If sec θ = 13/5, then what is the value of (2sin θ - 3cos θ)/(4sin θ - 9cos θ) ?

Question

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

(a) 1 

(b) 3 

(c) 2 

(d) 4

Answer 1

(b) 3

Given, sec θ = \(\frac{13}{5}\) ⇒ sec² θ = \(\frac{169}{25}\)

⇒ 1 + tan²θ = \(\frac{169}{25}\)                 (∵ sec²θ – tan²θ = 1) 

⇒ tan² θ = \(\frac{169}{25}\) - 1 = \(\frac{144}{25}\) ⇒ tan θ = \(\frac{12}{5}\)

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

(On dividing each term of numerator and denominator by cos θ)

\(= \frac{2\,tan\,\theta-3}{4\,tan\,\theta-9}=\frac{2\times\frac{12}{5}-3}{4\times\frac{12}{5}-9}\)

= \(\frac{24-15}{48-45}=\frac93=3.\)