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Question: If \(\cos ecx - \cot x = \dfrac{1}{3}\), where \(x \ne 0\), then the value of \({\cos ^2}x - {\sin ^...

If cos⁡ecx−cot⁡x=13\cos ecx - \cot x = \dfrac{1}{3}, where x≠0x \ne 0, then the value of cos⁡2x−sin⁡2x{\cos ^2}x - {\sin ^2}x is
A. 1625\dfrac{{16}}{{25}}
B. 925\dfrac{9}{{25}}
C. 825\dfrac{8}{{25}}
D. 725\dfrac{7}{{25}}

Explanation

Solution

To solve this question, we will use some basic trigonometric identities and algebraic identities to evaluate the given expression. We have to remember cos⁡ec2x−cot⁡2x=1\cos e{c^2}x - {\cot ^2}x = 1, also a2−b2=(a−b)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)

Complete step-by-step answer:
Given that,
cos⁡ecx−cot⁡x=13\cos ecx - \cot x = \dfrac{1}{3} …….. (i)
We know that,
cos⁡ec2x−cot⁡2x=1\cos e{c^2}x - {\cot ^2}x = 1
Using the identity, a2−b2=(a−b)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right), we will expand the L.H.S,
⇒cos⁡ec2x−cot⁡2x=(cos⁡ecx−cot⁡x)(cos⁡ecx+cot⁡x)\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=13\cos ecx - \cot x = \dfrac{1}{3},
⇒cos⁡ec2x−cot⁡2x=13(cos⁡ecx+cot⁡ax)\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
⇒13(cos⁡ecx+cot⁡x)=1\Rightarrow \dfrac{1}{3}\left( {\cos ecx + \cot x} \right) = 1
⇒cos⁡ecx+cot⁡x=3\Rightarrow \cos ecx + \cot x = 3 ……… (ii)
Adding equation (i) and (ii), we will get
⇒cos⁡ecx+cot⁡x+cos⁡ecx−cot⁡x=13+3\Rightarrow \cos ecx + \cot x + \cos ecx - \cot x = \dfrac{1}{3} + 3
⇒2cos⁡ecx=103\Rightarrow 2\cos ecx = \dfrac{{10}}{3}
⇒cos⁡ecx=53\Rightarrow \cos ecx = \dfrac{5}{3}
Putting this value in equation (ii), we will get
⇒53+cot⁡x=3\Rightarrow \dfrac{5}{3} + \cot x = 3
⇒cot⁡x=3−53\Rightarrow \cot x = 3 - \dfrac{5}{3}
⇒cot⁡x=43\Rightarrow \cot x = \dfrac{4}{3}
Now, we know that
⇒sin⁡x=1cos⁡ecx\Rightarrow \sin x = \dfrac{1}{{\cos ecx}}
Putting the value of cosec x, we will get
⇒sin⁡x=153\Rightarrow \sin x = \dfrac{1}{{\dfrac{5}{3}}}
⇒sin⁡x=35\Rightarrow \sin x = \dfrac{3}{5}
Similarly, we know that
⇒cot⁡x=cos⁡xsin⁡x\Rightarrow \cot x = \dfrac{{\cos x}}{{\sin x}}
⇒cos⁡x=cot⁡xsin⁡x\Rightarrow \cos x = \cot x\sin x
Again, putting the values of sin x and cot x, we will get
⇒cos⁡x=43×35\Rightarrow \cos x = \dfrac{4}{3} \times \dfrac{3}{5}
Solving this, we will get
⇒cos⁡x=45\Rightarrow \cos x = \dfrac{4}{5}
We have to find out the value of cos⁡2x−sin⁡2x{\cos ^2}x - {\sin ^2}x
So,
Putting the values of sin x and cos x, we will get
⇒(45)2−(35)2\Rightarrow {\left( {\dfrac{4}{5}} \right)^2} - {\left( {\dfrac{3}{5}} \right)^2}
⇒1625−925\Rightarrow \dfrac{{16}}{{25}} - \dfrac{9}{{25}}
⇒725\Rightarrow \dfrac{7}{{25}}
Hence, the value of cos⁡2x−sin⁡2x{\cos ^2}x - {\sin ^2}x is 725\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.