Question 1

We will substitute the standard trigonometric values into each expression.

Step 1 — Evaluate (i)

Standard Trigonometric Values (used throughout this solution):

| Angle | $\sin$ | $\cos$ | $\tan$ | |-------|--------|--------|--------| | $0°$ | $0$ | $1$ | $0$ | | $30°$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ | | $45°$ | $\frac{1}{\sqrt{2}}$ | $\frac{1}{\sqrt{2}}$ | $1$ | | $60°$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ | | $90°$ | $1$ | $0$ | undefined |

Let's find the values for each term. We know $\sin 60^\circ = \mathbf{\frac{\sqrt{3}}{2}}$. We know $\cos 30^\circ = \mathbf{\frac{\sqrt{3}}{2}}$. We know $\sin 30^\circ = \mathbf{\frac{1}{2}}$. We know $\cos 60^\circ = \mathbf{\frac{1}{2}}$.

Now, we substitute these values into the expression. $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$

$= \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)$

$= \frac{3}{4} + \frac{1}{4}$

$= \frac{4}{4}$

$\boxed{1}$

Step 2 — Evaluate (ii)

Let's find the values for each term. We know $\tan 45^\circ = \mathbf{1}$. We know $\cos 30^\circ = \mathbf{\frac{\sqrt{3}}{2}}$. We know $\sin 60^\circ = \mathbf{\frac{\sqrt{3}}{2}}$.

Now, we substitute these values into the expression. $2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$

$= 2 (\mathbf{1})^2 + \left(\mathbf{\frac{\sqrt{3}}{2}}\right)^2 - \left(\mathbf{\frac{\sqrt{3}}{2}}\right)^2$

$= 2(1) + \frac{3}{4} - \frac{3}{4}$

$= 2 + 0$

$\boxed{2}$

Step 3 — Evaluate (iii)

Let's find the values for each term. We know $\cos 45^\circ = \mathbf{\frac{1}{\sqrt{2}}}$. We know $\sec 30^\circ = \mathbf{\frac{2}{\sqrt{3}}}$. We know $\text{cosec } 30^\circ = \mathbf{2}$.

Now, we substitute these values into the expression. $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$

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

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

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

$= \frac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}}$

Let's rationalize the denominator. $= \frac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}} \times \frac{2\sqrt{6} - 2\sqrt{2}}{2\sqrt{6} - 2\sqrt{2}}$

$= \frac{2\sqrt{18} - 2\sqrt{6}}{(2\sqrt{6})^2 - (2\sqrt{2})^2}$

$= \frac{2 \times 3\sqrt{2} - 2\sqrt{6}}{24 - 8}$

$= \frac{6\sqrt{2} - 2\sqrt{6}}{16}$

$= \frac{2(3\sqrt{2} - \sqrt{6})}{16}$

$\boxed{\frac{3\sqrt{2} - \sqrt{6}}{8}}$

Step 4 — Evaluate (iv)

Let's find the values for each term. We know $\sin 30^\circ = \mathbf{\frac{1}{2}}$. We know $\tan 45^\circ = \mathbf{1}$. We know $\text{cosec } 60^\circ = \mathbf{\frac{2}{\sqrt{3}}}$. We know $\sec 30^\circ = \mathbf{\frac{2}{\sqrt{3}}}$. We know $\cos 60^\circ = \mathbf{\frac{1}{2}}$. We know $\cot 45^\circ = \mathbf{1}$.

Now, we substitute these values into the expression. $\frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$

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

Let's simplify the numerator first. $= \frac{\frac{3}{2} - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{3}{2}}$

$= \frac{\frac{3\sqrt{3} - 4}{2\sqrt{3}}}{\frac{4 + 3\sqrt{3}}{2\sqrt{3}}}$

$= \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$

Let's rationalize the denominator. $= \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \times \frac{3\sqrt{3} - 4}{3\sqrt{3} - 4}$

$= \frac{(3\sqrt{3})^2 - 2(3\sqrt{3})(4) + (4)^2}{(3\sqrt{3})^2 - (4)^2}$

$= \frac{27 - 24\sqrt{3} + 16}{27 - 16}$

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

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

Step 5 — Evaluate (v)

Let's find the values for each term. We know $\cos 60^\circ = \mathbf{\frac{1}{2}}$. We know $\sec 30^\circ = \mathbf{\frac{2}{\sqrt{3}}}$. We know $\tan 45^\circ = \mathbf{1}$. We know $\sin 30^\circ = \mathbf{\frac{1}{2}}$. We know $\cos 30^\circ = \mathbf{\frac{\sqrt{3}}{2}}$.

Now, we substitute these values into the expression. $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$

Let's simplify the numerator. $5 \left(\frac{1}{2}\right)^2 + 4 \left(\frac{2}{\sqrt{3}}\right)^2 - (1)^2$

$= 5 \left(\frac{1}{4}\right) + 4 \left(\frac{4}{3}\right) - 1$

$= \frac{5}{4} + \frac{16}{3} - 1$

$= \frac{15}{12} + \frac{64}{12} - \frac{12}{12}$

$= \frac{15 + 64 - 12}{12}$

$= \frac{67}{12}$

Now, let's simplify the denominator. $\sin^2 30^\circ + \cos^2 30^\circ$

$= \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$

$= \frac{1}{4} + \frac{3}{4}$

$= \frac{4}{4}$

$= 1$

Now, we divide the numerator by the denominator. $= \frac{\frac{67}{12}}{1}$

$\boxed{\frac{67}{12}}$

Answer

(i) $1$ (ii) $2$ (iii) $\frac{3\sqrt{2} - \sqrt{6}}{8}$ (iv) $\frac{43 - 24\sqrt{3}}{11}$ (v) $\frac{67}{12}$