Question

Prove that

tan (55° − θ) − cot (35° + θ) = 0

Answer 1

\[\begin{array}{l} LHS =\tan( {55}^0 - \theta) - \cot( {35}^0 + \theta) \\ \end{array}\]

\[\begin{array}{l}=\tan{ {90}^0  - ( {35}^0 + \theta)} - \cot( {35}^0 + \theta) \\ \end{array}\]

\[\begin{array}{l}=\cot( {35}^0 + \theta) - \cot( {35}^0 + \theta) \\ \end{array}\]

= 0

= RHS