Question

How do you find the exact value of \[\csc \left[ \arctan \left( -\dfrac{5}{12} \right) \right]\]?

Answer 1

Hint: Write \[\arctan \left( -\dfrac{5}{12} \right)={{\tan }^{-1}}\left( \dfrac{-5}{12} \right)\] and use the properties \[{{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x\] and \[\csc \left( -\theta \right)=-\csc \theta \] to simplify the given expression. Now, assume a right-angle triangle with its perpendicular as 5 units and base as 12 units. Apply the Pythagoras theorem given as: - \[{{h}^{2}}={{p}^{2}}+{{b}^{2}}\] to determine the hypotenuse. Here, p = perpendicular, b = base and h = hypotenuse. Convert \[{{\tan }^{-1}}\left( \dfrac{5}{12} \right)\] into \[{{\csc }^{-1}}\] function and then apply the formula: - \[\csc \left( {{\csc }^{-1}}x \right)=x\] to get the answer.

Complete step-by-step solution:
Here, we have been provided with the expression \[\csc \left[ \arctan \left( -\dfrac{5}{12} \right) \right]\] and we are asked to find its value. So, let us assume the value of this expression as ‘E’.
\[\Rightarrow E=\csc \left[ \arctan \left( -\dfrac{5}{12} \right) \right]\]
Here, \[\arctan \] function means inverse tangent function, so we have,
\[\Rightarrow E=\csc \left[ {{\tan }^{-1}}\left( \dfrac{-5}{12} \right) \right]\]
We know that \[{{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x\], we get,
\[\Rightarrow E=\csc \left[ -{{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right]\]
Now, using the property \[\csc \left( -\theta \right)=-\csc \theta \], we get,
\[\Rightarrow E=-\csc \left[ {{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right]\]
We know that \[\tan \theta \] = (perpendicular / base) = \[\dfrac{p}{b}\], so we have,
\[\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{p}{b} \right)\]
On comparing the above relation with \[{{\tan }^{-1}}\left( \dfrac{5}{12} \right)\], we have,
\[\Rightarrow \] p = 5 units and b = 12 units
So, applying the Pythagoras theorem given as: - where p = perpendicular, b = base and h = hypotenuse, we get,
\[\Rightarrow {{h}^{2}}={{p}^{2}}+{{b}^{2}}\]
Substituting the values of p and b, we have,
\[\begin{align}
  & \Rightarrow {{h}^{2}}={{5}^{2}}+{{12}^{2}} \\
 & \Rightarrow {{h}^{2}}=25+144 \\
 & \Rightarrow {{h}^{2}}=169 \\
\end{align}\]
Taking square root both the sides, we get,
\[\Rightarrow h=\sqrt{169}\]
\[\Rightarrow h=13\] units
Now, we know that \[\csc \theta \] = (hypotenuse / perpendicular) = \[\dfrac{h}{p}\], so we have,
\[\Rightarrow \theta ={{\csc }^{-1}}\left( \dfrac{h}{p} \right)\]
Therefore, converting \[{{\tan }^{-1}}\left( \dfrac{5}{12} \right)\] into \[{{\csc }^{-1}}\] function, we get,
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{5}{12} \right)={{\csc }^{-1}}\left( \dfrac{13}{5} \right)\]
Therefore, the given expression becomes,
\[\Rightarrow E=-\csc \left[ {{\csc }^{-1}}\left( \dfrac{13}{5} \right) \right]\]
Using the identity, \[{{\csc }^{-1}}\left( {{\csc }^{-1}}x \right)=x\], we get,
\[\Rightarrow E=-\dfrac{13}{5}\]
Hence, the value of the given expression is \[-\dfrac{13}{5}\].

Note: One may note that there is no option other than converting the given inverse tangent function into cosecant inverse function. We cannot convert it into any other inverse function of trigonometry because then we would not be able to apply the required formula. So, it is necessary to check which function is outside. Here, it was a cosecant function. You must remember the Pythagoras theorem to solve the question.