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23. Five mark example problems III

If \(\sec \ \theta = \frac{13}{5}\), then show that \(\frac{2 \sin \ \theta - 3 c\os \ \theta}{4 \sin \ \theta - 9 \cos \ \theta} = 3\).

Proof:

\(2 \sin \ \theta =\)

\frac{i}{i}

\(3 \cos \ \theta =\)

\frac{i}{i}

\(4 \sin \ \theta =\)

\frac{i}{i}

\(9 \cos \ \theta =\)

\frac{i}{i}

\(\frac{2 \sin \ \theta - 3 \cos \ \theta}{4 \sin \ \theta - 9 \cos \ \theta} =\)

Since LHS \(=\) RHS, then \(\frac{2 \sin \ \theta - 3 \cos \ \theta}{4 \sin \ \theta - 9 \cos \ \theta} = 3\).

Hence, we proved.

Did you find an error?