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If $\cos ecx - \cot x = \dfrac{1}{3}$, where $x \ne 0$, then the value of ${\cos ^2}x - {\sin ^2}x$ is
A. $\dfrac{{16}}{{25}}$
B. $\dfrac{9}{{25}}$
C. $\dfrac{8}{{25}}$
D. $\dfrac{7}{{25}}$

Last updated date: 20th Jun 2024
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Hint: To solve this question, we will use some basic trigonometric identities and algebraic identities to evaluate the given expression. We have to remember $\cos e{c^2}x - {\cot ^2}x = 1$, also ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$

Complete step-by-step answer:
Given that,
$\cos ecx - \cot x = \dfrac{1}{3}$ …….. (i)
We know that,
$\cos e{c^2}x - {\cot ^2}x = 1$
Using the identity, ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, we will expand the L.H.S,
$ \Rightarrow \cos e{c^2}x - {\cot ^2}x = \left( {\cos ecx - \cot x} \right)\left( {\cos ecx + \cot x} \right)$
Put the value of $\cos ecx - \cot x = \dfrac{1}{3}$,
$ \Rightarrow \cos e{c^2}x - {\cot ^2}x = \dfrac{1}{3}\left( {\cos ecx + \cot ax} \right)$
Equating this L.H.S with R.H.S, we will get
$ \Rightarrow \dfrac{1}{3}\left( {\cos ecx + \cot x} \right) = 1$
$ \Rightarrow \cos ecx + \cot x = 3$ ……… (ii)
Adding equation (i) and (ii), we will get
$ \Rightarrow \cos ecx + \cot x + \cos ecx - \cot x = \dfrac{1}{3} + 3$
$ \Rightarrow 2\cos ecx = \dfrac{{10}}{3}$
$ \Rightarrow \cos ecx = \dfrac{5}{3}$
Putting this value in equation (ii), we will get
$ \Rightarrow \dfrac{5}{3} + \cot x = 3$
\[ \Rightarrow \cot x = 3 - \dfrac{5}{3}\]
\[ \Rightarrow \cot x = \dfrac{4}{3}\]
Now, we know that
$ \Rightarrow \sin x = \dfrac{1}{{\cos ecx}}$
Putting the value of cosec x, we will get
$ \Rightarrow \sin x = \dfrac{1}{{\dfrac{5}{3}}}$
$ \Rightarrow \sin x = \dfrac{3}{5}$
Similarly, we know that
\[ \Rightarrow \cot x = \dfrac{{\cos x}}{{\sin x}}\]
\[ \Rightarrow \cos x = \cot x\sin x\]
Again, putting the values of sin x and cot x, we will get
\[ \Rightarrow \cos x = \dfrac{4}{3} \times \dfrac{3}{5}\]
Solving this, we will get
\[ \Rightarrow \cos x = \dfrac{4}{5}\]
We have to find out the value of ${\cos ^2}x - {\sin ^2}x$
Putting the values of sin x and cos x, we will get
$ \Rightarrow {\left( {\dfrac{4}{5}} \right)^2} - {\left( {\dfrac{3}{5}} \right)^2}$
$ \Rightarrow \dfrac{{16}}{{25}} - \dfrac{9}{{25}}$
$ \Rightarrow \dfrac{7}{{25}}$
Hence, the value of ${\cos ^2}x - {\sin ^2}x$ is $\dfrac{7}{{25}}$

So, the correct answer is “Option D”.

Note: Whenever we are asked such types of questions, we have to remember the trigonometric ratios of cos x and sin x. First, we have to simplify the given expression in terms of sin x and cos x and then by solving it we will get their values. After that, we will put those values in the required expression and we will get the correct answer.