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Question
Prove that
cosec (65 °+ θ) sec (25° − θ) − tan (55° − θ) + cot (35° + θ) = 0
Solution
\[\begin{array}{l}\text{ L.H.S}=cosec( {65}^0 + \theta) - \sec( {25}^0 - \theta) - \tan( {55}^0 - \theta) + \cot( {35}^0 + \theta) \\ \end{array}\]
\[\begin{array}{l}= \ cosec{ {90}^0 - ( {25}^0 - \theta)} - \sec( {25}^0 - \theta) - \tan( {55}^0 - \theta) + \cot{ {90}^0 -( {55}^0 - \theta)} \\ \end{array}\]
\[\begin{array}{l}= \sec( {25}^0 - \theta) - \sec( {25}^0 - \theta) - \tan( {55}^0 - \theta) + \tan( {55}^0 -\theta) \\ \end{array}\]
= 0
= RHS
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