In a , =90^ and =1 3 . Prove that: 1. + =1 2. - =0 - Vidyadip

Question

In a $\triangle\text{ABC},\angle\text{B}=90^\circ$ and $\tan\text{A}=\frac{1}{\sqrt{3}}.$ Prove that:

$\sin\text{A}\cdot\cos\text{C}+\cos\text{A}\cdot\sin\text{C}=1$
$\cos\text{A}\cdot\cos\text{C}-\sin\text{A}\cdot\sin\text{C}=0$

✓

Answer

$\tan\text{A}=\frac{1}{\sqrt{3}}$
Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$
Then, $\tan\text{A}=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac{1}{\sqrt{3}}$
Let BC = 1 and $\text{AB}=\sqrt{3}$
Then, by Pythagoras theoram,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$=\big(\sqrt{3}\big)^2+1^2=3+1=4$
$\Rightarrow\text{AC}=2$
Now,
$\sin\text{A}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac12$ and $\cos\text{A}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}}{2}$
$\sin\text{C}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}}{2}$ and $\cos\text{C}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{1}{2}$

$\text{L.H.S.}=\sin\text{A}\cos\text{C}+\cos\text{A}\sin\text{C}$

$=\frac12\times\frac12+\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}$

$=\frac14+\frac{3}{4}$

$=\frac44$

$=1$

$=\text{R.H.S.}$

$\text{L.H.S.}=\cos\text{A}\cos\text{C}-\sin\text{A}\sin\text{C}$

$=\frac{\sqrt{3}}{2}\times\frac12-\frac12\times\frac{\sqrt{3}}{2}$

$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

$=0$

$=\text{R.H.S.}$

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