Question

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

Answer 1

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.