Question

If $\tan A=\frac{3}{4}$, then $\sin A\cos A=\frac{12}{25}$

Answer 1

$\tan A=\frac{3}{4}$

$\frac{\sin A}{\cos A}=\frac{3}{4}$

$\sin A=\frac{3}{4}\cos A$

${\sin }^{2}A=\frac{9}{16}{\cos }^{2}A$

$\Rightarrow {\sin }^{2}A=\frac{9}{16}(1-{\sin }^{2}A)$

$\Rightarrow 16{\sin }^{2}A=9(1-{\sin }^{2}A)$

$\Rightarrow 16{\sin }^{2}A=9-9{\sin }^{2}A$

$\Rightarrow 16{\sin }^{2}A+9{\sin }^{2}A=9$

$\Rightarrow 25{\sin }^{2}A=9$

$\Rightarrow {\sin }^{2}A=\frac{9}{25}$

$\therefore \sin A=\pm \frac{3}{5}$

$\Rightarrow \cos A=\sqrt{1-{\sin }^{2}A}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{9}}=\pm \frac{4}{5}$

$\therefore \sin A\cos A=\pm \frac{3}{5}\times \pm \frac{4}{5}=\frac{12}{25}$

Hence proved.