Question

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

Answer 1

Hint: We solve this question by first considering the LHS of the given equation and then we consider the trigonometric formulas $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$. Then we use them to simplify it to write the equation in terms of sine and cosines. Then we multiply the numerator and denominator of the equation with $\sin A+1-\cos A$. Then we consider the formulas, $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ and ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Using them we can simplify the equation. Then we use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to expand the terms in the equation. Then we use the trigonometric identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ and simplify the equation. Then we factorise the numerator and denominator and cancel the terms that are common in both numerator and denominator to find the answer.

Complete step-by-step solution
We are asked to prove that $\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1+\sin A}{\cos A}$.
So, now let us consider the left-hand side of the above expression.
$\Rightarrow \dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}$
Now let us use the formula for trigonometric identities.
$\tan A=\dfrac{\sin A}{\cos A}$
$\sec A=\dfrac{1}{\cos A}$
Using them we can write the above expression as,
$\Rightarrow \dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{\dfrac{\sin A}{\cos A}+\dfrac{1}{\cos A}-1}{\dfrac{\sin A}{\cos A}-\dfrac{1}{\cos A}+1}$
$\Rightarrow \dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{\dfrac{\sin A+1-\cos A}{\cos A}}{\dfrac{\sin A-1+\cos A}{\cos A}}$
$\Rightarrow \dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{\sin A+1-\cos A}{\sin A-1+\cos A}................\left( 1 \right)$
Now, let us consider the expression $\dfrac{\sin A+1-\cos A}{\sin A-1+\cos A}$.
We can write it as,
$\Rightarrow \dfrac{\sin A+1-\cos A}{\sin A-\left( 1-\cos A \right)}$
Now let us multiply the numerator and the denominator with $\sin A+1-\cos A$.
$\begin{align}
  & \Rightarrow \dfrac{\sin A+1-\cos A}{\sin A-\left( 1-\cos A \right)}\times \dfrac{\sin A+1-\cos A}{\sin A+\left( 1-\cos A \right)} \\
 & \Rightarrow \dfrac{{{\left( \sin A+\left( 1-\cos A \right) \right)}^{2}}}{\left( \sin A-\left( 1-\cos A \right) \right)\times \left( \sin A+\left( 1-\cos A \right) \right)} \\
\end{align}$
Now let us consider the formulas,
$\begin{align}
  & \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} \\
 & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
\end{align}$
Using it we can convert the denominator and numerator and write it as,
$\Rightarrow \dfrac{{{\sin }^{2}}A+{{\left( 1-\cos A \right)}^{2}}+2\sin A\left( 1-\cos A \right)}{{{\sin }^{2}}A-{{\left( 1-\cos A \right)}^{2}}}$
Now let us consider the formula,
${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
Using it we can write the above equation as,
$\begin{align}
  & \Rightarrow \dfrac{{{\sin }^{2}}A+\left( 1+{{\cos }^{2}}A-2\cos A \right)+2\sin A-2\sin A\cos A}{{{\sin }^{2}}A-\left( 1+{{\cos }^{2}}A-2\cos A \right)} \\
 & \Rightarrow \dfrac{{{\sin }^{2}}A+1+{{\cos }^{2}}A-2\cos A+2\sin A-2\sin A\cos A}{{{\sin }^{2}}A-1-{{\cos }^{2}}A+2\cos A} \\
 & \Rightarrow \dfrac{1+{{\sin }^{2}}A+{{\cos }^{2}}A-2\cos A+2\sin A-2\sin A\cos A}{{{\sin }^{2}}A-1-{{\cos }^{2}}A+2\cos A} \\
\end{align}$
Now let us consider the formula,
$\begin{align}
  & {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\
 & {{\sin }^{2}}A-1=-{{\cos }^{2}}A \\
\end{align}$
Using it we can write the above equation as,
$\begin{align}
  & \Rightarrow \dfrac{1+1-2\cos A+2\sin A-2\sin A\cos A}{-{{\cos }^{2}}A-{{\cos }^{2}}A+2\cos A} \\
 & \Rightarrow \dfrac{2-2\cos A+2\sin A-2\sin A\cos A}{-2{{\cos }^{2}}A+2\cos A} \\
\end{align}$
Now we can take terms common in them and factorize them as,
\[\begin{align}
  & \Rightarrow \dfrac{2\left( 1-\cos A \right)+2\sin A\left( 1-\cos A \right)}{2\cos A\left( -\cos A+1 \right)} \\
 & \Rightarrow \dfrac{\left( 2+2\sin A \right)\left( 1-\cos A \right)}{2\cos A\left( 1-\cos A \right)} \\
 & \Rightarrow \dfrac{2\left( 1+\sin A \right)\left( 1-\cos A \right)}{2\cos A\left( 1-\cos A \right)} \\
\end{align}\]
Now let us cancel the term $2\left( 1-\cos A \right)$ from the numerator and denominator. Then we get,
\[\Rightarrow \dfrac{\left( 1+\sin A \right)}{\cos A}\]
So, we get that
$\Rightarrow \dfrac{\sin A+1-\cos A}{\sin A-1+\cos A}=\dfrac{1+\sin A}{\cos A}$
Substituting this value in equation (1) we get,
$\Rightarrow \dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1+\sin A}{\cos A}$
Hence Proved.

Note: We can also solve it in an alternative method. Instead of converting them into sine and cosine in the starting and simplifying it, we can first simplify it and then convert it into sine and cosine. We solve it similarly by multiplying the numerator and denominator with $\tan A+\sec A-1$. Then we simplify it using the identity, ${{\sec }^{2}}A-{{\tan }^{2}}A=1$ and follow a similar process as above and then convert the tan and secant functions into sin and cosine.