Question

If $x = \tan {15^ \circ }$ , $y = \cos ec{75^ \circ }$ and $z = 4\sin {18^ \circ }$
A. $x < y < z$
B. $y < z < x$
C. $z < x < y$
D. $x < z < y$

Answer 1

First, we will try to split the internal angles of trigonometric ratios given. For tan15 we can split the angle as tan(60-45) and then solve it. For cosec75, first, we will convert it into sine and after that, we will split the internal angle as $\left( {45 + 30} \right)$ to solve it. We can find the value of $\sin 18$ by assuming $A = 18$
To get $5A = 90$ . We can see this as $2A = 90 - 3A$ , then we will multiply sin to both sides and solve it to evaluate sin18. After that, we can compare the three terms given in the question to get to the final answer.

Complete step by step solution:
Let us take the terms one by one,
$x = \tan {15^ \circ }$

Put 15 as 60-45, we get
$\Rightarrow \tan {15^ \circ } = \tan {(60 - 45)^ \circ }$ … (1)

We know that,
$\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \times \tan B}}$ … (2)

Using (2) in (1), we get

$\Rightarrow \tan (60 - 45) = \dfrac{{\tan 60 - \tan 45}}{{1 + \tan 60 \times \tan 45}}$ … (3)

Put $\tan {60^ \circ } = \sqrt 3$ and $\tan {45^ \circ } = 1$ in (3), we get
$\Rightarrow \tan {15^ \circ } = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }}$

Hence,
$\Rightarrow x = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }}$
We know that $\sqrt 3 = 1.732$
$\Rightarrow x = \dfrac{{1.732 - 1}}{{1 + \sqrt 3 }}$
$\Rightarrow x = \dfrac{{0.732}}{{1 + \sqrt 3 }}$ … (4)

Now,
$\Rightarrow y = \cos ec{75^ \circ }$ … (5)

We know that
$\cos ecx = \dfrac{1}{{\sin x}}$ … (6)

Using (6) in (5), we get
$\Rightarrow \cos ec{75^ \circ } = \dfrac{1}{{\sin {{75}^ \circ }}}$

Put 75 as 45+30, we get
$\Rightarrow \dfrac{1}{{\sin {{75}^ \circ }}} = \dfrac{1}{{\sin {{(45 + 30)}^ \circ }}}$ … (7)

We know that,
$\sin (x + y) = (\sin x \times \cos y) + (\cos x \times \sin y)$ … (8)

Using (8) in (7), we get
$\Rightarrow \dfrac{1}{{\sin {{(45 + 30)}^ \circ }}} = \dfrac{1}{{\sin {{45}^ \circ }\cos {{30}^ \circ } + \cos {{45}^ \circ }\sin {{30}^ \circ }}}$

Put $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ , $\sin {30^ \circ } = \dfrac{1}{2}$ , $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ and $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$ , we get
$\Rightarrow \dfrac{1}{{\sin {{(45 + 30)}^ \circ }}} = \dfrac{1}{{\left( {\dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2}} \right) + \left( {\dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}} \right)}}$

Hence, after simplification, it will evaluate to
$\Rightarrow \cos ec{75^ \circ } = \dfrac{1}{{\sin {{75}^ \circ }}} = \dfrac{{2\sqrt 2 }}{{1 + \sqrt 3 }}$

Hence,
$\Rightarrow y = \dfrac{{2\sqrt 2 }}{{1 + \sqrt 3 }}$
We know that $\sqrt 2 = 1.414$
$\Rightarrow y = \dfrac{{2 \times 1.414}}{{1 + \sqrt 3 }}$
$\Rightarrow y = \dfrac{{2.828}}{{1 + \sqrt 3 }}$ … (9)

Now,
$\Rightarrow z = 4\sin {18^ \circ }$

Let, $A = {18^ \circ }$
$\Rightarrow 5 \times A = 5 \times {18^ \circ }$
$\Rightarrow 2 \times A = {90^ \circ } - 3 \times A$

Now put sin on both sides, we get
$\Rightarrow \sin 2A = \sin ({90^ \circ } - 3A)$

