Question

How do you find the exact value \[{\csc ^{ - 1}}\left( 2 \right)\] ?

Answer 1

Hint: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. Also, we need to know the trigonometric conditions to make easy calculations. We need to know the trigonometric table values to solve this question. We need to know the relation between \[\sin ,\cos ,\tan \] with \[\csc ,\sec ,\cot \] respectively.

Complete step-by-step answer:
The given problem is shown below,
 \[{\csc ^{ - 1}}\left( 2 \right) = ?\]
It also can be written as,
 \[{\csc ^{ - 1}}\left( {\dfrac{2}{1}} \right) = ?\]
We know that,
 \[\sin \theta = \dfrac{1}{{\csc \theta }}\]
By using this condition we get,
 \[{\csc ^{ - 1}}\left( {\dfrac{2}{1}} \right) = {\sin ^{ - 1}}\left( {\dfrac{1}{{\left( {\dfrac{2}{1}} \right)}}} \right)\]
So, we get
 \[{\csc ^{ - 1}}\left( {\dfrac{2}{1}} \right) = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \arcsin \left( {\dfrac{1}{2}} \right)\] \[ \to \left( 1 \right)\]
By using trigonometric table values, we know that
 \[
  \sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2} \\
  \dfrac{\pi }{6} = \arcsin \left( {\dfrac{1}{2}} \right) \;
 \]
Let’s substitute these value in the equation \[\left( 1 \right)\] , we get
 \[\left( 1 \right) \to {\csc ^{ - 1}}\left( {\dfrac{2}{1}} \right) = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \arcsin \left( {\dfrac{1}{2}} \right)\]
 \[{\csc ^{ - 1}}\left( 2 \right) = \arcsin \left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}\]
It \[\sin \theta \] is in the interval of \[\left( {0,2\pi } \right)\] , the above equation can also be written as,
 \[{\csc ^{ - 1}}\left( 2 \right) = \arcsin \left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}or\dfrac{{5\pi }}{6}\]
So, the final answer is,
 \[{\csc ^{ - 1}}\left( 2 \right) = \dfrac{\pi }{6}or\dfrac{{5\pi }}{6}\]
So, the correct answer is “$\dfrac{{5\pi }}{6}$”.

Note: Note that \[\sin \theta \] is the inverse form of \[\csc \theta \] \[\left( {\sin \theta = \dfrac{1}{{\csc \theta }}} \right)\] , \[\cos \theta \] is the inverse form of \[\sec \theta \] \[\left( {\cos \theta = \dfrac{1}{{\sec \theta }}} \right)\] , and \[\tan \theta \] is the inverse form of \[\cot \theta \] \[\left( {\tan \theta = \dfrac{1}{{\cot \theta }}} \right)\] . This relation will be used to solve these types of questions. Also, remember the trigonometric table values to find the final answer for these types of questions. Also, remember that the basic trigonometric formulae and conditions to make easy calculations. Also, these types of questions involve arithmetic operations like addition/ subtraction/ multiplication/ division. Note that if the denominator term is zero the answer will be infinity. The inverse form of infinity is equal to \[1\] . Also, note that if the fraction term is present in the denominator we can take the denominator of the fraction term as the numerator of the main term \[\left( {\dfrac{a}{{\left( {\dfrac{b}{c}} \right)}} = \dfrac{{ac}}{b}} \right)\] . Note that every whole number has the denominator as \[1\] .