Question

If sinA-cosA=3/5, then sinAcosA=

Answer 1

Step-by-step explanation:

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Answer 2

The answer is sinAcosA= $\frac{8}{25}$

Step-by-step explanation:

$\Rightarrow \left ( \sin A - \cos A \right )=\frac{3}{5}$

Now squaring on both side,

$\Rightarrow \left ( \sin A - \cos A \right )^{2}=\frac{9}{25}$

$\Rightarrow \left ( \sin^{2} A + \cos^{2} A -2\sin A \cos A\right )=\frac{9}{25}$

$\Rightarrow \left ( 1 -2\sin A \cos A\right )=\frac{9}{25}$

$\Rightarrow \left ( 1 -\frac{9}{25}\right )=2\sin A \cos A$

$\Rightarrow \left ( \frac{16}{25}\right )=2\sin A \cos A$

$\Rightarrow \sin A \cos A= \frac{8}{25}$

The answer is sinAcosA= $\frac{8}{25}$