Option 4 : 0

Given:

\(\frac{{\left( {\tan \frac{\pi }{4} - \cot \frac{{2\pi }}{9}} \right)}}{{\left( {\tan \frac{\pi }{4} + \cot \frac{{2\pi }}{9}} \right)}} - \frac{{\left( {\cot \frac{{5\pi }}{{18}} - 1} \right)}}{{\left( {\cot \frac{{5\pi }}{{18}} + 1} \right)}}\)

Concept:

tan (90° - A) = cot A

tan A = 1/cot A

Calculation:

\(\frac{{\left( {\tan \frac{\pi }{4} - \cot \frac{{2\pi }}{9}} \right)}}{{\left( {\tan \frac{\pi }{4} + \cot \frac{{2\pi }}{9}} \right)}} - \frac{{\left( {\cot \frac{{5\pi }}{{18}} - 1} \right)}}{{\left( {\cot \frac{{5\pi }}{{18}} + 1} \right)}}\)

⇒ \(\frac{{\left( {1 - \frac{1}{{tan40^\circ }}} \right)}}{{\left( {1 + \frac{1}{{tan40^\circ }}} \right)}} - \frac{{\left( {\cot 50^\circ - 1} \right)}}{{\left( {\cot 50^\circ + 1} \right)}}\)

⇒ \(\frac{{tan40^\circ - 1}}{{tan40^\circ + 1}} - \frac{{\left( {\cot 50^\circ - 1} \right)}}{{\left( {\cot 50^\circ + 1} \right)}}\)

⇒ \(\frac{{tan40^\circ - 1}}{{tan40^\circ + 1}} - \frac{{\left( {tan40^\circ - 1} \right)}}{{\left( {tan40^\circ + 1} \right)}}\) (As we know cot 50° = tan 40°)

⇒ [(tan40° -1) - (tan40° + 1)]/(tan40° + 1)

⇒ 0/(tan40° + 1)

⇒ 0