
Find the principal value of ${\tan ^{ - 1}}\left( { - 1} \right)$.
Answer
510k+ views
Hint: The principal value of an inverse trigonometric function at a point x is the value of the inverse function at the point x, which lies in the range of the principal branch. We have to find the principal value of $tan^{-1}(-1)$. We know that the principal value of $tan^{-1}(x)$ is given by $\left( { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right)$ . So, here first we have to find the value of $tan^{-1}(-1)$ using the identity $tan^{-1}(-x) = -(tan^{-1}(x))$. Then ,we can find the principal value.
Complete step-by-step answer:
The principal values of the trigonometric functions are-
In the given question we need to find the principal value of ${\tan ^{ - 1}}\left( { - 1} \right)$. We know that for tangent function to be negative, the angle should be negative, that is less than $0^o$.
We know that ${\tan ^{ - 1}}\left( 1 \right) = {45^o}$. This means that the value of $\tan {45^{\text{o}}} = 1$. So the value of the given expression can be calculated as-
$tan^{-1}(-1) = - tan^{-1}(1)$
${\tan ^{ - 1}}\left( { - 1} \right) = - \left( {{{45}^{\text{o}}}} \right) = - {45^{\text{o}}}$
We know that ${{\pi }}$ rad = $180^o$, so
${\tan ^{ - 1}}\left( { - 1} \right) = - \dfrac{{{\pi }}}{4}$
This is the required value.
Note: In such types of questions, we need to strictly follow the range of the principal values that have been specified. The principal value of tangent in the question should always lie between $-90^o$ and $90^o$. This is because there can be infinite values of any inverse trigonometric function. A common mistake is that students neglect the negative sign, often leading to the wrong answer.
Complete step-by-step answer:
The principal values of the trigonometric functions are-
Function | Principal Value |
$sin^{-1}$ | $\left[ { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right]$ |
$cos^{-1}$ | $\left[ {0,\;{{\pi }}} \right]$ |
$tan^{-1}$ | $\left( { - \dfrac{{{\pi }}}{2},\dfrac{{\;{{\pi }}}}{2}} \right)$ |
$cot^{-1}$ | $\left( {0,\;{{\pi }}} \right)$ |
$sec^{-1}$ | $\left[ {0,\;{{\pi }}} \right] - \left\{ {\dfrac{{{\pi }}}{2}} \right\}$ |
$cosec^{-1}$ | $\left[ { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right] - \left\{ 0 \right\}$ |
In the given question we need to find the principal value of ${\tan ^{ - 1}}\left( { - 1} \right)$. We know that for tangent function to be negative, the angle should be negative, that is less than $0^o$.
We know that ${\tan ^{ - 1}}\left( 1 \right) = {45^o}$. This means that the value of $\tan {45^{\text{o}}} = 1$. So the value of the given expression can be calculated as-
$tan^{-1}(-1) = - tan^{-1}(1)$
${\tan ^{ - 1}}\left( { - 1} \right) = - \left( {{{45}^{\text{o}}}} \right) = - {45^{\text{o}}}$
We know that ${{\pi }}$ rad = $180^o$, so
${\tan ^{ - 1}}\left( { - 1} \right) = - \dfrac{{{\pi }}}{4}$
This is the required value.
Note: In such types of questions, we need to strictly follow the range of the principal values that have been specified. The principal value of tangent in the question should always lie between $-90^o$ and $90^o$. This is because there can be infinite values of any inverse trigonometric function. A common mistake is that students neglect the negative sign, often leading to the wrong answer.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
The gas that burns in oxygen with a green flame is class 12 chemistry CBSE

Most of the Sinhalaspeaking people in Sri Lanka are class 12 social science CBSE

Give 10 examples of unisexual and bisexual flowers

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Write a short note on Franklands reaction class 12 chemistry CBSE
