Question

If $\tan A=\dfrac{5}{12}$, find the value of $\left( \sin A+\cos A \right)\sec A$.
(a) $\dfrac{12}{5}$
(b) $\dfrac{17}{12}$
(c) $\dfrac{7}{12}$
(d) none of these

Answer 1

Hint: We will simplify the given expression. For this we will use the fact that $\sec A=\dfrac{1}{\cos A}$. We will substitute this in the given expression and then obtain an expression that has the tangent function. Since we are given the value of $\tan A$, we will be able to calculate the value of the given expression after simplification.

Complete step by step answer:
Let us denote the given expression as $E=\left( \sin A+\cos A \right)\sec A$. Now, we know the fact that $\sec A=\dfrac{1}{\cos A}$. Substituting this definition of the secant function in the given expression, we get
$E=\left( \sin A+\cos A \right)\dfrac{1}{\cos A}$
Dividing each term in the expression separately by the cosine function, we get
$E=\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\cos A}$
We know that the first term in the above expression is the definition of the tangent function, that is $\tan A=\dfrac{\sin A}{\cos A}$ and also, we know that $\dfrac{\cos A}{\cos A}=1$. Substituting these values in the above expression, we get
$E=\tan A+1$
Now, we are given that $\tan A=\dfrac{5}{12}$. Substituting this value in the above expression we get the following,
$\begin{align}
  & E=\dfrac{5}{12}+1 \\
 & \Rightarrow E=\dfrac{5+12}{12} \\
 & \therefore E=\dfrac{17}{12} \\
\end{align}$

So, the correct answer is “Option b”.

Note: There is an alternate method to solve this question. We know that $\tan A=\dfrac{\sin A}{\cos A}$ and $\tan A=\dfrac{5}{12}$. Now, we can construct a right angled triangle and use the definitions of the trigonometric functions to obtain the values of lengths of the sides. For example, $\tan A=\dfrac{\text{Opposite}}{\text{Adjacent}}=\dfrac{5}{12}$. Now, we can use the Pythagoras theorem to find the length of the third side of the triangle. After that we can use the definitions of the trigonometric functions like $\sin A=\dfrac{\text{Opposite}}{\text{Hypotenuse}}$ and substitute these values in the given expression to find its value.