If $\tan\theta=\dfrac{3}{4}$ and $0< \theta, < 90^0,$ then the value of $\sin\theta \cos \theta$ is

Given, $\tan \theta=\dfrac{3}{4}$

$\therefore \sec \theta=\sqrt{1+\tan^2\theta}$

$=\sqrt{1+\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}$

or

$\cos \theta=\dfrac{4}{5}$

$\therefore \sin \theta \cos \theta = \tan \theta \cos^2 \theta$

$=\dfrac{3}{4}\times \dfrac{16}{25}=\dfrac{12}{25}$