Question
Prove the following.
If $\tan\text{A}=\frac{3}{4},$ then $\sin\text{A}\cos\text{A}=\frac{12}{25}$
If $\tan\text{A}=\frac{3}{4},$ then $\sin\text{A}\cos\text{A}=\frac{12}{25}$
✓
Answer
Given, $\tan\text{A}=\frac{3}{4}=\frac{\text{P}}{\text{B}}=\frac{\text{perpendicular}}{\text{Base}}$
Let $P = 3k$ and $B = 4k$
By pythagoras theorem,
$H^2 = P^2 + B^2 = (3k)^2 + (4k)^2$
$= 9k^2 + 16k^2 = 25k^2$
$\Rightarrow H = 5k$ [since, side cannot be negative]
$\therefore\ \sin\text{A}=\frac{\text{P}}{\text{H}}=\frac{3\text{k}}{5\text{k}}=\frac{3}{5}\text{ and }\cos\text{A}=\frac{\text{B}}{\text{H}}=\frac{4\text{k}}{5\text{k}}=\frac{4}{5}$
Now, $\sin\text{A}\cos\text{A}=\frac{3}{5}\cdot\frac{4}{5}=\frac{12}{25}$
Hence proved.
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