Question

If \( \tan A = \cot B \), prove that \( A + B = 90^\circ \).

Hint:

To prove trigonometric identities, use fundamental identities like \( \tan (90^\circ - \theta) = \cot \theta \) to relate different angles.

Answer 1

We are given that: \[ \tan A = \cot B \] We know that: \[ \cot B = \frac{1}{\tan B} \] So, the equation becomes: \[ \tan A = \frac{1}{\tan B} \] This implies: \[ \tan A \cdot \tan B = 1 \] Now, using the identity for the tangent of complementary angles: \[ \tan (90^\circ - A) = \cot A \] Thus, we have: \[ \tan A = \cot B \implies A + B = 90^\circ \] Hence, we have proved that \( A + B = 90^\circ \).