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If $ \cos A=\frac{\sqrt{3}}{2} $, find the value of $ \frac{1}{\tan A}+\frac{\sin A}{1+\cos A} $
Given:
\( \cos A=\frac{\sqrt{3}}{2} \).
To do:
We have to find the value of \( \frac{1}{\tan A}+\frac{\sin A}{1+\cos A} \).
Solution:
Let, in a triangle $ABC$ right angled at $B$, $cos\ A = \frac{\sqrt3}{2}$.
We know that,
In a right-angled triangle $ABC$ with right angle at $B$,
By Pythagoras theorem,
$AC^2=AB^2+BC^2$
By trigonometric ratios definitions,
$sin\ A=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$
$cos\ A=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$
$tan\ A=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$
Here,
$AC^2=AB^2+BC^2$
$\Rightarrow (2)^2=(\sqrt3)^2+BC^2$
$\Rightarrow BC^2=4-3$
$\Rightarrow BC=\sqrt{1}=1$
Therefore,
$sin\ A=\frac{BC}{AC}=\frac{1}{2}$
$tan\ A=\frac{BC}{AB}=\frac{1}{\sqrt3}$This implies,
$\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=\frac{1}{\frac{1}{\sqrt3}}+\frac{\frac{1}{2}}{1+\frac{\sqrt3}{2}}$
$=\sqrt{3}+\frac{\frac{1}{2}}{\frac{2+\sqrt3}{2}}$
$=\frac{\sqrt3}{1}+\frac{1}{2+\sqrt3}$
$=\frac{\sqrt3(2+\sqrt3)+1(1)}{2+\sqrt3}$
$=\frac{2\sqrt3+3+1}{2+\sqrt3}$
$=\frac{2(2+\sqrt3)}{2+\sqrt3}$
$=2$
The value of \( \frac{1}{\tan A}+\frac{\sin A}{1+\cos A} \) is \( 2 \).