Question

How do you use the fundamentals identities to simplify \[\dfrac{{{\cos }^{2}}y}{1-\sin y}\]?

Answer 1

Hint: To solve this question, we should know the trigonometric identity relation between \[\sin y\And \cos y\] which states that \[{{\sin }^{2}}y+{{\cos }^{2}}y=1\]. We should also know the expansion formula for subtraction of squares, \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]. We will use these properties to simplify the given expression.

Complete step by step answer:
We know the trigonometric identity which states that \[{{\sin }^{2}}y+{{\cos }^{2}}y=1\]. Subtracting \[{{\sin }^{2}}y\] from both sides of the identity, it can be expressed as
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}y+{{\cos }^{2}}y-{{\sin }^{2}}y=1-{{\sin }^{2}}y \\
 & \Rightarrow {{\cos }^{2}}y=1-{{\sin }^{2}}y \\
\end{align}\]
We know that the 1 is square of itself, so we can write the above expression as \[{{\cos }^{2}}y={{(1)}^{2}}-{{\sin }^{2}}y\]. The RHS of this identity of the form \[{{a}^{2}}-{{b}^{2}}\], is called the subtraction of squares. We know the expansion formula for \[{{a}^{2}}-{{b}^{2}}\] is \[\left( a+b \right)\left( a-b \right)\]. Using this expansion formula for the RHS of the above identity, we get
\[\begin{align}
  & \Rightarrow {{\cos }^{2}}y={{(1)}^{2}}-{{\sin }^{2}}y \\
 & \Rightarrow {{\cos }^{2}}y=\left( 1-\sin y \right)\left( 1+\sin y \right) \\
\end{align}\]
We are given the given expression \[\dfrac{{{\cos }^{2}}y}{1-\sin y}\]. The numerator of the expression is \[{{\cos }^{2}}y\]. As we have already shown, \[{{\cos }^{2}}y\] can also be expressed as \[\left( 1-\sin y \right)\left( 1+\sin y \right)\]. Using this in the numerator of the given expression, we get
\[\begin{align}
  & \Rightarrow \dfrac{{{\cos }^{2}}y}{1-\sin y} \\
 & \Rightarrow \dfrac{\left( 1-\sin y \right)\left( 1+\sin y \right)}{\left( 1-\sin y \right)} \\
\end{align}\]
As the numerator and denominator of the above expression have a common factor \[\left( 1-\sin y \right)\], we can cancel it out.
\[\begin{align}
  & \Rightarrow \dfrac{\left( 1-\sin y \right)\left( 1+\sin y \right)}{\left( 1-\sin y \right)} \\
 & \Rightarrow \left( 1+\sin y \right) \\
\end{align}\]

Hence, the expression \[\dfrac{{{\cos }^{2}}y}{1-\sin y}\] can be written as \[1+\sin y\] in simplified form.

Note: It should be noted that we can apply the above simplification only when \[\sin y\ne 1\]. Because If the \[\sin y=1\], this means that the \[\sin y-1=0\], in this case, the common factor of the denominator and the numerator is zero. We cannot cancel out zero as a common factor. The graph of the expression will have a hole in the graph on the values for which \[\sin y-1=0\].