Question

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

Answer 1

Hint- Here, we will be using the concept of trigonometric functions and standard values in trigonometric tables.

Given, ${\text{cosecA}} = 2$
As we know that the sine trigonometric function is the reciprocal of cosecant trigonometric function. $\sin {\text{A}} = \dfrac{1}{{{\text{cosecA}}}} = \dfrac{1}{2} \Rightarrow {\text{A}} = {30^ \circ }$
i.e., angle A has a measure of 30 degrees.
According to the trigonometric table, we have
$\therefore \tan {\text{A}} = \tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$ and $cos{\text{A}} = \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Now substituting the values of trigonometric functions (sine, cosine and tangent) in the expression whose value is needed.
i.e., \[
  \dfrac{1}{{\tan {\text{A}}}} + \dfrac{{\sin {\text{A}}}}{{1 + \cos {\text{A}}}} = \dfrac{1}{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}} + \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{1 + \left( {\dfrac{{\sqrt 3 }}{2}} \right)}} = \sqrt 3 + \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{{2 + \sqrt 3 }}{2}} \right)}} = \sqrt 3 + \dfrac{1}{{2 + \sqrt 3 }} = \dfrac{{\left( {\sqrt 3 } \right)\left( {2 + \sqrt 3 } \right) + 1}}{{2 + \sqrt 3 }} = \dfrac{{2\sqrt 3 + 3 + 1}}{{2 + \sqrt 3 }} \\
   \Rightarrow \dfrac{1}{{\tan {\text{A}}}} + \dfrac{{\sin {\text{A}}}}{{1 + \cos {\text{A}}}} = \dfrac{{2\sqrt 3 + 4}}{{2 + \sqrt 3 }} \\
 \]

Note- In these types of problems, the given trigonometric function is converted into other trigonometric functions whose values are further needed to find the value of the unknown expression.