Question

How do you simplify \[\dfrac{1+\sec \theta }{\sec \theta }=\dfrac{{{\sin }^{2}}\theta }{1-\cos \theta }\] ?

Answer 1

Hint: So, before we start off with any trigonometric equations, we need to know some of the very basic equations beforehand. These equations are described as:
\[\begin{align}
  & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
 & \cot \theta =\dfrac{1}{\tan \theta } \\
\end{align}\]
We also need to remember some of the basic defined relations, which are:
\[\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \\
 & \text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1 \\
\end{align}\]
In the solution method, we will follow the simple steps using all the known relations and equations we have in mind. We first replace \[\sec \theta =\dfrac{1}{\cos \theta }\] , then take L.C.M on the numerator to solve the problem further step by step.

Complete step by step answer:
Now, starting off to solve the given problem using Method 1, we first replace \[\sec \theta =\dfrac{1}{\cos \theta }\] then proceeding we get,
\[\begin{align}
  & \dfrac{1+\dfrac{1}{\cos \theta }}{\dfrac{1}{\cos \theta }}=\dfrac{{{\sin }^{2}}\theta }{1-\cos \theta } \\
 & \Rightarrow \dfrac{\dfrac{1+\cos \theta }{\cos \theta }}{\dfrac{1}{\cos \theta }}=\dfrac{{{\sin }^{2}}\theta }{1-\cos \theta } \\
\end{align}\]
Now, multiplying both the numerator and the denominator with \[\cos \theta \] , we get,
\[\Rightarrow 1+\cos \theta =\dfrac{{{\sin }^{2}}\theta }{1-\cos \theta }\]
Now, cross multiplying both sides of the above equation and multiplying we get,
\[\begin{align}
  & \left( 1+\cos \theta \right)\left( 1-\cos \theta \right)={{\sin }^{2}}\theta \\
 & \Rightarrow 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \\
\end{align}\]
We now arrange the terms in order, by bringing \[{{\cos }^{2}}\theta \] on the right and side, we get
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
If we remember, we see that this was the first relation that we stated above in the list of the relations. Thus the given problem in simplifying turns out to be \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] .

Note: We can solve the problem using another method which will yield the same result, but it would be in the form of an equation containing P (perpendicular), B (base) and H (hypotenuse). Now replacing these terms with the corresponding angles, we get our desired result.