Question

If sec $\theta =\frac{13}{5}$,show that $\frac{2sin\theta -3cos\theta }{4sin\theta -9cos\theta }=3$

Answer 1

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

cos $\theta$ = $\frac{1}{sec\theta }$ = $\frac{5}{13}$

$si{n}^2\theta$ = 1- $co{s}^2\theta$

= 1 - $(\frac{5}{13}{)}^2$ = 1 - $\frac{25}{169}$ = $\frac{169-25}{169}$ = $\frac{144}{169}$

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

Now, put the values in the equation

= 2*sin $\theta$ - $\frac{3\ast cos\theta }{4\ast sin\theta }$ - 9*cos $\theta$

=$\frac{\left(\frac{2\ast 12}{13}\right)-\left(\frac{3\ast 5}{13}\right)}{\left(\frac{4\ast 12}{13}\right)-\left(\frac{9\ast 5}{13}\right)}$

= $\frac{\left(\frac{24}{13}\right)-\left(\frac{15}{13}\right)}{\left(\frac{48}{13}\right)-\left(\frac{45}{13}\right)}$

= $\frac{\left(\frac{24-15}{13}\right)}{\left(\frac{48-45}{13}\right)}$

(Cancel out that 13.)

= $\frac{9}{3}$

= 3