Question

If \( \tan x^\circ \tan 2^\circ \tan 3^\circ \cdots \tan 88^\circ \tan y^\circ = 1 \), then \( \cot(x+y)= \)

• A. 0
• B. 1
• C. \( \frac{1}{2} \)
• D. Undefined

Hint:

Use the identity \( \tan\theta \tan(90^\circ-\theta)=1 \) to simplify products of tangent terms involving complementary angles.

Answer 1

The Correct Option is A

Step 1: Recall complementary angle identity.
We know that:
\[ \tan\theta \cdot \tan(90^\circ-\theta)=1. \]
This is because:
\[ \tan(90^\circ-\theta)=\cot\theta=\frac{1}{\tan\theta}. \]

Step 2: Pair the tangent terms.
The product contains tangent terms from \(2^\circ\) to \(88^\circ\), along with \(\tan x^\circ\) and \(\tan y^\circ\).
Complementary pairs are:
\[ \tan 2^\circ \tan 88^\circ=1, \] \[ \tan 3^\circ \tan 87^\circ=1, \] and so on.

Step 3: Understand the missing complementary pair.
For the whole product to be equal to 1, \(\tan x^\circ\) and \(\tan y^\circ\) must also form a complementary pair.
So:
\[ x+y=90^\circ. \]

Step 4: Use the required expression.
We need to find:
\[ \cot(x+y). \]

Step 5: Substitute \(x+y=90^\circ\).
\[ \cot(x+y)=\cot 90^\circ. \]

Step 6: Evaluate \(\cot 90^\circ\).
\[ \cot 90^\circ=\frac{\cos 90^\circ}{\sin 90^\circ}. \]
\[ =\frac{0}{1}=0. \]

Step 7: Final conclusion.
Thus,
\[ \cot(x+y)=0. \]
Final Answer:
\[ \boxed{0} \]