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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 = ii
 
3 cos \ \theta = ii
 
4 sin \ \theta = ii
 
9 cos \ \theta = ii
 
\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.