If $\theta$ is in the first quadrant and $\cos\theta=\dfrac{3}{5}$, then the value of $\displaystyle\frac{5\tan\theta-4 cosec\theta}{5 sec\theta-4\cot\theta}$ will be

$\because$ Since $\theta$ is in the first quadrant, all trigonometrical ratios will be positive.

and $\cos\theta=\dfrac{3}{5}$

then $\sin\theta=sqrt{1-\cos^2\theta}$

$=\sqrt{1-\displaystyle\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}$

and $\tan\theta=\dfrac{4}{3}$

$\therefore$ Given, Exp. =$\displaystyle\dfrac{5\tan\theta-4cosec\theta}{5sec\theta-4\cot\theta}$

$=\displaystyle\dfrac{5\times \dfrac{4}{3}-4\times \dfrac{5}{4}}{5\times \displaystyle \dfrac{5}{3}-4\times \dfrac{3}{4}}$

$=\displaystyle\dfrac{5\left(\dfrac{4}{3}-1\right)}{\displaystyle\dfrac{25}{3}-3}$

$=\displaystyle\dfrac{\dfrac{5}{3}}{\dfrac{16}{3}}=\dfrac{5}{16}$