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If \(sec \ \theta = \frac{13}{5}\), then show that \(\frac{2 sin \ \theta - 3 cos \ \theta}{4 sin \ \theta - 9 cos \ \theta} = 3\).
Proof:
\(2 sin \ \theta =\)
\(3 cos \ \theta =\)
\(4 sin \ \theta =\)
\(9 cos \ \theta =\)
\(\frac{2 sin \ \theta - 3 cos \ \theta}{4 sin \ \theta - 9 cos \ \theta} =\)
Since LHS \(=\) RHS, then \(\frac{2 sin \ \theta - 3 cos \ \theta}{4 sin \ \theta - 9 cos \ \theta} = 3\).
Hence, we proved.
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