Question

Prove the following
$\dfrac{\sin A-\cos A+1}{\sin A+\cos A-1}=\dfrac{1}{\sec A-\tan A}$

Answer 1

Hint:Consider the LHS of the equation given in the question. Now, replace 1 in the numerator or denominator by $\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)$ and use ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. Now, take all the common terms and cancel the like terms in the LHS to prove the desired result.

Complete step-by-step answer:
In this question, we have to prove that,
$\dfrac{\sin A-\cos A+1}{\sin A+\cos A-1}=\dfrac{1}{\sec A-\tan A}$
Let us consider the LHS of the given equation.
$LHS=\dfrac{\sin A-\cos A+1}{\sin A+\cos A-1}$
By dividing the numerator and denominator by cos A in the above expression, we get,
$LHS=\dfrac{\dfrac{\sin A}{\cos A}-\dfrac{\cos A}{\cos A}+\dfrac{1}{\cos A}}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\cos A}-\dfrac{1}{\cos A}}$
We know that, $\dfrac{\sin \theta }{\cos \theta }=\tan \theta$ and $\dfrac{1}{\cos \theta }=\sec \theta$. By using these in the above expression, we get,
$LHS=\dfrac{\tan A-1+\sec A}{\tan A+1-\sec A}$
We know that,
${{\sec }^{2}}A-{{\tan }^{2}}A=1$
So, by replacing 1 in the numerator of the above expression by ${{\sec }^{2}}A-{{\tan }^{2}}A$, we get,
$LHS=\dfrac{\left( \tan A+\sec A \right)-\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)}{\left( \tan A+1-\sec A \right)}$
We know that, ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. By using this in the numerator of the above expression, we get,
$LHS=\dfrac{\left( \sec A+\tan A \right)-\left[ \left( \sec A+\tan A \right)\left( \sec A-\tan A \right) \right]}{\left( \tan A+1-\sec A \right)}$
By taking out (sec A + tan A) common from the numerator in the above expression, we get,
$LHS=\dfrac{\left( \sec A+\tan A \right)\left[ 1-\left( \sec A-\tan A \right) \right]}{\left( \tan A+1-\sec A \right)}$
$LHS=\dfrac{\left( \sec A+\tan A \right)\left( 1-\sec A+\tan A \right)}{\left( 1-\sec A+\tan A \right)}$
Now, by canceling the like terms that is (1 – sec A + tan A) from the numerator and denominator of the above expression, we get,
$LHS=\left( \sec A+\tan A \right)$
By multiplying and diving (sec A – tan A) in the above expression, we get,
$LHS=\dfrac{\left( \sec A+\tan A \right)\left( \sec A-\tan A \right)}{\left( \sec A-\tan A \right)}$
By using $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get,
$LHS=\dfrac{{{\sec }^{2}}A-{{\tan }^{2}}A}{\sec A-\tan A}$
We know that, ${{\sec }^{2}}A-{{\tan }^{2}}A=1$. By using this, we get,
$LHS=\dfrac{1}{\sec A-\tan A}=RHS$
Hence proved

Note: In this question, students can easily prove the given equation by replacing 1 by ${{\sec }^{2}}A-{{\tan }^{2}}A$ in the denominator as well but students must keep in mind that we only need to replace 1 either in the numerator or the denominator but not at both the places in order to prove the required result.