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Trigonometry Practice Test 2

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Question : 8 of 22

Marks: +1, -0

If

\sec \theta = \frac{13}{5}

, then what is the value of

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

?

1
2
3
4
None of these

Solution:

Concept:

Use

\sec^2\theta = 1 + \tan^2\theta

and divide numerator and denominator by

\cos\theta

.

Explanation:

Given

\sec\theta=\frac{13}{5}

, so

\cos\theta=\frac{5}{13}

.

For the standard acute-angle case,

\sin\theta=\frac{12}{13}

and

\tan\theta=\frac{12}{5}

.

Now,

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

Divide every term by

\cos\theta

:

=\frac{2\tan\theta-3}{4\tan\theta-9}

.

Substitute

\tan\theta=\frac{12}{5}

:

=\frac{2\cdot\frac{12}{5}-3}{4\cdot\frac{12}{5}-9}

=\frac{\frac{24}{5}-\frac{15}{5}}{\frac{48}{5}-\frac{45}{5}}

=\frac{\frac{9}{5}}{\frac{3}{5}}

=3

.

Answer:

The value is

3

.

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