Question

If $4\tan \theta = 3$, evaluate$\left( {\dfrac{{4\sin \theta - \cos \theta + 1}}{{4\sin \theta + \cos \theta - 1}}} \right)$.

Answer 1

Hint: The given question is related to the concept of trigonometric functions. Here in this question, we have to evaluate a trigonometric function using the given value i.e.,$4\tan \theta = 3$. In order to start solving this question, we will draw a diagram of a triangle in which we will show the value of trigonometric ratio as we already know that tan is the ratio of perpendicular to base of right-angle triangle. By using Pythagoras theorem, we will get the value of the hypotenuse and then we will use the values of perpendicular, base and hypotenuse to find the values of sin, cos and tan.

Formula used: ${\left( {Hypotenuse} \right)^2} = {\left( {Perpendicular} \right)^2} + {\left( {Base} \right)^2}$

Complete step-by-step solution:
Given is$4\tan \theta = 3$
From the given, we find that$\tan \theta = \dfrac{3}{4}$
Let us assume that a right-angle triangle ABC is right-angled at B, with one of the acute angles$\theta $.
Since, we know that tan$\theta $is the ratio of perpendicular of $\theta $ to base of $\theta $ i.e.,$\tan \theta = \dfrac{{perpendicular}}{{base}}$
So, we can conclude that the perpendicular =$3$ and base =$4$. This gives us,

Using Pythagoras theorem, we will find the value of the hypotenuse.
$
   \Rightarrow {\left( {Hypotenuse} \right)^2} = {\left( {Perpendicular} \right)^2} + {\left( {Base} \right)^2} \\
   \Rightarrow A{C^2} = B{C^2} + A{B^2} \\
   \Rightarrow {x^2} = {4^2} + {3^2} \\
   \Rightarrow {x^2} = 16 + 9 \\
   \Rightarrow {x^2} = 25 \\
   \Rightarrow x = \pm 5 \\
 $
As length can never be negative so, we ignore $ - 5$. Thus, hypotenuse =$x = 5$. So,
$
 \Rightarrow \tan \theta = \dfrac{{AB}}{{AB}} = \dfrac{3}{4} \\
   \Rightarrow \sin \theta = \dfrac{{BC}}{{AC}} = \dfrac{3}{5} \\
   \Rightarrow \cos \theta = \dfrac{{AB}}{{AC}} = \dfrac{4}{5} \\
 $
Putting values in the given trigonometric function, we get,
$
   \Rightarrow \left( {\dfrac{{4\sin \theta - \cos \theta + 1}}{{4\sin \theta + \cos \theta - 1}}} \right) = \left( {\dfrac{{4 \times \dfrac{3}{5} - \dfrac{4}{5} + 1}}{{4 \times \dfrac{3}{5} + \dfrac{4}{5} - 1}}} \right) \\
   \Rightarrow \left( {\dfrac{{\dfrac{{12 - 4 + 5}}{5}}}{{\dfrac{{12 + 4 - 5}}{5}}}} \right) \\
   \Rightarrow \dfrac{{13}}{{11}} \\
 $

Therefore, the required answer is $\dfrac{{13}}{{11}}$.

Note: In this above question, we drew a figure of a triangle. Drawing a figure really helped to solve the question. While solving any trigonometric ratio related question, it is highly recommended to draw a diagram which will not only help in clearing the confusion but will also help in easily solving the question.