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If x=$\tan {15^ \circ }$ ,y=${\text{cosec}}{75^ \circ }$ and z=$4\sin {18^ \circ }$ then,
A) $x < y < z$
B) $y < z < x$
C) $z < x < y$
D) $x < z < y$

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Last updated date: 25th Jun 2024
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Answer
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Hint: Use the following trigonometric identities to get the values of x, y and z.
$\tan \left( {A + B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
$\sin \left( {A + B} \right) = \sin A\cos B + \operatorname{Cos} A\sin B$
And we know that $\sin {18^ \circ } = \dfrac{{\sqrt 5 - 1}}{4}$

Complete step-by-step answer:
 We are given that x=$\tan {15^ \circ }$ -- (i)
y=${\text{cosec}}{75^ \circ }$ --- (ii)
and z=$4\sin {18^ \circ }$--- (iii)
We can write x=$\tan {15^ \circ }$$ = \tan \left( {45 - 30} \right)$
And we know that
$\tan \left( {A + B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
On substituting the given values we get,
$ \Rightarrow $ x=$\dfrac{{\tan {{45}^ \circ } - \tan {{30}^ \circ }}}{{1 + \tan {{45}^ \circ }\tan {{30}^ \circ }}}$
We know that $\tan {45^ \circ } = 1$ and $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
On substituting these values in the equation we get,
$ \Rightarrow $ x=$\dfrac{{1 - \dfrac{1}{{\sqrt 3 }}}}{{1 + \dfrac{1}{{\sqrt 3 }}}}$
$ \Rightarrow $ x=$\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$
On rationalization we get,
$ \Rightarrow $ x= $\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}$
On solving the equation, we get-
$ \Rightarrow $ x=$\dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}$
We know that $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ and${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
On using the formulae we get
 $ \Rightarrow $ x=$\dfrac{{{{\left( {\sqrt 3 } \right)}^2} + 1 - 2 \times \sqrt 3 }}{{{{\left( {\sqrt 3 } \right)}^2} - 1}}$
On solving we get-
$ \Rightarrow $ x= $\dfrac{{3 + 1 - 2\sqrt 3 }}{{3 - 1}}$
$ \Rightarrow $ x=$\dfrac{{4 - 2\sqrt 3 }}{2} = \dfrac{{2\left( {2 - \sqrt 3 } \right)}}{2}$
$ \Rightarrow $ x=$2 - \sqrt 3 $
We know the value of $\sqrt 3 = 1.732$ then,
$ \Rightarrow $ x=$2 - 1.732 = 0.268$ --- (iv)
Now we know that ${\text{cosec}}\theta {\text{ = }}\dfrac{1}{{\sin \theta }}$ and $\sin \left( {A + B} \right) = \sin A\cos B + \operatorname{Cos} A\sin B$,
So we can write eq. (ii) as
$ \Rightarrow $ y=\[{\text{cosec7}}{5^ \circ } = \dfrac{1}{{\sin {{75}^ \circ }}}\]
$ \Rightarrow $ y=$\dfrac{1}{{\sin \left( {40 + 30} \right)}}$
On using the formulae we get,
$ \Rightarrow $ y=$\dfrac{1}{{\sin 45\cos 30 + \cos 45\sin 30}}$
We know that $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ ,$\cos 30 = \dfrac{{\sqrt 3 }}{2}$ ,$\sin 30 = \dfrac{1}{2}$ and $\cos 45 = \dfrac{1}{{\sqrt 2 }}$
On substituting these values we get,
$ \Rightarrow $ y=$\dfrac{1}{{\dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}}}$
On simplifying we get,
$ \Rightarrow $ y=$\dfrac{1}{{\dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }}}}$
On taking LCM,
$ \Rightarrow $ y=$\dfrac{1}{{\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}}}$
$ \Rightarrow $ y=$\dfrac{{2\sqrt 2 }}{{\sqrt 3 + 1}}$
On rationalization, we get-
$ \Rightarrow $ y=$\dfrac{{2\sqrt 2 }}{{\sqrt 3 + 1}} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}$ $ = \dfrac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - 1}}$
$ \Rightarrow $ y=$\dfrac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{3 - 1}}$
On further simplifying we get-
$ \Rightarrow $ y=$\dfrac{{2\sqrt 6 - 2}}{2} = \dfrac{{2\left( {\sqrt 6 - \sqrt 2 } \right)}}{2}$
$ \Rightarrow $ y=$\sqrt 6 - \sqrt 2 $
We know that $\sqrt 2 = 1.414$ and $\sqrt 6 = 2.449$. On substituting these values we get,
 $ \Rightarrow $ y=$2.449 - 1.414 = 1.035$ --- (v)
Now, in eq. (ii) on putting the value $\sin {18^ \circ } = \dfrac{{\sqrt 5 - 1}}{4}$, we get
z=$4\sin {18^ \circ }$=$4 \times \dfrac{{\sqrt 5 - 1}}{4} = \sqrt 5 - 1$
we know that $\sqrt 5 = 2.23$
On putting this value we get,
z=$2.23 - 1 = 1.23$ -- (vi)
From eq. (iv), (v), (vi), it is clear that $x < y < z$

Hence, option A is correct.

Note: We write $\tan {15^ \circ }$$ = \tan \left( {45 - 30} \right)$ because we can easily calculate using the values of $\tan {45^ \circ }$ and $\tan {30^ \circ }$ . Similarly we can easily calculate the value of y using the values of $\sin {45^ \circ }$ and $\sin {30^ \circ }$ as we know the values of $\sin {45^ \circ }$ and $\sin {30^ \circ }$.