If cosec A = 2, find the value of 1tan A+sin A1+cos A.

Given:

$\operatorname{cosec} A=2 \ldots \ldots$ (1)

To find:

$\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}$

Now we know $\operatorname{cosec} A$ is defined as below

$\operatorname{cosec} A=\frac{1}{\sin A}$

Therefore,

$\sin A=\frac{1}{\operatorname{cosec} A}$

Now by substituting the value of $\operatorname{cosec} A$ from equation (1)

We get,

$\sin A=\frac{1}{2}$

Now by substituting the value of $\sin A$ in the following identity of trigonometry

$\sin ^{2} A+\cos ^{2} A=1$

We get,

$\cos ^{2} A=1-\sin ^{2} A$

$=1-\left(\frac{1}{2}\right)^{2}$

$=1-\frac{1}{4}$

Now by taking L.C.M we get

$\cos ^{2} A=\frac{4-1}{4}$

$=\frac{3}{4}$

Now by taking square root on both sides

We get,

$\cos A=\sqrt{\frac{3}{4}}$

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

$=\frac{\sqrt{3}}{2}$

Therefore,

$\cos A=\frac{\sqrt{3}}{2}$....(3)

Now $\tan A$ is defined as follows

$\tan A=\frac{\sin A}{\cos A}$

Now by substituting the value of $\sin A$ and $\cos A$ from equation $(2)$ and $(3)$ respectively we get,

$\tan A=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$=\frac{1}{2} \times \frac{2}{\sqrt{3}}$

$=\frac{1}{\sqrt{3}}$

Therefore,

$\tan A=\frac{1}{\sqrt{3}}$....(4)

Now by substituting the value of $\sin A, \cos A$ and $\tan A$ from equation ( 2$),(3)$ and (4) respectively we get,

$\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=\frac{1}{\frac{1}{\sqrt{3}}}+\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}$

$=\frac{\sqrt{3}}{1}+\frac{1}{2\left(1+\frac{\sqrt{3}}{2}\right)}$

Now by taking L.C.M we get

$\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=\frac{\sqrt{3}}{1}+\frac{1}{2\left(\frac{2+\sqrt{3}}{2}\right)}$

Now 2 gets cancelled and we get

$\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=\frac{\sqrt{3}}{1}+\frac{1}{(2+\sqrt{3})}$

Now by taking L.C.M, we get,

$\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=\frac{\sqrt{3} \times(2+\sqrt{3})}{1 \times(2+\sqrt{3})}+\frac{1}{(2+\sqrt{3})}$

$=\frac{\sqrt{3} \times(2+\sqrt{3})+1}{(2+\sqrt{3})}$

Now by opening the brackets in the numerator

We get,

$\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=\frac{2 \sqrt{3}+\sqrt{3} \sqrt{3}+1}{(2+\sqrt{3})}$

Since $\sqrt{3} \sqrt{3}=3$

Therefore,

$\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=\frac{2 \sqrt{3}+3+1}{(2+\sqrt{3})}$

$=\frac{2 \sqrt{3}+4}{(2+\sqrt{3})}$

Now by taking 2 common

We get,

$\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=\frac{2(\sqrt{3}+2)}{(2+\sqrt{3})}$

$=\frac{2(2+\sqrt{3})}{(2+\sqrt{3})}$

Now as $(2+\sqrt{3})$ is present in both numerator as well as denominator, it gets cancelled

Therefore,

$\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=2$