If $\sin A + {\left( {\sin A} \right)^2} = 1$, then the value of the expression \[\left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right]\] is
$
{\text{A}}{\text{. 1}} \\
{\text{B}}{\text{. }}\dfrac{1}{2} \\
{\text{C}}{\text{. 2}} \\
{\text{D}}{\text{. 3}} \\
$
Last updated date: 17th Mar 2023
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Answer
305.1k+ views
Hint: Here, we will be using the formula ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1$ in order to determine the values of \[{\left( {\cos A} \right)^2}\] and \[{\left( {\cos A} \right)^4}\] from the given equation which is $\sin A + {\left( {\sin A} \right)^2} = 1$ and then ultimately the expression whose value is required will appear as the LHS of the given equation.
Complete step-by-step answer:
Given, $
\sin A + {\left( {\sin A} \right)^2} = 1{\text{ }} \to {\text{(1)}} \\
\Rightarrow \sin A = 1 - {\left( {\sin A} \right)^2}{\text{ }} \to {\text{(2)}} \\
$
As we know that
$
{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1 \\
\Rightarrow {\left( {\cos \theta } \right)^2} = 1 - {\left( {\sin \theta } \right)^2}{\text{ }} \to {\text{(3)}} \\
$
Replacing the angle $\theta $ with angle $A$ in equation (3), we get
$ \Rightarrow {\left( {\cos A} \right)^2} = 1 - {\left( {\sin A} \right)^2}{\text{ }} \to {\text{(4)}}$
Clearly, the RHS of both the equations (2) and (4) are the same so the LHS of both the equations will also be equal.
\[ \Rightarrow \sin A = {\left( {\cos A} \right)^2}{\text{ }} \to {\text{(5)}}\]
So, the value of the expression \[\left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right]\] can be determined by little modification as under.
\[\left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right] = \left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^2} \times {{\left( {\cos A} \right)}^2}} \right]\]
Using equation (5), we get
\[
\Rightarrow \left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right] = \left[ {\sin A + \left( {\sin A} \right) \times \left( {\sin A} \right)} \right] \\
\Rightarrow \left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right] = \left[ {\sin A + {{\left( {\sin A} \right)}^2}} \right] \\
\]
Finally using the given equation (1), we get
\[ \Rightarrow \left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right] = 1\]
Therefore, the value of the expression \[\left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right]\] is 1.
Hence, option A is correct.
Note: In this particular problem, we obtained the value of \[{\left( {\cos A} \right)^2}\] in terms of \[\sin A\]using the given equation and some trigonometric formula. From there we represented the expression whose value is required in terms of \[{\left( {\cos A} \right)^2}\] which is ultimately converted in terms of \[\sin A\].
Complete step-by-step answer:
Given, $
\sin A + {\left( {\sin A} \right)^2} = 1{\text{ }} \to {\text{(1)}} \\
\Rightarrow \sin A = 1 - {\left( {\sin A} \right)^2}{\text{ }} \to {\text{(2)}} \\
$
As we know that
$
{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1 \\
\Rightarrow {\left( {\cos \theta } \right)^2} = 1 - {\left( {\sin \theta } \right)^2}{\text{ }} \to {\text{(3)}} \\
$
Replacing the angle $\theta $ with angle $A$ in equation (3), we get
$ \Rightarrow {\left( {\cos A} \right)^2} = 1 - {\left( {\sin A} \right)^2}{\text{ }} \to {\text{(4)}}$
Clearly, the RHS of both the equations (2) and (4) are the same so the LHS of both the equations will also be equal.
\[ \Rightarrow \sin A = {\left( {\cos A} \right)^2}{\text{ }} \to {\text{(5)}}\]
So, the value of the expression \[\left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right]\] can be determined by little modification as under.
\[\left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right] = \left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^2} \times {{\left( {\cos A} \right)}^2}} \right]\]
Using equation (5), we get
\[
\Rightarrow \left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right] = \left[ {\sin A + \left( {\sin A} \right) \times \left( {\sin A} \right)} \right] \\
\Rightarrow \left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right] = \left[ {\sin A + {{\left( {\sin A} \right)}^2}} \right] \\
\]
Finally using the given equation (1), we get
\[ \Rightarrow \left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right] = 1\]
Therefore, the value of the expression \[\left[ {{{\left( {\cos A} \right)}^2} + {{\left( {\cos A} \right)}^4}} \right]\] is 1.
Hence, option A is correct.
Note: In this particular problem, we obtained the value of \[{\left( {\cos A} \right)^2}\] in terms of \[\sin A\]using the given equation and some trigonometric formula. From there we represented the expression whose value is required in terms of \[{\left( {\cos A} \right)^2}\] which is ultimately converted in terms of \[\sin A\].
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