Since, $\sin (90 - \theta ) = \cos \theta$
$\Rightarrow \sin 2A = \cos 3A$ … (10)

We know that,
$\sin 2\theta = 2\sin \theta \cos \theta$ … (11)
$\cos 2\theta = 4{\cos ^3}\theta - 3\cos \theta$ … (12)
${\cos ^2}\theta = 1 - {\sin ^2}\theta$ … (13)

Using (11) and (12) in (10), we get
$\Rightarrow 2\sin A\cos A = 4{\cos ^3}A - 3\cos A$

Canceling out $\cos A$ form both sides, we get
$\Rightarrow 2\sin A = 4{\cos ^2}A - 3$

Using (13), we get
$\Rightarrow 2\sin A = 4 - 4{\sin ^2}A - 3$

Forming the quadratic equation, we get
$\Rightarrow 4{\sin ^2}A + 2\sin A - 1 = 0$
Let $\sin A = t$ , we get
$\Rightarrow 4{t^2} + 2t - 1 = 0$

This equation can be solved by using a quadratic formula which states that,
If a quadratic equation is given as $a{x^2} + bx + c = 0$ then the solutions for x are given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Now, solving the above equation using the quadratic formula, we get
$\Rightarrow t = \dfrac{{ - 2 \pm 2\sqrt 5 }}{8}$

Taking 2 common, we get
$\Rightarrow t = \dfrac{{ - 1 \pm 1\sqrt 5 }}{4}$

Since, $\sin {18^ \circ }$ fall in the first quadrant hence, $\sin {18^ \circ }$ can not be negative, therefore
$\Rightarrow t \ne \dfrac{{ - 1 - 1\sqrt 5 }}{4}$
Hence, one option is eliminated. Therefore, we get
$\Rightarrow t = \dfrac{{ - 1 + 1\sqrt 5 }}{4}$
Since $t = \sin A$ and $A = 18$ , we get
$\Rightarrow \sin {18^ \circ } = \dfrac{{ - 1 + \sqrt 5 }}{4}$ … (14)

we are given $z = 4\sin {18^ \circ }$ , hence
$\Rightarrow z = 4\dfrac{{( - 1 + \sqrt 5 )}}{4}$
$\Rightarrow z = - 1 + \sqrt 5$
Multiplying and dividing RHS by $1 + \sqrt 3$ , we get
$\Rightarrow z = \dfrac{{( - 1 + \sqrt 5 ) \times (1 + \sqrt 3 )}}{{1 + \sqrt 3 }}$
$\Rightarrow z = \dfrac{{\sqrt {15} - \sqrt 3 + \sqrt 5 - 1}}{{1 + \sqrt 3 }}$ … (15)
We know that, $\sqrt {15} \approx 3.87$ , $\sqrt 5 \approx 2.23$ , $\sqrt 3 \approx 1.73$
Hence, put the values in (15), we get
$\Rightarrow z = \dfrac{{3.87 - 1.73 + 2.23 - 1}}{{1 + \sqrt 3 }}$
$\Rightarrow z = \dfrac{{3.37}}{{1 + \sqrt 3 }}$ … (16)

Now, by comparing (4),(9), and (16), Since, denominators at RHS are the same, we can say that
$\Rightarrow \dfrac{{0.732}}{{1 + \sqrt 3 }} < \dfrac{{2.828}}{{1 + \sqrt 3 }} < \dfrac{{3.37}}{{1 + \sqrt 3 }}$ … (17)

Hence, from (17), it is clear that
$\Rightarrow x < y < z$

Hence, the final answer is A.

Note:
It is not advisable to derive each small step in question during the paper, in the above question we should remember the value of trigonometric ratio $\sin 18$ . Similarly, there are some more angles which should be remembered
(1) $\sin {36^ \circ } = \dfrac{{\sqrt {10 - 2\sqrt 5 } }}{4}$
(2) $\sin {54^ \circ } = \dfrac{{\sqrt 5 + 1}}{4}$
(3) $\sin {72^ \circ } = \dfrac{{\sqrt {10 + 2\sqrt 5 } }}{4}$