Answer
Verified
460.8k+ views
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 ${\cos ^2}x - {\sin ^2}x$
So,
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.
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 ${\cos ^2}x - {\sin ^2}x$
So,
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.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE