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# Prove that: $\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$  - Hint: In this given question we can solve the question by converting $\sec A$in the Left Hand Side (LHS) into $\dfrac{1}{\cos A}$. Then using the identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ in place of 1 we can prove that the Left Hand Side (LHS) is equal to the Right Hand Side (RHS).

Complete step-by-step solution -

In this question, we are asked to prove that the Left Hand Side (LHS) $\dfrac{1+\sec A}{\sec A}$ is equal to the Right Hand Side (RHS) $\dfrac{{{\sin }^{2}}A}{1-\cos A}$, that is $\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$.
Let us start the proving from the LHS and arrive at the RHS.
\begin{align} & LHS=\dfrac{1+\sec A}{\sec A} \\ & \\ \end{align}
$=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}$ (as $\sec A=\dfrac{1}{\cos A}$)
$=\dfrac{\cos A+1}{\cos A}\times \cos A$
$=\cos A+1.............(1.1)$
Multiplying $\left( 1-\cos A \right)$to both the numerator and the denominator in equation (1.1), we obtain

$LHS=\dfrac{\left( 1+\cos A \right)\left( 1-\cos A \right)}{\left( 1-\cos A \right)}$
$=\dfrac{1-{{\cos }^{2}}A}{1-\cos A}................(1.2)$
Now, we know that
\begin{align} & {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\ & \Rightarrow 1-{{\cos }^{2}}A={{\sin }^{2}}A................(1.3) \\ \end{align}
Putting the value of $1-{{\cos }^{2}}A$ as obtained in equation (1.3) in equation (1.2), we get
$LHS=\dfrac{1-{{\cos }^{2}}A}{1-\cos A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}=RHS$
Therefore, we arrive at the required condition of equalized Left Hand Side (LHS) and Right Hand Side (RHS). That is
$\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$.

Note: Note that in equation (1.1), we need to multiply and divide the expression by a term involving $\cos A$ so that the numerator and denominator become quadratic in cos term and therefore we can convert it in terms of sin terms by using equation (1.3).

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