Question

If we have a trigonometric expression as $\sin x={{\cos }^{2}}x$, then ${{\cos }^{2}}x\left( 1+{{\cos }^{2}}x \right)$ is equal to
a) 0
b) 1
c) 2
d) None of these

Answer 1

Hint: We are given a trigonometric equation as: ${{\cos }^{2}}x\left( 1+{{\cos }^{2}}x \right)$. Simplify the equation by opening the brackets and multiply the terms. We also have a relation, $\sin x={{\cos }^{2}}x$, get the value of ${{\sin }^{2}}x$. Since, both the expressions square of have sine and cosine, so try to make the expression ${{\cos }^{2}}x\left( 1+{{\cos }^{2}}x \right)$ in the form of identity: ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ and substitute the values to get a finite value.

Complete step-by-step solution:
We have the following expression: ${{\cos }^{2}}x\left( 1+{{\cos }^{2}}x \right)......(1)$
Now, expanding the equation (1), we get:
$\left( {{\cos }^{2}}x+{{\cos }^{4}}x \right)......(2)$
We also have a relation as: $\sin x={{\cos }^{2}}x......(3)$
Now, square equation (3), we get:
${{\sin }^{2}}x={{\cos }^{4}}x......(4)$
Now, substitute value of ${{\cos }^{4}}x$ from equation (4) in equation (2), we get:
$\Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x$
Since, ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
So, the value of ${{\cos }^{2}}x\left( 1+{{\cos }^{2}}x \right)$ is equal to 1.
Hence, option (b) is the correct answer.

Note: We can solve the expression ${{\cos }^{2}}x\left( 1+{{\cos }^{2}}x \right)......(1)$ by another way.
 As we have the relation: $\sin x={{\cos }^{2}}x......(2)$
Now, substitute the value from equation (2) in equation (1), we get:
$\sin x\left( 1+\sin x \right)......(3)$
Now, expanding the equation (3), we get:
$\left( \sin x+{{\sin }^{2}}x \right)......(4)$
From the relation in equation (2), we have: $\sin x={{\cos }^{2}}x$
Now, substitute the value of $\sin x$ in equation (4), we get:
$\Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x$
Since, ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
So, the value of ${{\cos }^{2}}x\left( 1+{{\cos }^{2}}x \right)$ is equal to 1.
Hence, option (b) is the correct answer.