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# If ${\text{cosecA}} = 2$, find the value of $\dfrac{1}{{\tan {\text{A}}}} + \dfrac{{\sin {\text{A}}}}{{1 + \cos {\text{A}}}}$.

Last updated date: 16th Mar 2023
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Given, ${\text{cosecA}} = 2$
As we know that the sine trigonometric function is the reciprocal of cosecant trigonometric function. $\sin {\text{A}} = \dfrac{1}{{{\text{cosecA}}}} = \dfrac{1}{2} \Rightarrow {\text{A}} = {30^ \circ }$
$\therefore \tan {\text{A}} = \tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$ and $cos{\text{A}} = \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
i.e., $\dfrac{1}{{\tan {\text{A}}}} + \dfrac{{\sin {\text{A}}}}{{1 + \cos {\text{A}}}} = \dfrac{1}{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}} + \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{1 + \left( {\dfrac{{\sqrt 3 }}{2}} \right)}} = \sqrt 3 + \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{{2 + \sqrt 3 }}{2}} \right)}} = \sqrt 3 + \dfrac{1}{{2 + \sqrt 3 }} = \dfrac{{\left( {\sqrt 3 } \right)\left( {2 + \sqrt 3 } \right) + 1}}{{2 + \sqrt 3 }} = \dfrac{{2\sqrt 3 + 3 + 1}}{{2 + \sqrt 3 }} \\ \Rightarrow \dfrac{1}{{\tan {\text{A}}}} + \dfrac{{\sin {\text{A}}}}{{1 + \cos {\text{A}}}} = \dfrac{{2\sqrt 3 + 4}}{{2 + \sqrt 3 }} \\$