Question

Prove that if tanA=$\frac{3}{4}$, then sinAcosA=$\frac{12}{25}$

Answer 1

As ${\sec }^{2}A-{\tan }^{2}A=1\Rightarrow secA=\pm \sqrt{{\tan }^{2}A+1}$

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

$\Rightarrow \cos A=\frac{1}{secA}=\pm \frac{4}{5}$

And $si{n}^{2}A+co{s}^{2}A=1$

$\Rightarrow sinA=\pm \sqrt{1-{\cos }^{2}A}=\pm \sqrt{1-\frac{16}{25}}$

$=\pm \sqrt{\frac{9}{25}}=\pm \frac{3}{5}$

$\Rightarrow sinAcosA=\frac{12}{25}$