Answer
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Hint: Convert the given angles of the inverse trigonometric functions so that their value lies in their domain and range and then substitute these values into the given expression. After the simplification, we have the result required in the given problem.
Complete step-by-step answer:
Consider the given expression in the problem:
${\sin ^{ - 1}}(\sin 12) + {\cos ^{ - 1}}(\cos 12)$
We have to find the value of the given trigonometric expression.
We know that the principal value of the inverse of sine angle is given as:
\[{\sin ^{ - 1}} \in \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\], where the value of $x$ are \[x \in \left[ { - 1,1} \right]\].
Similarly, we know that the principal value of inverse of the cosine angle is given as:
${\cos ^{ - 1}}y \in \left[ {0,\pi } \right]$, where the value of $x$ are $y \in \left[ { - 1,1} \right]$
Now, in the term given to us is
${\sin ^{ - 1}}(\sin 12) \ne 12$, where $12 \notin \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$
Similarly,${\cos ^{ - 1}}(\cos 12) \ne 12$, where$12 \notin \left[ {0,\pi } \right]$
Therefore, we can rewrite the given expression as:
${\sin ^{ - 1}}\left( {\sin 12} \right) = {\sin ^{ - 1}}\left( {\sin \left( {12 - 4\pi } \right)} \right)$
Let us rewrite the given function by changing the angles.
${\sin ^{ - 1}}(\sin 12) + {\cos ^{ - 1}}(\cos 12) = {\sin ^{ - 1}}\left( {\sin \left( {12 - 4\pi } \right)} \right) + {\cos ^{ - 1}}\left( {\cos \left( {4\pi - 12} \right)} \right)$
Then this obtained expression can be given as:
\[{\sin ^{ - 1}}(\sin 12) + {\cos ^{ - 1}}(\cos 12) = (12 - 4\pi ) + (4\pi - 12)\]
Usual formulas in the above expression as:
$\sin (2n\pi - x) = - \sin (x);$
$\sin ( - x) = - \sin x$
\[{\sin ^{ - 1}}(\sin 12) + {\cos ^{ - 1}}(\cos 12) = 0\]
So, we have the conclusion that:
\[{\sin ^{ - 1}}(\sin 12) + {\cos ^{ - 1}}(\cos 12) = 0\]
Therefore, the option (a) is correct.
Note: The domain of the function consists of all possible values of the independent variable where the function is defined and the range is the set which is obtained by the substitution of all values of the domain into the function.
The principal value of the inverse trigonometric function is the least absolute value of the angle.
Complete step-by-step answer:
Consider the given expression in the problem:
${\sin ^{ - 1}}(\sin 12) + {\cos ^{ - 1}}(\cos 12)$
We have to find the value of the given trigonometric expression.
We know that the principal value of the inverse of sine angle is given as:
\[{\sin ^{ - 1}} \in \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\], where the value of $x$ are \[x \in \left[ { - 1,1} \right]\].
Similarly, we know that the principal value of inverse of the cosine angle is given as:
${\cos ^{ - 1}}y \in \left[ {0,\pi } \right]$, where the value of $x$ are $y \in \left[ { - 1,1} \right]$
Now, in the term given to us is
${\sin ^{ - 1}}(\sin 12) \ne 12$, where $12 \notin \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$
Similarly,${\cos ^{ - 1}}(\cos 12) \ne 12$, where$12 \notin \left[ {0,\pi } \right]$
Therefore, we can rewrite the given expression as:
${\sin ^{ - 1}}\left( {\sin 12} \right) = {\sin ^{ - 1}}\left( {\sin \left( {12 - 4\pi } \right)} \right)$
Let us rewrite the given function by changing the angles.
${\sin ^{ - 1}}(\sin 12) + {\cos ^{ - 1}}(\cos 12) = {\sin ^{ - 1}}\left( {\sin \left( {12 - 4\pi } \right)} \right) + {\cos ^{ - 1}}\left( {\cos \left( {4\pi - 12} \right)} \right)$
Then this obtained expression can be given as:
\[{\sin ^{ - 1}}(\sin 12) + {\cos ^{ - 1}}(\cos 12) = (12 - 4\pi ) + (4\pi - 12)\]
Usual formulas in the above expression as:
$\sin (2n\pi - x) = - \sin (x);$
$\sin ( - x) = - \sin x$
\[{\sin ^{ - 1}}(\sin 12) + {\cos ^{ - 1}}(\cos 12) = 0\]
So, we have the conclusion that:
\[{\sin ^{ - 1}}(\sin 12) + {\cos ^{ - 1}}(\cos 12) = 0\]
Therefore, the option (a) is correct.
Note: The domain of the function consists of all possible values of the independent variable where the function is defined and the range is the set which is obtained by the substitution of all values of the domain into the function.
The principal value of the inverse trigonometric function is the least absolute value of the angle.
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