Question

The value of ${\text{cosec}}{15^ \circ }$ is equal to
A) $2\sqrt 3 $
B) $\sqrt 6 $
C) $2\sqrt 6 $
D) $\sqrt 6 + \sqrt 2 $

Answer 1

Hint:
Calculate the value of $\sin {15^ \circ }$ because $\sin x = \dfrac{1}{{{\text{cosec }}x}}$. Write $\sin {15^ \circ } = \sin \left( {{{45}^ \circ } - {{30}^ \circ }} \right)$. Then, solve the value of $\sin {15^ \circ }$ using the identity $\sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y$Substitute the values \[\sin {30^ \circ } = \dfrac{1}{2},\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\] and then take reciprocal of the $\sin {15^ \circ }$.

Complete step by step solution:
We will first calculate the value of $\sin {15^ \circ }$ because $\sin x = \dfrac{1}{{{\text{cosec }}x}}$
We can write $\sin {15^ \circ } = \sin \left( {{{45}^ \circ } - {{30}^ \circ }} \right)$
We will calculate $\sin \left( {{{45}^ \circ } - {{30}^ \circ }} \right)$ using the formula, $\sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y$
On substituting $x = {45^ \circ },y = {30^ \circ }$in the formula, $\sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y$, we get,
$\sin \left( {{{45}^ \circ } - {{30}^ \circ }} \right) = \sin {45^ \circ }\cos {30^ \circ } - \cos {45^ \circ }\sin {30^ \circ }$
On substituting the values of \[\sin {30^ \circ } = \dfrac{1}{2},\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\], we have,
$
  \sin \left( {{{45}^ \circ } - {{30}^ \circ }} \right) = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\sqrt 3 }}{2}} \right) - \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{1}{2}} \right) \\
   = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} \\
$
Value of $\sin {15^ \circ }$ is $\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
Calculate the value of ${\text{cosec}}{15^ \circ }$by taking reciprocal of the terms on both sides.
$
  \sin {15^ \circ } = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} \\
  {\text{cosec}}{15^ \circ } = \dfrac{{2\sqrt 2 }}{{\sqrt 3 - 1}} \\
$
Rationalise the above expression by multiplying and dividing by $\sqrt 3 + 1$
$
  {\text{cosec}}{15^ \circ } = \dfrac{{2\sqrt 2 }}{{\sqrt 3 - 1}} \times \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} \\
  {\text{cosec}}{15^ \circ } = \dfrac{{2\sqrt 2 \left( {\sqrt 3 + 1} \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - {1^2}}} \\
  {\text{cosec}}{15^ \circ } = \dfrac{{2\sqrt 6 + 2\sqrt 2 }}{2} \\
  {\text{cosec}}{15^ \circ } = \dfrac{{2\left( {\sqrt 6 + \sqrt 2 } \right)}}{2} \\
  {\text{cosec}}{15^ \circ } = \sqrt 6 + \sqrt 2 \\
$

Hence, option D is correct.

Note:
The value of the $\sin \left( {x - y} \right)$ is equal to the $\sin x\cos y - \cos x\sin y$. Also the multiplication of the $\sin $function and the ${\text{cosec}}$ function is equal to 1,mathematically it is represented as $\sin x = \dfrac{1}{{{\text{cosec }}x}}$. The values \[\sin {30^ \circ } = \dfrac{1}{2},\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\] should be known.