# 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 $2$.

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Updated on: 10-Oct-2022

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