Question

If $\sin x + {\sin ^2}x = 1$ then the value of ${\cos ^2}x + {\cos ^4}x + {\cot ^4}x - {\cot ^2}x$ is?
A) \[1\]
B) \[0\]
C)\[2\]
D) None of these

Answer 1

Hint: The given question deals with finding the value of trigonometric expression doing basic simplification of trigonometric functions by using some of the trigonometric identities such as ${\sin ^2}x + {\cos ^2}x = 1$ and ${\cot ^2}x + 1 = \cos e{c^2}x$. We will solve the question in two parts and then add both the portions at the end to get to the final answer.

Complete step-by-step answer:
In the given problem we have to find the value of trigonometric expression given to us in the problem itself.
So, we have, ${\cos ^2}x + {\cos ^4}x + {\cot ^4}x - {\cot ^2}x$.
We are given that $\sin x + {\sin ^2}x = 1$.
Shifting the ${\sin ^2}x$ term to the right side of the equation, we get,
$ \Rightarrow \sin x = 1 - {\sin ^2}x$
Now we know the trigonometric identity ${\cos ^2}x = 1 - {\sin ^2}x$, we get,
$ \Rightarrow \sin x = {\cos ^2}x$
Now squaring both the sides of the equation. We get,
$ \Rightarrow {\sin ^2}x = {\cos ^4}x$
Substitute the value of ${\sin ^2}x$ by $1 - {\cos ^2}x$. We get,
$ \Rightarrow 1 - {\cos ^2}x = {\cos ^4}x$
Shifting the ${\cos ^2}x$ term to the right side of the equation. We get,
$ \Rightarrow 1 = {\cos ^2}x + {\cos ^4}x - - - - - \left( 1 \right)$
Again we have
$\sin x + {\sin ^2}x = 1$
Divide both sides of the equation by ${\sin ^2}x$.We get,
$ \Rightarrow \dfrac{{\sin x}}{{{{\sin }^2}x}} + \dfrac{{{{\sin }^2}x}}{{{{\sin }^2}x}} = \dfrac{1}{{{{\sin }^2}x}}$
$ \Rightarrow \cos ecx + 1 = \cos e{c^2}x$
Shifting 1 to the right side of the equation. We get,
$ \Rightarrow \cos ecx = \cos e{c^2}x - 1$
Now we know the trigonometric identity${\cot ^2}x = 1 - \cos e{c^2}x$. So, we get,
$ \Rightarrow \cos ecx = {\cot ^2}x$
Squaring both the sides of the equation. We get,
$ \Rightarrow \cos ec{x^2}x = {\cot ^4}x$
Substitute the value of $\cos e{c^2}x$ by $1 + {\cot ^2}x$. We get,
$ \Rightarrow 1 + {\cot ^2}x = {\cot ^4}x$
Shifting the ${\cot ^2}x$term to the right side of the equation. We get,
$ \Rightarrow {\cot ^4}x - {\cot ^2}x = 1 - - - - - \left( 2 \right)$
Now adding the equations \[1\] and \[2\]. We get,
$ \Rightarrow {\cos ^2}x + {\cos ^4}x + {\cot ^4}x - {\cot ^2}x = 1 + 1 = 2$
Therefore, the value of trigonometric expression ${\cos ^2}x + {\cos ^4}x + {\cot ^4}x - {\cot ^2}x$ is equal to \[2\].
So, the correct answer is “Option C”.

Note: There are six trigonometric ratios: $\sin \theta $, $\cos \theta $, $\tan \theta $, $\cos ec\theta $, $\sec \theta $and $\cot \theta $. Basic trigonometric identities include ${\sin ^2}\theta + {\cos ^2}\theta = 1$, ${\sec ^2}\theta = {\tan ^2}\theta + 1$ and $\cos e{c^2}\theta = {\cot ^2}\theta + 1$. These identities are of vital importance for solving any question involving trigonometric functions and identities.