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Find the principal value of ${\tan ^{ - 1}}\left[ {\sin \left( { - \dfrac{{{\pi }}}{2}} \right)} \right]$.

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Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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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 $\sin \left( { - \dfrac{{{\pi }}}{2}} \right)$ using the identity $sin(-x)=-sinx$. Then ,we can find the principal value.

Complete step-by-step answer:
The values of the sine and tangent functions are-

Function$0^o$$30^o$$45^o$$60^o$$90^o$
tan0\[\dfrac{1}{{\sqrt 3 }}\]1\[\sqrt 3 \]Not defined
sin0$\dfrac{1}{2}$$\dfrac{1}{{\sqrt 2 }}$$\dfrac{{\sqrt 3 }}{2}$1


In the given question we need to find the principal value of ${\tan ^{ - 1}}\left[ {\sin \left( { - \dfrac{{{\pi }}}{2}} \right)} \right]$. We know that for tangent function to be negative, the angle should be negative, that is less than $0^o$.
We know that $sin(-A) = -sinA$
$\sin \left( { - \dfrac{{{\pi }}}{2}} \right) = - \sin \left( {\dfrac{{{\pi }}}{2}} \right) = - 1\;$
So the expression can be simplified as-
${\tan ^{ - 1}}\left( {\sin \left( { - \dfrac{{{\pi }}}{2}} \right)} \right) = {\tan ^{ - 1}}\left( { - 1} \right)$
But we know that $tan45^o = -1$, that is $tan^{-1}(1) = 45^o$. So,
$tan^{-1}(-1) = -45^o$
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. In this case, the principal value of both tangent and sine function ranges from $-90^o$ to $90^o$. This is because there can be infinite values of any inverse trigonometric functions. While solving the questions with multiple types of trigonometric functions, we should start solving from the innermost function first and try to convert that function into a form of the outer function.