Question : If $\sin t+\cos t=\frac{4}{5}$, then find $\sin t.\cos t$.
Option 1: $\frac{9}{50}$
Option 2: $-\frac{9}{50}$
Option 3: $\frac{9}{25}$
Option 4: $-\frac{9}{25}$
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Correct Answer: $-\frac{9}{50}$
Solution :
Given: $\sin t+\cos t=\frac{4}{5}$
Squaring both sides, we have,
⇒ $(\sin t+\cos t)^{2}=(\frac{4}{5})^{2}$
⇒ $\sin^{2} t+\cos^{2} t+2\sin t.\cos t=\frac{16}{25}$
We know that $\sin^{2} t+\cos^{2}=1$
⇒ $1+2\sin t.\cos t=\frac{16}{25}$
⇒ $2\sin t.\cos t=\frac{16}{25}–1$
⇒ $2\sin t.\cos t=-\frac{9}{25}$
⇒ $\sin t.\cos t=-\frac{9}{50}$
Hence, the correct answer is $-\frac{9}{50}$.
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