Question

Given a trigonometric ratio as $\tan A=\dfrac{4}{3}$, then find other trigonometric ratios of angle A.

Answer 1

Hint: First, we should know that the above condition of trigonometry is only valid for the right-angled triangle. Then, before we construct the triangle, we need to calculate the hypotenuse value also which could be found by using the Pythagoras theorem which is given by ${{h}^{2}}={{p}^{2}}+{{b}^{2}}$. Then, by using that triangle, we can get the value of all remaining trigonometric functions.

Complete step-by-step solution:
In this question, we are supposed to find the other trigonometric ratios of angle A with the condition given as $\tan A=\dfrac{4}{3}$.
Firstly, we should know that the above condition of trigonometry is only valid for the right-angled triangle.
So, let us suppose the right-angled triangle ABC and tan are defined for the ratio of the perpendicular to the base of the triangle.
Now, before we construct the triangle, we need to calculate the hypotenuse value also which could be found by using the Pythagoras theorem.
So, let x be the proportioning ratio of the sides of the triangle which gives the values as:
Base(b)=3x and Perpendicular(p)=4x
Now, by using the Pythagoras theorem where h is hypotenuse as:
$\begin{align}
  & {{h}^{2}}={{p}^{2}}+{{b}^{2}} \\
 & \Rightarrow {{h}^{2}}={{\left( 4x \right)}^{2}}+{{\left( 3x \right)}^{2}} \\
 & \Rightarrow {{h}^{2}}=16{{x}^{2}}+9{{x}^{2}} \\
 & \Rightarrow {{h}^{2}}=25{{x}^{2}} \\
 & \Rightarrow h=\sqrt{25{{x}^{2}}} \\
 & \Rightarrow h=5x \\
\end{align}$
So, we get the hypotenuse as 5x.
Now, we can construct a right-angled triangle with the base as 3x, perpendicular as 4x, and hypotenuse as 5x as:

Now, by using the above figure and the rules of the trigonometry which gives the various trigonometric functions as:
$\sin A=\dfrac{p}{h}$
So, the value of perpendicular from the triangle ABC is 4x and the value of the hypotenuse is 5x.
The, we get the value of sin A as:
$\begin{align}
  & \sin A=\dfrac{4x}{5x} \\
 & \Rightarrow \sin A=\dfrac{4}{5} \\
\end{align}$
Similarly, by using another trigonometric rule as:
$\cos A=\dfrac{b}{h}$
So, the value of base from the triangle ABC is 3x and the value of the hypotenuse is 5x.
The, we get the value of cos A as:
$\begin{align}
  & \cos A=\dfrac{3x}{5x} \\
 & \Rightarrow \cos A=\dfrac{3}{5} \\
\end{align}$
Now, we know from trigonometry, that cosec A is reciprocal of sin A, so:
$\begin{align}
  & \cos ecA=\dfrac{1}{\sin A} \\
 & \Rightarrow \cos ecA=\dfrac{1}{\dfrac{4}{5}} \\
 & \Rightarrow \cos ecA=\dfrac{5}{4} \\
\end{align}$
Now, we know from trigonometry, that sec A is reciprocal of cos A, so:
$\begin{align}
  & secA=\dfrac{1}{\cos A} \\
 & \Rightarrow secA=\dfrac{1}{\dfrac{3}{5}} \\
 & \Rightarrow secA=\dfrac{5}{3} \\
\end{align}$
Now, we know from trigonometry, that cot A is reciprocal of tan A, so:
$\begin{align}
  & \cot A=\dfrac{1}{\tan A} \\
 & \Rightarrow \cot A=\dfrac{1}{\dfrac{4}{3}} \\
 & \Rightarrow \cot A=\dfrac{3}{4} \\
\end{align}$
Hence, all the trigonometric values with $\tan A=\dfrac{4}{3}$are as $\sin A=\dfrac{4}{5}$, $\cos A=\dfrac{3}{5}$, $\cos ecA=\dfrac{5}{4}$, $secA=\dfrac{5}{3}$and $\cot A=\dfrac{3}{4}$.

Note: Now, to solve these types of the questions we can use an alternative approach in which trigonometric identities are used. So, the following trigonometric identities are:
$\begin{align}
  & {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\
 & 1+{{\tan }^{2}}A={{\sec }^{2}}A \\
 & 1+{{\cot }^{2}}A=\cos e{{c}^{2}}A \\
\end{align}$
So, we can use these identities to get the other trigonometric functions.