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.