Question
In a right $\triangle\text{ABC},$ right-angled at B, if $\tan\text{A}=1$ then verify that $2\sin\text{A}\cdot\cos\text{A}=1.$
✓
Answer
$\tan\text{A}=1$Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$
Then, $\tan\text{A}=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac11$
Let BC = 1 and AB = 1
Then, by Pythagoras theoram,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$=2^2+1^2=1=2$
$\Rightarrow\text{AC}=\sqrt{2}$
Now,
$\sin\text{A}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{1}{\sqrt{2}}$
$\cos\text{A}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{1}{\sqrt{2}}$
$\therefore\text{L.H.S.}=2\sin\text{A}\cdot\cos\text{A}$
$=2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$
$=\frac22$
$=1$
$=\text{R.H.S.}$
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