Question

If $a\sec \theta +b\tan \theta =1$ and $a\sec \theta -b\tan \theta =5$, then the value of ${{a}^{2}}\left( {{b}^{2}}+4 \right)$ is
[a] $3{{b}^{2}}$
[b] $9{{b}^{2}}$
[c] ${{b}^{2}}$
[d] $4{{b}^{2}}$

Answer 1

Hint:Solve the two systems of the equation using any of the known methods like elimination method, substitution method. Hence find the value of $\sec \theta $ and $\tan \theta $. Use the trigonometric identity ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ to get the relation between a and b. Hence find the value of ${{a}^{2}}\left( {{b}^{2}}+4 \right)$

Complete step by step answer:
We have
$\begin{align}
  & a\sec \theta +b\tan \theta =1\text{ }\left( i \right) \\
 & a\sec \theta -b\tan \theta =5\text{ }\left( ii \right) \\
\end{align}$
Adding equation (i) and equation (ii), we get
$2a\sec \theta =6$
Dividing both sides by 2a, we get
$\sec \theta =\dfrac{3}{a}$
Subtracting equation (ii) from equation (i), we get
$2b\tan \theta =1-5=-4$
Dividing both sides by 2b, we get
$\tan \theta =\dfrac{-2}{b}$
Now, we know that
${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $
Hence, we have
$\begin{align}
  & {{\left( \dfrac{3}{a} \right)}^{2}}=1+{{\left( \dfrac{-2}{b} \right)}^{2}} \\
 & \Rightarrow \dfrac{9}{{{a}^{2}}}=1+\dfrac{4}{{{b}^{2}}} \\
\end{align}$
Multiplying both sides by ${{a}^{2}}{{b}^{2}}$, we get
$9{{b}^{2}}={{a}^{2}}{{b}^{2}}+4{{a}^{2}}$
Taking ${{a}^{2}}$ common from the terms on RHS, we get
$9{{b}^{2}}={{a}^{2}}\left( {{b}^{2}}+4 \right)$
Hence the value of ${{a}^{2}}\left( {{b}^{2}}+4 \right)$ is $9{{b}^{2}}$
Hence option [b] is correct.

Note: In the questions of these types we usually eliminate the value of $\theta $ by the use of Pythagorean identities as done in the above solution. Sometimes squaring and adding the equations eliminate the variable $\theta $ from the equations more easily. The method of squaring and adding should be preferred in case of sines and cosines.