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Question
Prove that
cosec (67° + θ) − sec (23° − θ) = 0
Solution
\[\begin{array}{l} \text{L.H.S}=cosec( {67}^\circ + \theta) - \sec( {23}^\circ- \theta) \\ \end{array}\]
\[\begin{array}{l}=cosec{ {90}^\circ - ( {23}^\circ-\theta)}-\sec( {23}^\circ - \theta) \\ \end{array}\]
\[\begin{array}{l}=\sec( {23}^\circ -\theta) - \sec( {23}^\circ - \theta) \\ \\ \end{array}\]
\[=0=R.H.S.\]
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