Question

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

Answer 1

Hint: In order to prove this relation, we need to have some basic knowledge of the trigonometric identities like, ${{\tan }^{2}}A+1={{\sec }^{2}}A$ and some trigonometric ratios transformations like $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$. We should also know a few algebraic identities like, ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. By using these identities, we can prove the required relation.
Complete step-by-step answer:
In this question, we are asked to prove that, $\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1+\sin A}{\cos A}$. For the same, we will first consider the left hand side or the LHS of the equation. So, we can write it as, $LHS=\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}$
Now, we know that ${{\tan }^{2}}A+1={{\sec }^{2}}A$, so we will write $1={{\sec }^{2}}A-{{\tan }^{2}}A$. So, we get the LHS as,
$LHS=\dfrac{\left( \tan A+\sec A \right)-\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)}{\left( \tan A-\sec A+1 \right)}$
Now, we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. So, we can write ${{\sec }^{2}}A-{{\tan }^{2}}A$ as $\left( \sec A-\tan A \right)\left( \sec A+\tan A \right)$. Therefore, we can write the LHS as follows,
$LHS=\dfrac{\left( \tan A+\sec A \right)-\left( \sec A-\tan A \right)\left( \sec A+\tan A \right)}{\left( \tan A-\sec A+1 \right)}$
We can see that $\left( \tan A+\sec A \right)$ is common in both the terms of the numerator of LHS, so we get,
 $LHS=\dfrac{\left( \tan A+\sec A \right)\left[ 1-\left( \sec A-\tan A \right) \right]}{\left( \tan A-\sec A+1 \right)}$
And after simplifying further, we get,
$\begin{align}
  & LHS=\dfrac{\left( \tan A+\sec A \right)\left[ 1-\sec A+\tan A \right]}{\left( \tan A-\sec A+1 \right)} \\
 & \Rightarrow LHS=\dfrac{\left( \tan A+\sec A \right)\left[ \tan A-\sec A+1 \right]}{\left( \tan A-\sec A+1 \right)} \\
\end{align}$
We can see here that $\left( \tan A-\sec A+1 \right)$ is common in both the numerator and denominator, so cancelling these common terms, we get,
$LHS=\tan A+\sec A$
We also know that $\tan A$ and $\sec A$ can be written in terms of $\sin A$ and $\cos A$. So, $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$. By applying these identities, we get,
$LHS=\dfrac{\sin A}{\cos A}+\dfrac{1}{\cos A}$
Taking the LCM of the terms of the LHS, we get,
$LHS=\dfrac{\sin A+1}{\cos A}$, which is the same as, $LHS=\dfrac{1+\sin A}{\cos A}$
We know the right hand side or the RHS of the given statement is $RHS=\dfrac{1+\sin A}{\cos A}$. So, we can say, $LHS=RHS$
Hence, we have proved that, $\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1+\sin A}{\cos A}$.

Note: In this question, one can think of converting $\tan A$ and $\sec A$in terms of $\sin A$ and $\cos A$ to prove the expression or we can think of rationalizing the denominator, but that will take us to a point where we may get stuck and confused. So, it is better to write $1={{\sec }^{2}}A-{{\tan }^{2}}A$, by using the property ${{\tan }^{2}}A+1={{\sec }^{2}}A$ in the LHS and then applying ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ and simplifying to get the desired result.