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

In this problem we have to prove $\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{1}{\sec \theta -\tan \theta }$ by using ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$. In the given equations we will consider LHS and convert it in terms of $\sec \theta$ , $\tan \theta$ by taking $\cos \theta$ common from both numerator and denominator applying the know formulas $\sec \theta =\dfrac{1}{\cos \theta }$ ,$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . now the LHS part is converted in terms of $\sec \theta$ , $\tan \theta$. In the problem they have suggested to use the identity ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$ ,from this identity we will find the value of $\sec \theta +\tan \theta$ by using the know algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . now we can use the aloe of $\sec \theta +\tan \theta$ in the LHS part which is in terms of $\sec \theta$ , $\tan \theta$ and simplify the LHS part by doing L.C.M then we will get the required solution.

FORMULAS USED:
1. $\sec \theta =\dfrac{1}{\cos \theta }$
2. $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
3. ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$

Complete step by step solution:
Given that
$\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\dfrac{1}{\sec \theta -\tan \theta }$
Considering the LHS part
LHS $=\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}$
Now taking common $\cos \theta$ from both numerator and denominator, then we will get
$LHS=\dfrac{\cos \theta \left( \dfrac{\sin \theta }{\cos \theta }-\dfrac{\cos \theta }{\cos \theta }+\dfrac{1}{\cos \theta } \right)}{\cos \theta \left( \dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\cos \theta }-\dfrac{1}{\cos \theta } \right)}$
Now using the known formulas $\sec \theta =\dfrac{1}{\cos \theta }$,$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, then we will get
$\begin{aligned} & LHS=\dfrac{\cos \theta (\tan \theta -1+\sec \theta )}{\cos \theta (\tan \theta -1+\sec \theta )} \\\ & \Rightarrow LHS=\dfrac{\tan \theta +\sec \theta -1}{1-\left( \sec \theta -\tan \theta \right)} \\\ \end{aligned}$
Now the LHS parts are completely converted in terms of $\sec \theta$ , $\tan \theta$.
We have the identity,
$\begin{aligned} & {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \\\ & {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \\\ \end{aligned}$
Using the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the above equation, then we will get
$\Rightarrow \left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)=1$
From the above equation the value of $\sec \theta +\tan \theta$ can be written as
$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }$
Now substituting above value in the LHS part, then we will get
$\begin{aligned} & LHS=\dfrac{\sec \theta +\tan \theta -1}{1-\left( \sec \theta -\tan \theta \right)} \\\ & \Rightarrow LHS=\dfrac{\dfrac{1}{\sec \theta -\tan \theta }-1}{1-\left( \sec \theta -\tan \theta \right)} \\\ \end{aligned}$
Simplify the above part by taking L.C.M in numerator
$\begin{aligned} & LHS=\dfrac{1-(\sec \theta -\tan \theta )}{\left( \sec \theta -\tan \theta \right)\left( 1-\left( \sec \theta -\tan \theta \right) \right)} \\\ & \Rightarrow LHS=\dfrac{1}{\left( \sec \theta -\tan \theta \right)} \\\ & \therefore LHS=RHS \\\ \end{aligned}$
Hence proved.

Note: We have some other trigonometric identity’s that will help us in solving this type of numerical. They are
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
${{\cot }^{2}}\theta -{{\csc }^{2}}\theta =1$