Question

How do you simplify the expression \[\dfrac{1-{{\cos }^{2}}t}{1+\cos t}\]?

Answer 1

Hint: To simplify the given expression we will need to use an algebraic expansion property. We will use the expansion of \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\] this is known as the subtraction of squares. Also, we should know the general solution of \[{{\cos }^{-1}}(a)\] is \[2n\pi \pm \theta \], where \[\theta \] is the solution in the principal range of cosine inverse function. We will use these to simplify the expression.

Complete step by step answer:
We are given the expression \[\dfrac{1-{{\cos }^{2}}t}{1+\cos t}\], we need to simplify this.
The numerator of the given expression is \[1-{{\cos }^{2}}t\]. Here the first term is 1, as we know that 1 is square of itself. We can also express the numerator of the expression as \[{{1}^{2}}-{{\cos }^{2}}t\]. This expression is of the form \[{{a}^{2}}-{{b}^{2}}\] it is known as subtraction of squares. We know that the expansion of this is \[(a+b)(a-b)\]. Using this expansion in the numerator of the expression we get
\[\Rightarrow \dfrac{1-{{\cos }^{2}}t}{1+\cos t}\]
\[\Rightarrow \dfrac{\left( 1+\cos t \right)\left( 1-\cos t \right)}{1+\cos t}\]
Canceling out \[\left( 1+\cos t \right)\] as a common factor of the numerator and denominator, we get
\[\Rightarrow \left( 1-\cos t \right)\]

Hence the simplified form of the given expression is \[\left( 1-\cos t \right)\].

Note: It should be noted that we can cancel out a factor from numerator and denominator only if it is non-zero. Hence to cancel out the factor \[\left( 1+\cos t \right)\] we must make sure that it is non-zero.
\[\Rightarrow 1+\cos t\ne 0\]
Subtracting 1 from both sides of the above expression, we get
\[\Rightarrow 1+\cos t-1\ne 0-1\]
\[\Rightarrow \cos t\ne -1\]
Taking \[{{\cos }^{-1}}\] of both sides of the above expression, we get
\[\Rightarrow t\ne {{\cos }^{-1}}\left( -1 \right)\]
We know that the general solution of \[{{\cos }^{-1}}(a)\] is \[2n\pi \pm \theta \], where \[\theta \] is the solution in the principal range of cosine inverse function. The principal solution for \[-1\] is \[\pi \], so \[\theta =\pi \].
Hence, we get \[t\ne 2n\pi \pm \pi \] here \[n\] is an integer.