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Find the value of $\frac{1}{tanA}+\frac{sinA}{( 1+cosA)}$, if $cosec A=2$.
Given: $cosec A=2$.
To do: To find the value of $\frac{1}{tanA}+\frac{sinA}{( 1+cosA)}$.
Solution:
$cosecA=2$
$\Rightarrow sinA = \frac{1}{2}$ ...... $( i)$
Therefore, $cosA=\sqrt{( 1-sin²A)}$
$\Rightarrow \sqrt{( 1-\frac{1}{4})}$
$\Rightarrow \sqrt{( \frac{3}{4})}$
$\Rightarrow \frac{\sqrt{3}}{2}$ ...... $( ii)$
So, $tanA=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$
$=( \frac{1}{2})\times ( \frac{2}{\sqrt{3}})$
$=\frac{1}{\sqrt{3}}$ ....... $( iii)$
Now in $\frac{1}{tanA}+\frac{sinA}{( 1+cosA)}$
By $( i),\ ( ii)$ & $( iii)$, put the value
$=\frac{1}{\frac{1}{\sqrt{3}}}+\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}$
$=\sqrt{3}+\frac{( \frac{1}{2})}{ ( \frac{2+\sqrt{3}}{2})}$
$= \sqrt{3}+\frac{1}{(2+\sqrt{3})}$
$= \frac{( 2\sqrt{3}+3+1)}{( 2+\sqrt{3})}$
$=2\frac{( \sqrt{3}+2)}{( \sqrt{3}+2)}$
$= 2$
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