Q.Find the value of:

We use standard compound-angle formulas and known values of $30^\circ, 45^\circ$.

Step 1 — Express as a sum/difference of known angles

$\sin 75^\circ = \sin(45^\circ + 30^\circ)$

$\tan 15^\circ = \tan(45^\circ - 30^\circ)$

Step 2 — Apply formulas

$\sin(A+B) = \sin A \cos B + \cos A \sin B$

$\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Step 3 — Substitute known values …

The key insight here is that $75^\circ$ and $15^\circ$ are not standard angles whose trig values you memorise (like $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$). But both can be written as sums or differences of those standard angles:

• $75^\circ = 45^\circ + 30^\circ$
• $15^\circ = 45^\circ - 30^\circ$ (or $60^\circ - 45^\circ$ — either works)

Once you see that, the problem reduces to applying the sum and difference formulas for sine and tangent. These formulas are your friends — they let you break unfamiliar angles into familiar pieces.

Let’s work through each part.

(i) $\sin 75^\circ$

1. Write $75^\circ$ as $45^\circ + 30^\circ$.

We know the exact values:

$\sin 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 45^\circ = \frac{\sqrt{2}}{2}$

$\sin 30^\circ = \frac{1}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$

2. Apply the sine sum formula:

$\sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$

3. Substitute the known values:

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

4. Simplify:

$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$

(ii) $\tan 15^\circ$

1. Write $15^\circ$ as $45^\circ - 30^\circ$.

We know: $\tan 45^\circ = 1$, $\tan 30^\circ = \frac{1}{\sqrt{3}}$

2. Apply the tangent difference formula: …