# 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

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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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