Show that :
(i) `tan 10^(@) tan 15^(@) tan 75^(@) tan 80^(@) = 1`
(ii) `sin 42^(@) sec 48^(@) + cos 42^(@) "cosec 48"^(@) = 2`

Answer

Let's solve the given problems step by step.

### (i) Show that:
\[ \tan 10^\circ \tan 15^\circ \tan 75^\circ \tan 80^\circ = 1 \]

**Step 1: Use the identity for complementary angles.**
We know that:
\[ \tan(90^\circ - \theta) = \cot \theta \]
Thus, we can rewrite \(\tan 75^\circ\) and \(\tan 80^\circ\):
\[ \tan 75^\circ = \cot 15^\circ \quad \text{(since } 75^\circ = 90^\circ - 15^\circ\text{)} \]
\[ \tan 80^\circ = \cot 10^\circ \quad \text{(since } 80^\circ = 90^\circ - 10^\circ\text{)} \]

**Step 2: Substitute the identities into the equation.**
Now substituting these identities into the original equation:
\[ \tan 10^\circ \tan 15^\circ \tan 75^\circ \tan 80^\circ = \tan 10^\circ \tan 15^\circ \cot 15^\circ \cot 10^\circ \]

**Step 3: Simplify the expression.**
Using the identity \(\tan \theta \cdot \cot \theta = 1\):
\[ = (\tan 10^\circ \cot 10^\circ) \cdot (\tan 15^\circ \cot 15^\circ) = 1 \cdot 1 = 1 \]

**Conclusion:**
Thus, we have shown that:
\[ \tan 10^\circ \tan 15^\circ \tan 75^\circ \tan 80^\circ = 1 \]