$

For the denominator: $\frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{3}{2}$
Finding a common denominator ($2\sqrt{3}$):
$\frac{4 + 3\sqrt{3}}{2\sqrt{3}}$

Step 4: Perform the division of the fractions.
$\text{Expression} = \frac{\frac{3\sqrt{3} - 4}{2\sqrt{3}}}{\frac{3\sqrt{3} + 4}{2\sqrt{3}}}$
Since the denominators are identical, they cancel out:
$\text{Expression} = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$

Step 5: Rationalize the denominator.
To rationalize, multiply the numerator and denominator by the conjugate of the denominator, which is $(3\sqrt{3} - 4)$:
$\frac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \times \frac{3\sqrt{3} - 4}{3\sqrt{3} - 4} = \frac{(3\sqrt{3} - 4)^2}{(3\sqrt{3})^2 - (4)^2}$

Applying the algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$ in the numerator and $a^2 - b^2$ in the denominator:
Numerator: $(3\sqrt{3})^2 - 2(3\sqrt{3})(4) + (4)^2 = (9 \times 3) - 24\sqrt{3} + 16 = 27 - 24\sqrt{3} + 16 = 43 - 24\sqrt{3}$
Denominator: $(3\sqrt{3})^2 - (4)^2 = 27 - 16 = 11$

Thus, the expression simplifies to:
$\frac{43 - 24\sqrt{3}}{11}$

Final Answer: \frac{43 - 24\sqrt{3}}{11}