Question

If $\sec A=\dfrac{17}{8}$ , show that $\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}$.

Answer 1

Hint:For solving this question, we will use the formulas like ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$ , $\cos \theta =\dfrac{1}{\sec \theta }$ and ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ to calculate the value of ${{\tan }^{2}}A$ , ${{\cos }^{2}}A$ and ${{\sin }^{2}}A$ separately. After that, we will calculate the value of the term on the left-hand side and right-hand side separately and show that, if $\sec A=\dfrac{17}{8}$ , then $\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}$.

Complete step-by-step answer:
Given:
It is given that, if $\sec A=\dfrac{17}{8}$ and we have to show that $\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}$ .
Now, we will calculate the value of the term on the left-hand side and prove that it is equal to the value of the term on the right-hand side.
Now, before we proceed we should know the following formulas:
$\begin{align}
  & {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1................\left( 1 \right) \\
 & \cos \theta =\dfrac{1}{\sec \theta }.......................\left( 2 \right) \\
 & {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta ................\left( 3 \right) \\
\end{align}$
Now, we will use the above three formulas to calculate the value of ${{\tan }^{2}}A$ , ${{\cos }^{2}}A$ and ${{\sin }^{2}}A$ separately.
Calculation for ${{\tan }^{2}}A$ :
Now, we will use the formula from the equation (1) to write ${{\tan }^{2}}A={{\sec }^{2}}A-1$ and we can put $\sec A=\dfrac{17}{8}$ in it. Then,
$\begin{align}
  & {{\tan }^{2}}A={{\sec }^{2}}A-1 \\
 & \Rightarrow {{\tan }^{2}}A={{\left( \dfrac{17}{8} \right)}^{2}}-1 \\
 & \Rightarrow {{\tan }^{2}}A=\dfrac{289}{64}-1 \\
 & \Rightarrow {{\tan }^{2}}A=\dfrac{225}{64}................\left( 4 \right) \\
\end{align}$

Calculation for ${{\cos }^{2}}A$ :
Now, we will use the formula from the equation (2) to write $\cos A=\dfrac{1}{\sec A}$ and we can put $\sec A=\dfrac{17}{8}$ in it. Then,
$\begin{align}
  & \cos A=\dfrac{1}{\sec A} \\
 & \Rightarrow \cos A=\dfrac{8}{17} \\
 & \Rightarrow {{\cos }^{2}}A={{\left( \dfrac{8}{17} \right)}^{2}} \\
 & \Rightarrow {{\cos }^{2}}A=\dfrac{64}{289}................\left( 5 \right) \\
\end{align}$
Calculation for ${{\sin }^{2}}A$ :
Now, we will use the formula from the equation (3) to write ${{\sin }^{2}}A=1-{{\cos }^{2}}A$ and we can put ${{\cos }^{2}}A=\dfrac{64}{289}$ in it from equation (5). Then,
$\begin{align}
  & {{\sin }^{2}}A=1-{{\cos }^{2}}A \\
 & \Rightarrow {{\sin }^{2}}A=1-\dfrac{64}{289} \\
 & \Rightarrow {{\sin }^{2}}A=\dfrac{225}{289}................\left( 6 \right) \\
\end{align}$
Now, we will use the above results to calculate the value of the term on the left-hand side and right-hand side separately.
Calculation for $\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}$ :
Now, we will directly substitute ${{\sin }^{2}}A=\dfrac{225}{289}$ from equation (6) and ${{\cos }^{2}}A=\dfrac{64}{289}$ from equation (5) in the term $\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}$ . Then,
$\begin{align}
  & \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3} \\
 & \Rightarrow \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-4\times \dfrac{225}{289}}{4\times \dfrac{64}{289}-3} \\
\end{align}$
Now, we will multiply the numerator and denominator by $289$ in the above equation. Then,
$\begin{align}
  & \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-4\times \dfrac{225}{289}}{4\times \dfrac{64}{289}-3} \\
 & \Rightarrow \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3\times 289-4\times 225}{4\times 64-3\times 289} \\
 & \Rightarrow \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{867-900}{256-867} \\
 & \Rightarrow \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{-33}{-611} \\
 & \Rightarrow \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{33}{611}...................\left( 7 \right) \\
\end{align}$
Calculation for $\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}$ :
Now, we will directly substitute ${{\tan }^{2}}A=\dfrac{225}{64}$ from equation (4) in the term $\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}$ . Then,
$\begin{align}
  & \dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A} \\
 & \Rightarrow \dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}=\dfrac{3-\dfrac{225}{64}}{1-3\times \dfrac{225}{64}} \\
\end{align}$
Now, we will multiply the numerator and denominator by $64$ in the above equation. Then,
$\begin{align}
  & \dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}=\dfrac{3-\dfrac{225}{64}}{1-3\times \dfrac{225}{64}} \\
 & \Rightarrow \dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}=\dfrac{3\times 64-225}{64-3\times 225} \\
 & \Rightarrow \dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}=\dfrac{192-225}{64-675} \\
 & \Rightarrow \dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}=\dfrac{-33}{-611} \\
 & \Rightarrow \dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}=\dfrac{33}{611}...................\left( 8 \right) \\
\end{align}$
Now, we can equate the result of the equation (7) and (8). Then,
$\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}$
Thus, if $\sec A=\dfrac{17}{8}$ , then $\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}$ .
Hence, proved.

Note: Here, the student should apply every trigonometric formula correctly and proceed in a stepwise manner to prove the desired result easily. Moreover, we could have proved it without any given data also by multiplying the numerator and denominator of the term $\dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}$ by ${{\sec }^{2}}A$ as given below:
$\begin{align}
  & \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3} \\
 & \Rightarrow \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3{{\sec }^{2}}A-4{{\sec }^{2}}A{{\sin }^{2}}A}{4{{\sec }^{2}}A{{\cos }^{2}}A-3{{\sec }^{2}}A} \\
\end{align}$
Now, we can write ${{\sec }^{2}}A{{\sin }^{2}}A={{\tan }^{2}}A$ and ${{\sec }^{2}}A{{\cos }^{2}}A=1$ in the above equation. Then,
$\begin{align}
  & \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3{{\sec }^{2}}A-4{{\sec }^{2}}A{{\sin }^{2}}A}{4{{\sec }^{2}}A{{\cos }^{2}}A-3{{\sec }^{2}}A} \\
 & \Rightarrow \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3{{\sec }^{2}}A-4{{\tan }^{2}}A}{4-3{{\sec }^{2}}A} \\
 & \Rightarrow \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)-{{\tan }^{2}}A}{1-3\left( {{\sec }^{2}}A-1 \right)} \\
\end{align}$
Now, we will write ${{\sec }^{2}}A-{{\tan }^{2}}A=1$ and ${{\sec }^{2}}A-1={{\tan }^{2}}A$ in the above equation. Then,
$\begin{align}
  & \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)-{{\tan }^{2}}A}{1-3\left( {{\sec }^{2}}A-1 \right)} \\
 & \Rightarrow \dfrac{3-4{{\sin }^{2}}A}{4{{\cos }^{2}}A-3}=\dfrac{3-{{\tan }^{2}}A}{1-3{{\tan }^{2}}A} \\
\end{align}$