Question

The expression $\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}$ reduces to:
(a) $\dfrac{1+\sin A}{\cos A}$
(b) $\dfrac{1-\sin A}{\cos A}$
(c) $\dfrac{1+\cos A}{\sin A}$
(d) $\dfrac{1+\cos A}{\cos A}$

Answer 1

Hint: Rationalize the denominator of the given expression and use the trigonometric identity given by ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ to simplify the given expression and get the reduced the form.

Complete step by step answer:

We have been provided with the expression, $\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}$. Let us assume that its reduced form is $E$. Therefore,
$E=\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}$
This can be written as:
$E=\dfrac{\tan A+(\sec A-1)}{\tan A-(\sec A-1)}$
Rationalizing the denominator we get,
$\begin{aligned} & E=\dfrac{\tan A+(\sec A-1)}{\tan A-(\sec A-1)}\times \dfrac{\tan A+(\sec A-1)}{\tan A+(\sec A-1)} \\\ & =\dfrac{{{\left[ \tan A+(\sec A-1) \right]}^{2}}}{\left( \tan A-(\sec A-1) \right)\left( \tan A+(\sec A-1) \right)} \\\ \end{aligned}$
Expanding the whole square in the numerator and using the algebraic identity $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ in the denominator, we get,

& E=\dfrac{{{\tan }^{2}}A+{{(\sec A-1)}^{2}}+2\tan A(\sec A-1)}{{{\tan }^{2}}A-{{(\sec A-1)}^{2}}} \\\ & =\dfrac{{{\tan }^{2}}A+{{\sec }^{2}}A+1-2\sec A+2\tan A(\sec A-1)}{{{\tan }^{2}}A-\left( {{\sec }^{2}}A+1-2\sec A \right)} \\\ & =\dfrac{({{\tan }^{2}}A+1)+{{\sec }^{2}}A-2\sec A+2\tan A(\sec A-1)}{{{\tan }^{2}}A-{{\sec }^{2}}A-1+2\sec A} \\\ \end{aligned}$$ Now, we know that, $1+{{\tan }^{2}}A={{\sec }^{2}}A$, applying this identity in the above expression we get, $\begin{aligned} & E=\dfrac{{{\sec }^{2}}A+{{\sec }^{2}}A-2\sec A+2\tan A(\sec A-1)}{-1-1+2\sec A} \\\ & =\dfrac{2{{\sec }^{2}}A-2\sec A+2\tan A(\sec A-1)}{-2+2\sec A} \\\ & =\dfrac{2\sec A(\sec A-1)+2\tan A(\sec A-1)}{2(\sec A-1)} \\\ \end{aligned}$ Cancelling the common term we get, $E=\dfrac{\sec A+\tan A}{1}$ Changing, $\sec A=\dfrac{1}{\cos A}\text{ and tanA=}\dfrac{\sin A}{\cos A}$ we have, $E=\dfrac{\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}}{1}$ Multiplying numerator and denominator by $\cos A$, we get, $E=\dfrac{1+\sin A}{\cos A}$ Hence, option (a) is the correct answer. Note: It is important to note that this expression can also be simplified by changing $\sec A=\dfrac{1}{\cos A}\text{ and tanA=}\dfrac{\sin A}{\cos A}$ at the initial step. If we are provided with the options just like in the above question, we can directly check the correct option by some value to angle A, like angle A can be substituted ${{0}^{\circ }}$. Remember that, do not substitute angle A equal to ${{90}^{\circ }}$ because at ${{90}^{\circ }}$ tangent and secant of the angle is undefined.