Question

If $sin\theta +cos\theta =\frac{9}{7}$, then find the value of $sin\theta cos\theta$.

Answer 1

$sin\theta +cos\theta =\frac{9}{7}$

Squaring on both sides, we get
$\Rightarrow (sin\theta +cos\theta {)}^2=(\frac{9}{7}{)}^2$
$\Rightarrow si{n}^2\theta +co{s}^2\theta +2sin\theta cos\theta =\left(\frac{81}{49}\right)$
$\Rightarrow 1+2sin\theta cos\theta =\left(\frac{81}{49}\right)$
[since, $si{n}^2\theta +co{s}^2\theta =1$]

$\Rightarrow 2sin\theta cos\theta =\left(\frac{81}{49}\right)-1$
$\Rightarrow 2sin\theta cos\theta =\left(\frac{81-49}{49}\right)$
$\Rightarrow sin\theta cos\theta =\left(\frac{32}{49\times 2}\right)$
$\Rightarrow sin\theta cos\theta =\left(\frac{16}{49}\right)$