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Question
If sec 4 A = cosec (A − 15°), where 4 A is an acute angle, find the value of A.
Solution
\[\begin{array}{l}\sec4A = \ cosec(A -{15}^\circ ) \\ \end{array}\]
\[\begin{array}{l}=> cosec( {90}^\circ - 4A) = \ cosec(A- {15}^\circ )[ \because \sec\theta = \ cosec ( {90}^\circ -\theta)] \\ \end{array}\]
\[\begin{array}{l}{=>90}^\circ - 4A = A - {15}^\circ \\ \end{array}\]
\[\begin{array}{l}{=>105}^\circ = 5A \\ \\ \end{array}\]
\[=>A = \frac{{105}^\circ}{5} = {21}^\circ\]
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