Question

How do you find the exact value of $ \tan \left( \arccos \left( \dfrac{3}{5} \right) \right) $

Answer 1

Hint: We will first recall the concept of the inverse trigonometric function and then we will use the geometrical method to find the exact value of $ \tan \left( \arccos \left( \dfrac{3}{5} \right) \right) $ . So, we will draw a right triangle which is making an angle $ \theta $ between base and the hypotenuse of the triangle. Then, let us take $ \theta =\arccos \left( \dfrac{3}{5} \right) $ , so the hypotenuse of triangle will be equal to 5 and will be equal to 3 because $ {{\cos }^{-1}}\left( \dfrac{3}{5} \right)=\theta \Rightarrow \cos \theta =\dfrac{3}{5} $ .

Complete step by step answer:
We know that the inverse trigonometric also know as arc functions which perform the inverse operations of the trigonometric functions.
We will use the geometrical method to solve the above question.
We know that if h is the hypotenuse, p is the perpendicular, and b is the base of a right angle triangle and $ \theta $ is the angle between base and the hypotenuse of the triangle then $ \tan \theta =\dfrac{p}{b} $ , $ \cos \theta =\dfrac{b}{h} $

So, in $ \tan \left( \arccos \left( \dfrac{3}{5} \right) \right) $ let us take $ \theta =\arccos \left( \dfrac{3}{5} \right) $ ,
So, we will find the value of $ \tan \theta $ to get the value of $ \tan \left( \arccos \left( \dfrac{3}{5} \right) \right) $ .
Also, we know that $ \arccos \left( \dfrac{3}{5} \right)={{\cos }^{-1}}\left( \dfrac{3}{5} \right)=\theta $
Now, we will take $ \cos $ on both the side of above expression, we will get:
 $ \cos \left( {{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right)=\cos \theta $
We know that $ \cos \left( {{\cos }^{-1}}\alpha \right)=\alpha $
So, we can write $ \cos \theta =\dfrac{3}{5} $
And, from triangle law of trigonometric function we know that $ \cos \theta =\dfrac{b}{h} $ , where $ \theta $ is the angle between base (b) and the hypotenuse (h) of the triangle.
So, from $ \cos \theta =\dfrac{3}{5} $ , we can say that b = 3, and h = 5.
From, Pythagoras Theorem, we know that if base and hypotenuse are known then, perpendicular is given by:
 $ p=\sqrt{{{h}^{2}}-{{b}^{2}}} $
So, $ p=\sqrt{{{5}^{2}}-{{3}^{2}}} $
= $ \sqrt{25-9} $
= $ \sqrt{16}=4 $
Now, we will again use the triangle rule to find the value of $ \tan \theta $ , as from triangle rule we know that $ \tan \theta =\dfrac{p}{b} $
And, we know that p = 4 and b = 3, so $ \tan \theta =\dfrac{4}{3} $ .
Since, we have assumed that $ \theta =\arccos \left( \dfrac{3}{5} \right) $ , so $ \tan \left( \arccos \left( \dfrac{3}{5} \right) \right)=\dfrac{4}{3} $ .
This is our required solution.

Note:
We can also solve the above question alternatively by using the identity $ {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 $
Since, we know that $ \sec \theta =\dfrac{1}{\cos \theta } $ , so we write $ \tan \theta =\sqrt{\dfrac{1}{{{\cos }^{2}}\theta }-1} $
So, when $ \theta =\arccos \left( \dfrac{3}{5} \right) $ , then $ \tan \left( \arccos \left( \dfrac{3}{5} \right) \right)=\sqrt{\dfrac{1}{{{\cos }^{2}}\left( \arccos \left( \dfrac{3}{5} \right) \right)}-1} $
Since, we know that $ {{\cos }^{2}}\left( \arccos \left( \alpha \right) \right)={{\alpha }^{2}} $ .
\[\Rightarrow \tan \left( \arccos \left( \dfrac{3}{5} \right) \right)=\sqrt{\dfrac{1}{{{\left( \dfrac{3}{5} \right)}^{2}}}-1}\]
\[\Rightarrow \tan \left( \arccos \left( \dfrac{3}{5} \right) \right)=\sqrt{\dfrac{{{5}^{2}}}{{{3}^{2}}}-1}\]
\[\Rightarrow \tan \left( \arccos \left( \dfrac{3}{5} \right) \right)=\sqrt{\dfrac{{{5}^{2}}-{{3}^{2}}}{{{3}^{2}}}}\]
\[\Rightarrow \tan \left( \arccos \left( \dfrac{3}{5} \right) \right)=\sqrt{\dfrac{{{4}^{2}}}{{{3}^{2}}}}=\dfrac{4}{3}\]