Question

Prove that \(\tan 9^\circ \cdot \tan 27^\circ = \cot 63^\circ \cdot \cot 81^\circ\).

Hint:

Whenever you see a trigonometric expression with angles that seem unrelated, check if any pairs add up to \(90^\circ\). If they do, using complementary angle identities is almost always the correct approach.

Answer 1

Step 1: Understanding the Concept:
This problem involves proving a trigonometric equality by using the relationship between trigonometric functions of complementary angles (angles that add up to \(90^\circ\)).

Step 2: Key Formula or Approach:
The key complementary angle identity we will use is: \[ \cot(90^\circ - \theta) = \tan\theta \]

Step 3: Detailed Explanation:
We can start from the Right Hand Side (RHS) and show that it is equal to the Left Hand Side (LHS).
\[ \text{RHS} = \cot 63^\circ \cdot \cot 81^\circ \] Notice that \(63^\circ = 90^\circ - 27^\circ\) and \(81^\circ = 90^\circ - 9^\circ\).
Apply the complementary angle identity to each term:
For the first term: \[ \cot 63^\circ = \cot(90^\circ - 27^\circ) = \tan 27^\circ \] For the second term: \[ \cot 81^\circ = \cot(90^\circ - 9^\circ) = \tan 9^\circ \] Now substitute these back into the RHS expression: \[ \text{RHS} = (\tan 27^\circ) \cdot (\tan 9^\circ) \] Rearranging the terms, we get: \[ \text{RHS} = \tan 9^\circ \cdot \tan 27^\circ \] This is exactly the expression on the LHS.

Step 4: Final Answer:
Since RHS = LHS, the identity is proved.