Question

How do you calculate \[\tan \left( \arccos \left( \dfrac{5}{13} \right) \right)\] ?

Answer 1

Hint: To solve \[\tan \left( \arccos \left( \dfrac{5}{13} \right) \right)\], we need to consider \[\arccos \left( \dfrac{5}{13} \right)\] as \[\alpha \]. We should take \[\cos \] on both the sides of the obtained function. Then use the trigonometric identity for inverse, that is, \[\cos \left( {{\cos }^{-1}}x \right)=x\] and find the value of \[\alpha \]. And then substitute these values in the given function and with the help of Pythagoras formula, we can find the value of the given function which is the required answer.

Complete step-by-step answer:
According to the question, we are asked to solve \[\tan \left( \arccos \left( \dfrac{5}{13} \right) \right)\].
We have been given the function is \[\tan \left( \arccos \left( \dfrac{5}{13} \right) \right)\] --------(1)
First, let us assume \[\arccos \left( \dfrac{5}{13} \right)=\alpha \].
We know that arccos x is nothing but the inverse of cosx.
Therefore, we get
\[co{{s}^{-1}}\left( \dfrac{5}{13} \right)=\alpha \]
Let us take cos on both the sides of α.
We get \[\cos \left( co{{s}^{-1}}\left( \dfrac{5}{13} \right) \right)=\cos \alpha \].
We know that \[\cos \left( {{\cos }^{-1}}x \right)=x\]. Using this identity in the above function, we get
\[\cos \alpha =\dfrac{5}{13}\]
We now have to find \[\tan \left( \alpha \right)\].
We know that, in a right-angled triangle, the ratio of base and hypotenuse is \[\cos \theta \].
Let us substitute these values in a right-angled triangle ABC.
Consider the base to be 5x and the hypotenuse to be 13x, where x is a constant.
That is, b=5x and h=13x

We have to find AB.
Let us use Pythagoras formula to find AB.
Here, \[{{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}\] by Pythagoras formula.
Substitute the values of BC and AC. We get,
\[{{\left( AB \right)}^{2}}+{{\left( 5x \right)}^{2}}={{\left( 13x \right)}^{2}}\]
On further simplification, we get
\[{{\left( AB \right)}^{2}}+25{{x}^{2}}=169{{x}^{2}}\]
\[\Rightarrow {{\left( AB \right)}^{2}}=169{{x}^{2}}-25{{x}^{2}}\]
Taking \[{{x}^{2}}\] common from the above expression, we get
\[{{\left( AB \right)}^{2}}=\left( 169-25 \right){{x}^{2}}\]
\[\Rightarrow {{\left( AB \right)}^{2}}=144{{x}^{2}}\]
To find AB, we have to take a square root on both the sides.
\[\sqrt{{{\left( AB \right)}^{2}}}=\sqrt{144{{x}^{2}}}\]
On further simplification, we get
\[\sqrt{{{\left( AB \right)}^{2}}}=\sqrt{144}\times \sqrt{{{x}^{2}}}\]
We know that \[{{\left( 12 \right)}^{2}}=144\].
\[\Rightarrow \sqrt{{{\left( AB \right)}^{2}}}=\sqrt{{{\left( 12 \right)}^{2}}}\times \sqrt{{{x}^{2}}}\]
Using the property \[\sqrt{{{a}^{2}}}=a\] in the above obtained expression, we get
\[AB=12\times x\]
\[\Rightarrow AB=12x\]
Now, we have to find \[\tan \left( \alpha \right)\].
We know that tan is the ratio of adjacent side and base.
Here, from the triangle ABC, the adjacent side is AB=12x and base=BC=5x.
Therefore, \[\tan \left( \alpha \right)=\dfrac{AB}{BC}\]
\[\Rightarrow \tan \left( \alpha \right)=\dfrac{12x}{5x}\]
Cancelling out the common term ‘x’, we get
\[\tan \left( \alpha \right)=\dfrac{12}{5}\]
But we have assumed \[\arccos \left( \dfrac{5}{13} \right)=\alpha \].
Therefore, \[\tan \left( \arccos \left( \dfrac{5}{13} \right) \right)=\dfrac{12}{5}\].

Note: For this type of question, we have to use trigonometric identities and Pythagoras theorem. We should not get confused with the term arc which means inverse. Also, we should be careful with the sign conventions.