Question

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

Answer 1

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:

& \sec \theta =\dfrac{1}{\cos \theta } \\\ & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\\ & \cot \theta =\dfrac{1}{\tan \theta } \\\ \end{aligned}$$ We also need to remember some of the basic defined relations, which are: $$\begin{aligned} & {{\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{aligned}$$ In the solution method, we will follow the simple steps using all the known relations and equations we have in mind. We first replace $$\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{aligned} & \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{aligned}$$ Now, multiplying both the numerator and the denominator with $$\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{aligned} & \left( 1+\cos \theta \right)\left( 1-\cos \theta \right)={{\sin }^{2}}\theta \\\ & \Rightarrow 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \\\ \end{aligned}$$ We now arrange the terms in order, by bringing $${{\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.