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Question
Prove that
sin (70° + θ) − cos (20° − θ) = 0
Solution
\[\begin{array}{l}(i) L.H.S=sin( {70}^0 + \theta) - \cos( {20}^0- \theta) \\ \end{array}\]
\[\begin{array}{l}=sin{ {90}^0 - ( {20}^0 - \theta)} - \cos( {20}^0 - \theta) \\ \end{array}\]
\[\begin{array}{l}=\cos( {20}^0 - \theta) -\cos( {20}^0 - \theta) \\ \end{array}\]
=0
= RHS
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