Question

Prove that \[\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{1}{\sec \theta -\tan \theta }\] using the identity \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta .\]

Answer 1

Hint: We are given \[\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}\] and we have to solve so that it becomes \[\dfrac{1}{\sec \theta -\tan \theta }.\] We will first start with the left-hand side then we will divide the numerator and denominator by \[\cos \theta .\] And then we will have the equation in terms of \[\sec \theta \] and \[\tan \theta .\] Once we have that we use \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\] to solve further we take the common term outside from the denominator. Once we simplify and cancel the common terms in the numerator and denominator, we will get our desired solution.

Complete step by step answer:
We are given that \[\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}\] we have to show that this is the same as \[\dfrac{1}{\sec \theta -\tan \theta }.\] This means we have to show that \[\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{1}{\sec \theta -\tan \theta }\] by using the identity \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta .\] To solve we should learn certain things. Firstly, \[\tan \theta \] is defined as \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }.\] And \[\sec \theta \] is required for \[\cos \theta .\]
\[\Rightarrow \sec \theta =\dfrac{1}{\cos \theta }\]
And also that since \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta ,\] so we get, \[1={{\sec }^{2}}\theta -{{\tan }^{2}}\theta .\]
Now, to solve the given problem, we will start by considering the left-hand side. So, we have
\[\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}\]
Now, we divide the numerator and denominator by \[\cos \theta \] so we get,
\[\Rightarrow \dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{\dfrac{\sin \theta -\cos \theta +1}{\cos \theta }}{\dfrac{\sin \theta +\cos \theta -1}{\cos \theta }}\]
On simplifying, we get,
\[\Rightarrow \dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{\dfrac{\sin \theta }{\cos \theta }-\dfrac{\cos \theta }{\cos \theta }+\dfrac{1}{\cos \theta }}{\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\cos \theta }-\dfrac{1}{\cos \theta }}\]
Now, as \[\dfrac{1}{\cos \theta }=\sec \theta ,\dfrac{\sin \theta }{\cos \theta }=\tan \theta ,\dfrac{\cos \theta }{\cos \theta }=1.\] So, we get,
\[\Rightarrow \dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{\tan \theta -1+\sec \theta }{\tan \theta +1-\sec \theta }\]
So,
\[\Rightarrow \dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{\tan \theta +\sec \theta -1}{\left( \tan \theta -\sec \theta +1 \right)}\]
Now as \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta ,\] so \[1={{\sec }^{2}}\theta -{{\tan }^{2}}\theta .\] Using this in the denominator, we get,
\[\Rightarrow \dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{\tan \theta +\sec \theta -1}{\left( \tan \theta -\sec \theta +\left( {{\sec }^{2}}\theta -{{\tan }^{2}}\theta \right) \right)}\]
Now as \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right),\] so,
\[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =\left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right)\]
So using this above, we get,
\[\Rightarrow \dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{\tan \theta +\sec \theta -1}{\left[ \left( \tan \theta -\sec \theta \right)+\left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right) \right]}\]
We take \[\sec \theta -\tan \theta \] common from the denominator, so we get,
\[\Rightarrow \dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{\tan \theta +\sec \theta -1}{\left( \sec \theta -\tan \theta \right)\left[ -1+\tan \theta +\sec \theta \right]}\]
Rearranging the terms, we get,
\[\Rightarrow \dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{\tan \theta +\sec \theta -1}{\left( \sec \theta -\tan \theta \right)\left[ \tan \theta +\sec \theta -1 \right]}\]
Cancelling the like terms in the numerator and denominator, we get,
\[\Rightarrow \dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{1}{\sec \theta -\tan \theta }\]
Hence proved.

Note:
Students need to remember that \[\tan \theta -\sec \theta \] can be written in terms of \[\sec \theta -\tan \theta .\] We take \[\tan \theta -\sec \theta \] and from this we take – 1 common, so we get,
\[\tan \theta -\sec \theta =-1\left( -\tan \theta +\sec \theta \right)\]
Rearranging the terms, we will get,
\[\Rightarrow \tan \theta -\sec \theta =-1\left( \sec \theta -\tan \theta \right)\]
So, we get,
\[\Rightarrow \tan \theta -\sec \theta =-\left( \sec \theta -\tan \theta \right)\]
We need to be very careful while simplifying or if we get missed in the initial state, it could lead to the wrong solution.