Question

Consider the given expression $\sin x+{{\sin }^{2}}x=1$ , then the value of ${{\cos }^{2}}x+{{\cos }^{4}}x$ is
(a) 0
(b) 2
(c) 1
(d) 3

Answer 1

Hint: In our question, we have to convert the term ${{\cos }^{2}}x+{{\cos }^{4}}x$ in terms of sine function and then we will have use $\sin x+{{\sin }^{2}}x=1$ in the modified sine function with necessary changes.

Complete step-by-step answer:
In our question, we are given that $\sin x+{{\sin }^{2}}x=1$ .We are going to use this while solving $\left[ {{\cos }^{2}}x+{{\cos }^{4}}x \right]$ part. Let us consider this equation as equation (i):
$\sin x+{{\sin }^{2}}x=1................\left( i \right)$
Now we have to solve the term ${{\cos }^{2}}x+{{\cos }^{4}}x$ and evaluate its value. Let its value be equal to y. Thus, we get:
$y=~co{{s}^{2}}x+{{\cos }^{4}}x..................\left( ii \right)$
Now, we know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ .So from this equation we will write ${{\cos }^{2}}x$ in terms of ${{\sin }^{2}}x$ i.e.
${{\cos }^{2}}x=1-{{\sin }^{2}}x................\left( iv \right)$
We will put the value of ${{\cos }^{2}}x$ from (iv) to (ii). Thus after putting we get,
$\begin{aligned} & y={{\cos }^{2}}x+{{\left( {{\cos }^{2}}x \right)}^{2}} \\\ & y=\left( 1-{{\sin }^{2}}x \right)+{{\left( 1-{{\sin }^{2}}x \right)}^{2}} \\\ \end{aligned}$
We will now simplify the above equation. We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ . Here a = 1, $b=-{{\sin }^{2}}x$ , thus we will get:
$\begin{aligned} & \Rightarrow y=\left( 1-{{\sin }^{2}}x \right)+\left[ {{\left( 1 \right)}^{2}}+{{\left( -{{\sin }^{2}}x \right)}^{2}}+2\left( 1 \right)\left( -{{\sin }^{2}}x \right) \right] \\\ & \Rightarrow y=1-{{\sin }^{2}}x+\left[ 1+{{\sin }^{4}}x-2{{\sin }^{2}}x \right] \\\ & \Rightarrow y=1-{{\sin }^{2}}x+1+{{\sin }^{4}}x-2{{\sin }^{2}}x \\\ & \Rightarrow y=2-3{{\sin }^{2}}x+{{\sin }^{4}}x \\\ & \Rightarrow y=2-4{{\sin }^{2}}x+{{\sin }^{2}}x+{{\sin }^{4}}x \\\ & \Rightarrow y=2\left( 1-2{{\sin }^{2}}x \right)+\left( {{\sin }^{2}}x+{{\sin }^{4}}x \right)................\left( v \right) \\\ \end{aligned}$ Now, we know that $\sin x+{{\sin }^{2}}x=1$ . We will square both the sides by the use of identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ where, $a=\sin x$ and $b={{\sin }^{2}}x$ .Therefore, we get
$\begin{aligned} & {{\left( \sin x+{{\sin }^{2}}x \right)}^{2}}={{\left( 1 \right)}^{2}} \\\ & {{\left( \sin x \right)}^{2}}+{{\left( {{\sin }^{2}}x \right)}^{2}}+2\left( \sin x \right)\left( {{\sin }^{2}}x \right)={{\left( 1 \right)}^{2}} \\\ & {{\sin }^{2}}x+{{\sin }^{4}}x+2{{\sin }^{3}}x=1 \\\ \end{aligned}$
Now, we will rearrange the terms. After rearranging we get:
${{\sin }^{2}}x+{{\sin }^{4}}x=1-2{{\sin }^{3}}x...............\left( vi \right)$
Now, we will put the value of $\left( {{\sin }^{2}}x+{{\sin }^{4}}x \right)$ from equation (vi) into equation (v). After doing this, we get:
$\begin{aligned} & \Rightarrow y=2\left( 1-2{{\sin }^{2}}x \right)+1-2{{\sin }^{3}}x \\\ & \Rightarrow y=2-4{{\sin }^{2}}x+1-2{{\sin }^{3}}x \\\ & \Rightarrow y=3-4{{\sin }^{2}}x-2{{\sin }^{3}}x \\\ & \Rightarrow y=3-2{{\sin }^{2}}x-2{{\sin }^{2}}x-2{{\sin }^{3}}x \\\ & \Rightarrow y=3-2{{\sin }^{2}}x-2\left( {{\sin }^{2}}x+{{\sin }^{3}}x \right) \\\ & \Rightarrow y=3-2{{\sin }^{2}}x-2\sin x\left( \sin x+{{\sin }^{2}}x \right).............\left( vii \right) \\\ \end{aligned}$
We will put the value of $\left( \sin x+{{\sin }^{2}}x \right)$ from equation (i) to equation (vii). After doing this, we get:
$\begin{aligned} & \Rightarrow y=3-2{{\sin }^{2}}x-2\sin x\left( 1 \right) \\\ & \Rightarrow y=3-2{{\sin }^{2}}x-2\sin x \\\ & \Rightarrow y=3-2\left( {{\sin }^{2}}x+\sin x \right)..............\left( viii \right) \\\ \end{aligned}$
Again, we will put the value of $\left( \sin x+{{\sin }^{2}}x \right)$ from equation (i) to equation (viii). After doing this, we will get:
$\begin{aligned} & \Rightarrow y=3-2\left( 1 \right) \\\ & \Rightarrow y=3-2 \\\ & \Rightarrow y=1 \\\ \end{aligned}$
Hence, we get the value of ${{\cos }^{2}}x+{{\cos }^{4}}x=1$ .When we compare with the options given, we find that our answer matches option (c).
Hence option (c) is correct.

Note: The alternate method of solving this question is as follows:
We know that $\begin{aligned} & \sin x+{{\sin }^{2}}x=1 \\\ & \Rightarrow \sin x=1-{{\sin }^{2}}x \\\ & \Rightarrow \sin x={{\cos }^{2}}x \\\ \end{aligned}$
We will put this value in ${{\cos }^{2}}x+{{\cos }^{4}}x$ . After putting the value, we get:
$y={{\cos }^{2}}x+{{\cos }^{4}}x=\sin x+{{\left( \sin x \right)}^{2}}=1$
Hence, by this method also, the answer coming is 1.