Question

If \[a+b\tan \theta =\sec \theta \] and \[b-a\tan \theta =3\sec \theta \], then the value of \[{{a}^{2}}+{{b}^{2}}\].
A. 15
B. 10
C. 12
D. 20

Answer 1

Hint: For the above question we will square each of the given equations and add them to find the required value. Also, we will use the trigonometric formula which is as follows:
\[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \]
Complete step-by-step answer:
We have been given \[a+b\tan \theta =\sec \theta \] and \[b-a\tan \theta =3\sec \theta \] and asked to find the value of \[{{a}^{2}}+{{b}^{2}}\]
\[a+b\tan \theta =\sec \theta .....(1)\]
\[b-a\tan \theta =3\sec \theta .....(2)\]
Now on squaring equation (1) we get as follows:
\[{{\left( a+b\tan \theta \right)}^{2}}={{\sec }^{2}}\theta \]
Using the identity \[{{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}\] we get as follows:
\[{{a}^{2}}+2ab\tan \theta +{{b}^{2}}{{\tan }^{2}}\theta ={{\sec }^{2}}\theta .....(4)\]
Now, on squaring the equation (2), we get as follows:
\[{{\left( b-a\tan \theta \right)}^{2}}=9{{\sec }^{2}}\theta \]
Using the identity \[{{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}\] we get as follows:
\[{{b}^{2}}-2ab\tan \theta +{{a}^{2}}{{\tan }^{2}}\theta =9{{\sec }^{2}}\theta ....(3)\]
After adding equations (3) and (4) we get a follows:
\[\begin{align}
  & {{a}^{2}}+{{b}^{2}}+2ab\tan \theta -2ab\tan \theta +{{b}^{2}}{{\tan }^{2}}\theta +{{a}^{2}}{{\tan }^{2}}\theta ={{\sec }^{2}}\theta +9{{\sec }^{2}}\theta \\
 & \Rightarrow \left( {{a}^{2}}+{{b}^{2}} \right)+{{\tan }^{2}}\theta \left( {{b}^{2}}+{{a}^{2}} \right)=10{{\sec }^{2}}\theta \\
\end{align}\]
Taking \[\left( {{a}^{2}}+{{b}^{2}} \right)\] as common, we get as follows:
\[\Rightarrow \left( {{a}^{2}}+{{b}^{2}} \right)\left( 1+{{\tan }^{2}}\theta \right)=10{{\sec }^{2}}\theta \]
Since we know the trigonometric identity,
\[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \]
\[\Rightarrow \left( {{a}^{2}}+{{b}^{2}} \right){{\sec }^{2}}\theta =10{{\sec }^{2}}\theta \]
On dividing the equation by \[{{\sec }^{2}}\theta \] we get as follows:
\[\begin{align}
  & \Rightarrow \left( {{a}^{2}}+{{b}^{2}} \right)\dfrac{{{\sec }^{2}}\theta }{{{\sec }^{2}}\theta }=\dfrac{10{{\sec }^{2}}\theta }{{{\sec }^{2}}\theta } \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}=10 \\
\end{align}\]
Therefore, the correct answer of the question is option B.

Note: Be careful while doing calculation and also check the sign in each step of calculation. In this type of questions we always use trigonometric identity after performing operation with the given equation according to what we have to find. Here, we have to find \[{{a}^{2}}+{{b}^{2}}\] and we know that by squaring the equation we will get these terms so we do that. There is a great possibility that students can substitute the value of \[\sec \theta =a+b\tan \theta \] in the second equation. Doing this will only make the solution more complex and time consuming.