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How do you evaluate $\csc \left( {\dfrac{\pi }{6}} \right)$?

Last updated date: 01st Mar 2024
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IVSAT 2024
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Hint: Here we can proceed by finding the $\sin $ of the same angle as given and we know that $\csc x = \dfrac{1}{{\sin x}}$ and therefore we can divide both the values of the $\sin $ of the same angle and get the exact value of the $\csc \left( {\dfrac{\pi }{6}} \right)$.

Complete step by step solution:
Now we are given to find the exact value of $\csc \left( {\dfrac{\pi }{6}} \right)$
We know that:
$\sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}$$ - - - - (1)$
Now we can find the relation between $\sin ,\csc $ to get the value of the $\csc \left( {\dfrac{\pi }{6}} \right)$
Let us consider the triangle $ABC$ right-angled at $B$
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We know that:
$\sin \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}} - - - - (2)$
We also know that:
$\csc \theta = \dfrac{{{\text{hypotenuse}}}}{{{\text{perpendicular}}}} - - - - (3)$
Now if we multiply the equation (2) and (3) we will get:
\[\sin \theta .\csc \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}} \times \dfrac{{{\text{hypotenuse}}}}{{{\text{perpendicular}}}} = 1\]
Hence we get that:
\[\sin \theta .\csc \theta = 1\]$ - - - (4)$
Now substituting the value of $\sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}$ we got in equation (1) in the above equation (4), we get:
\[\sin \theta .\csc \theta = 1\]
  \sin \dfrac{\pi }{6}.\csc \dfrac{\pi }{6} = 1 \\
  \dfrac{1}{2}.\csc \dfrac{\pi }{6} = 1 \\
So we know the value of $\sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}$
So putting it in above, we get:

  \dfrac{1}{2}.\csc \dfrac{\pi }{6} = 1 \\
  \csc \dfrac{\pi }{6} = 2 \\

Hence for this, we must know all the trigonometric relations between all trigonometric functions because due to this all the general values of all trigonometric functions can be found.

Here in these types of problems where we are asked to find the value of the tangent or cotangent of any angle, we must know the basic values of the sine and cosine of the angles like $0^\circ,30^\circ,45^\circ,60^\circ,90^\circ $ and then we can easily calculate the same angles of the tangent, cotangent, secant, and cosecant of that same angle.
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