Question

Which of the following formulas define functions equal to ${\cos ^2}2x$ wherever defined?
A) $\dfrac{1}{{1 - {{\tan }^2}(2x)}}$
B) ${\sin ^2}x \cdot {\cot ^2}x$
C) $\dfrac{1}{2}(1 + \cos 4x)$
D) $\dfrac{{{{\csc }^2}(2x) - 1}}{{{{\csc }^2}(2x)}}$

Answer 1

We have a trigonometric formula for the expression ${\cos ^2}x$. There is a relation connecting $\cos 2x$ and ${\cos ^2}x$. Using that we can find one equivalent of this expression from the options. Also there is another relation connecting ${\csc ^2}x$ and ${\cot ^2}x$. Using this we will get the next right option. This question has multiple right answers.

Formula used: For any angle $A$ we have,
$1 + \cos 2A = 2{\cos ^2}A$
${\csc ^2}A - {\cot ^2}A = 1$
${\cot ^2}A = \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}$
${\csc ^2}A = \dfrac{1}{{{{\sin }^2}A}}$

Complete step-by-step answer:
Given the expression ${\cos ^2}2x$.
We have to find its equivalent formula from the given options.
In Trigonometry we have,
For any angle $A$, $1 + \cos 2A = 2{\cos ^2}A$
So dividing both sides by $2$ we have,
$\Rightarrow$ ${\cos ^2}A = \dfrac{{1 + \cos 2A}}{2}$
Now put $A = 2x$
We have,
$\Rightarrow$ ${\cos ^2}2x = \dfrac{{1 + \cos (2 \times 2x)}}{2}$
Simplifying the equation we get,
$\Rightarrow$ ${\cos ^2}2x = \dfrac{{1 + \cos 4x}}{2} = \dfrac{1}{2}(1 + \cos 4x)$
This gives option C is right.
Moving on,
We have,
${\csc ^2}A - {\cot ^2}A = 1$
$\Rightarrow {\csc ^2}A - 1 = {\cot ^2}A$
So we can write,
$\Rightarrow$ ${\csc ^2}(2x) - 1 = {\cot ^2}2x$
$\Rightarrow$ $\dfrac{{{{\csc }^2}(2x) - 1}}{{{{\csc }^2}(2x)}} = \dfrac{{{{\cot }^2}2x}}{{{{\csc }^2}2x}}$
We know,
${\cot ^2}A = \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}$ and ${\csc ^2}A = \dfrac{1}{{{{\sin }^2}A}}$
$\Rightarrow$ $\dfrac{{{{\csc }^2}(2x) - 1}}{{{{\csc }^2}(2x)}} = \dfrac{{\dfrac{{{{\cos }^2}2x}}{{{{\sin }^2}2x}}}}{{\dfrac{1}{{{{\sin }^2}2x}}}}$
Cancelling ${\sin ^2}2x$ from numerator and denominator we have,
$\Rightarrow$ $\dfrac{{{{\csc }^2}(2x) - 1}}{{{{\csc }^2}(2x)}} = {\cos ^2}2x$
This gives option D is right.

$\therefore$ The answers are option C and option D.

Note: We have a similar formula for ${\sin ^2}A$.
For any angle $A$ we have,
$1 - \cos 2A = 2{\sin ^2}A$
Also we have $\cot A = \dfrac{{\cos A}}{{\sin A}}$.
So, ${\sin ^2}x \cdot {\cot ^2}x = {\sin ^2}x \times \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = {\cos ^2}x$
But we have to find an equivalent for ${\cos ^2}2x$. So this cannot be the answer. If we had $2x$ instead of $x$ in ${\sin ^2}x \cdot {\cot ^2}x$, it would be the right option.
Similarly we have another equation:
$1 + {\tan ^2}A = {\sec ^2}A$
So $\dfrac{1}{{1 + {{\tan }^2}(2x)}} = \dfrac{1}{{{{\sec }^2}2x}} = {\cos ^2}2x$
So in option A, If it was given $\dfrac{1}{{1 + {{\tan }^2}(2x)}}$ instead of $\dfrac{1}{{1 - {{\tan }^2}(2x)}}$, this option would also be true.