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Question
Prove that
tan (55° − θ) − cot (35° + θ) = 0
Solution
\[\begin{array}{l} LHS =\tan( {55}^0 - \theta) - \cot( {35}^0 + \theta) \\ \end{array}\]
\[\begin{array}{l}=\tan{ {90}^0 - ( {35}^0 + \theta)} - \cot( {35}^0 + \theta) \\ \end{array}\]
\[\begin{array}{l}=\cot( {35}^0 + \theta) - \cot( {35}^0 + \theta) \\ \end{array}\]
= 0
= RHS
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