Question

The value of the expression $\dfrac{\tan A}{1+\sec A}+\dfrac{1+\sec A}{\tan A}$ is equal to:

1. $2\sin A$
2. $2\cos A$
3. $2\csc A$
4. $2\sec A$

Answer 1

Here in this question we have been asked to solve the given expression $\dfrac{\tan A}{1+\sec A}+\dfrac{1+\sec A}{\tan A}$ . For answering this question we will use the following values $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$ and simplify the given expression.

Complete step by step answer:
Now considering from the question we have been asked to solve the given expression $\dfrac{\tan A}{1+\sec A}+\dfrac{1+\sec A}{\tan A}$ .
From the basic concepts of trigonometry we know that $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$ .
By using the above identities in the given expression we will have $\Rightarrow \dfrac{\dfrac{\sin A}{\cos A}}{1+\dfrac{1}{\cos A}}+\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{\sin A}{\cos A}}$ .
Now by further simplifying the given expression we will have
$\begin{aligned} & \Rightarrow \dfrac{\dfrac{\sin A}{\cos A}}{\dfrac{1+\cos A}{\cos A}}+\dfrac{\dfrac{1+\cos A}{\cos A}}{\dfrac{\sin A}{\cos A}} \\\ & \Rightarrow \dfrac{\sin A}{1+\cos A}+\dfrac{1+\cos A}{\sin A} \\\ & \Rightarrow \dfrac{{{\sin }^{2}}A+{{\left( 1+\cos A \right)}^{2}}}{\sin A\left( 1+\cos A \right)} \\\ \end{aligned}$ .
From the basic concepts of algebra we know that the valid identity of expansion is given as ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ .
Now by further expanding the given expression using the above identity we will have$\Rightarrow \dfrac{{{\sin }^{2}}A+1+{{\cos }^{2}}A+2\cos A}{\sin A\left( 1+\cos A \right)}$ .
From the basic concepts of trigonometry we know that the valid trigonometry identity is given as ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ .
Now we further simplify the given expression using the above trigonometry identity we will have $\Rightarrow \dfrac{2+2\cos A}{\sin A\left( 1+\cos A \right)}$ .
Now by simplifying this expression we will have $\Rightarrow \dfrac{2}{\sin A}$ .
From the basic concepts of trigonometry we know that $\csc A=\dfrac{1}{\sin A}$ .
Hence we can conclude that the value of the given expression is $2\csc A$ .

So, the correct answer is “Option 3”.

Note: During the process of answering questions of this type we should be sure with the concepts that we are going to apply in between the steps in order to simplify the given expression. We can also verify the expression $\dfrac{\tan A}{1+\sec A}+\dfrac{1+\sec A}{\tan A}=2\csc A$ by substituting a fixed value for $A$ . Let us take $A={{30}^{\circ }}$ then we will have
$\begin{aligned} & \Rightarrow \dfrac{\tan {{30}^{\circ }}}{1+\sec {{30}^{\circ }}}+\dfrac{1+\sec {{30}^{\circ }}}{\tan {{30}^{\circ }}} \\\ & \Rightarrow \dfrac{\dfrac{1}{\sqrt{3}}}{1+\dfrac{2}{\sqrt{3}}}+\dfrac{1+\dfrac{2}{\sqrt{3}}}{\dfrac{1}{\sqrt{3}}} \\\ & \Rightarrow \dfrac{1}{\sqrt{3}+2}+\dfrac{\sqrt{3}+2}{1} \\\ & \Rightarrow \dfrac{8+4\sqrt{3}}{\sqrt{3}+2} \\\ & \Rightarrow 4=2\left( 2 \right) \\\ \end{aligned}$ .
Now we can say that our evaluated expression is correct because $\csc {{30}^{\circ }}=2$ .