If \sec \theta = 13/5, then what is the value of \frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}?

Trigonometry Practice Test 2

Question : 8 of 22

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\sec θ = 13/5

, then what is the value of

{2\sinθ-3\cosθ}/{4\sinθ-9\cosθ}

Solution:

\sec θ = 13/5⇒\sec^2θ=169/25

⇒1+\tan^2θ=169/25

⇒\tan^2θ=144/25

⇒\tanθ=12/5

∴{2\sinθ-3\cosθ}/{4\sinθ-9\cosθ}={2{\sinθ}/{\cosθ}-3}/{4{\sinθ}/{\cosθ}-9}

={2\tanθ-3}/{4\tanθ-9}={2×12/5-3}/{4×12/5-9}=3

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