Questions & Answers

Question

Answers

$\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$

Answer
Verified

- 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).

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).

×

Sorry!, This page is not available for now to bookmark.