Question

Find the value of the following expression
\[\operatorname{cosec}\left\{ {{\cot }^{-1}}\left( -\dfrac{12}{5} \right) \right\}\]

Answer 1

Hint: We know that, \[{{\cot }^{-1}}\left( \dfrac{x}{y} \right)={{\operatorname{cosec}}^{-1}}\left( \dfrac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{y} \right)\], so we will convert \[{{\cot }^{-1}}\left( -\dfrac{12}{5} \right)\] in terms of \[{{\operatorname{cosec}}^{-1}}\lambda \] and simplify it by keeping in mind that \[{{\cot }^{-1}}\left( \dfrac{x}{y} \right)\] is defined in range of \[\left[ 0,\pi \right]\], so the sign of \[\left( \dfrac{x}{y} \right)\] is positive from \[\left[ 0,\dfrac{\pi }{2} \right]\] and negative from \[\left[ \dfrac{\pi }{2},\pi \right]\].

Complete step by step answer:

We have to evaluate \[\operatorname{cosec}\left\{ {{\cot }^{-1}}\left( -\dfrac{12}{5} \right) \right\}......\left( i \right)\]
To evaluate \[\operatorname{cosec}\left\{ {{\cot }^{-1}}\left( -\dfrac{12}{5} \right) \right\}\], first we will convert \[{{\cot }^{-1}}\left( -\dfrac{12}{5} \right)\] in terms of \[{{\operatorname{cosec}}^{-1}}\lambda \]. Now, let us consider \[\left( \dfrac{12}{5} \right)=\left( \dfrac{x}{y} \right)\]. Therefore, we can write \[{{\cot }^{-1}}\left( -\dfrac{12}{5} \right)={{\cot }^{-1}}\left( -\dfrac{x}{y} \right)......\left( ii \right)\]
As the sign of \[\left( \dfrac{x}{y} \right)\] is negative, so it will lie in the range of \[\left[ \dfrac{\pi }{2},\pi \right]\].
We know that, if \[\cot \theta =\dfrac{x}{y}\], then, we can write \[\cot \left( \pi -\theta \right)=-\dfrac{x}{y}\]. Therefore, we will get \[{{\cot }^{-1}}\left( \dfrac{x}{y} \right)=\theta \] and \[{{\cot }^{-1}}\left( -\dfrac{x}{y} \right)=\pi -\theta \]
From this we can conclude that, \[{{\cot }^{-1}}\left( -\dfrac{x}{y} \right)=\pi -{{\cot }^{-1}}\left( \dfrac{x}{y} \right)......\left( iii \right)\]
As we have assumed that \[\left( \dfrac{12}{5} \right)=\left( \dfrac{x}{y} \right)\] and from equation (ii) and (iii), we get that
\[{{\cot }^{-1}}\left( -\dfrac{12}{5} \right)=\pi -{{\cot }^{-1}}\left( \dfrac{12}{5} \right)\]
Now, we are putting the value of \[{{\cot }^{-1}}\left( -\dfrac{12}{5} \right)\] in (i), so we can write it as, \[\operatorname{cosec}\left\{ {{\cot }^{-1}}\left( -\dfrac{12}{5} \right) \right\}=\operatorname{cosec}\left\{ \pi -{{\cot }^{-1}}\left( \dfrac{12}{5} \right) \right\}......\left( iv \right)\]
As, we know that \[cosec\theta \] is positive in 1st as well as in 2nd quadrant that means positive in the domain of \[\left[ 0,\pi \right]\]. Therefore, we can write \[cosec\left( \pi -\theta \right)=cosec\left( \theta \right)\]
So, we can write equation (iv) as \[\operatorname{cosec}\left\{ \pi -{{\cot }^{-1}}\left( \dfrac{12}{5} \right) \right\}=\operatorname{cosec}\left\{ {{\cot }^{-1}}\left( \dfrac{12}{5} \right) \right\}......\left( v \right)\]
We know that, \[{{\cot }^{-1}}\left( \dfrac{x}{y} \right)={{\operatorname{cosec}}^{-1}}\left( \dfrac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{y} \right)\]
So, to simplify \[\operatorname{cosec}\left\{ {{\cot }^{-1}}\left( \dfrac{12}{5} \right) \right\}\], we will put \[{{\cot }^{-1}}\left( \dfrac{12}{5} \right)=cose{{c}^{-1}}\left( \dfrac{\sqrt{{{12}^{2}}+{{5}^{2}}}}{5} \right)\]
Now, after simplifying the above equation, we will get,
\[\Rightarrow {{\cot }^{-1}}\left( \dfrac{12}{5} \right)=cose{{c}^{-1}}\left( \dfrac{\sqrt{144+25}}{5} \right)\]
\[\Rightarrow {{\cot }^{-1}}\left( \dfrac{12}{5} \right)=cose{{c}^{-1}}\left( \dfrac{\sqrt{169}}{5} \right)\]
\[\Rightarrow {{\cot }^{-1}}\left( \dfrac{12}{5} \right)=cose{{c}^{-1}}\left( \dfrac{13}{5} \right)......\left( vi \right)\]
Now, we are putting the values of \[{{\cot }^{-1}}\left( \dfrac{12}{5} \right)\] from (vi) to (v). Therefore, we get \[\operatorname{cosec}\left\{ {{\cot }^{-1}}\left( \dfrac{12}{5} \right) \right\}=\operatorname{cosec}\left\{ cose{{c}^{-1}}\left( \dfrac{13}{5} \right) \right\}\]
\[\Rightarrow \operatorname{cosec}\left\{ {{\cot }^{-1}}\left( \dfrac{12}{5} \right) \right\}=\left( \dfrac{13}{5} \right)\]
Therefore, we conclude that on simplifying \[\operatorname{cosec}\left\{ {{\cot }^{-1}}\left( -\dfrac{12}{5} \right) \right\}\], we get \[\left( \dfrac{13}{5} \right)\] as an answer

Note: We can also convert \[\operatorname{cosec}\] in terms of \[\sin \] and \[\cot \] in terms of \[\tan \], if we don’t know the conversions from \[\cot \] to \[\operatorname{cosec}\]. It is necessary that we should know the domain of both the functions to get the answer correctly. The possible mistake one can commit while solving this question is not keeping in mind the negative sign of \[\left( -\dfrac{12}{5} \right)\], this might not change the answer but if we consider range and domain, then their values will be changed.