Question

If \[\tan (A - B) = 1,\] \[\sec (A + B) = \dfrac{2}{{\sqrt 3 }}\], then the smallest positive value of \[B\] is?
A.\[\dfrac{{25\pi }}{{24}}\]
B.\[\dfrac{{19\pi }}{{24}}\]
C. \[\dfrac{{13\pi }}{{24}}\]
D. \[\dfrac{{11\pi }}{{24}}\]

Answer 1

Hint: In this type we find the value of \[A + B\] and \[A - B\] and solving this system of equations we can find the value of \[B\]. In this particular question we will have cases, because they asked only the positive value. If we get a positive value it will be no problem. If we get negative value we move to further case which we have done below:

Complete step-by-step answer:
Case (1):
Given, \[\tan (A - B) = 1,\]we can be rewrite it as,
\[\tan (A - B) = \tan \left( {\dfrac{\pi }{4}} \right),\]
Cancelling on both side we get,
\[ \Rightarrow A - B = \dfrac{\pi }{4}\]
Similarly \[\sec (A + B) = \dfrac{2}{{\sqrt 3 }}\] we can be rewrite it as,
\[\sec (A + B) = \sec \left( {\dfrac{\pi }{6}} \right)\]
Cancelling on both side,
\[ \Rightarrow A + B = \dfrac{\pi }{6}\]
We subtracted because we need the value of \[B\] only, if added above we will get the value of \[A\].
So Subtracting, \[A - B = \dfrac{\pi }{4}\] with \[A + B = \dfrac{\pi }{6}\]we get
\[ \Rightarrow A - B - A - B = \dfrac{\pi }{4} - \dfrac{\pi }{6}\]
\[A\] Will get cancels out,
Taking L.C.M. and simplifying we get,
\[ \Rightarrow - 2B = \dfrac{\pi }{{12}}\]
Divided by \[ - 2\] on both sides,
\[ \Rightarrow B = - \dfrac{\pi }{{24}}\]
But \[B\] cannot be negative, we need only positive value, we change any one of the angles and continue the same procedure.
Case (2):
Changing the angle,
So we take \[\sec \left( {\dfrac{\pi }{6}} \right)\] as \[\sec \left( {2\pi - \dfrac{\pi }{6}} \right)\].
Because we know \[\sec \left( {2\pi - \theta } \right) = \sec \theta \].
Then, \[\sec (A + B) = \sec \left( {2\pi - \dfrac{\pi }{6}} \right)\]
Cancelling on both sides,
 \[ \Rightarrow A + B = 2\pi - \dfrac{\pi }{6}\]
\[ \Rightarrow A + B = \dfrac{{11\pi }}{6}\].
Now we have, \[A - B = \dfrac{\pi }{4}\] and \[A + B = \dfrac{{11\pi }}{6}\], subtracting we get,
\[ \Rightarrow A - B - A - B = \dfrac{\pi }{4} - \dfrac{{11\pi }}{6}\]
\[ \Rightarrow - 2B = - \dfrac{{19\pi }}{{12}}\]
\[ \Rightarrow B = \dfrac{{19\pi }}{{24}}\].
So, the correct answer is “Option B”.

Note: Since they asked us to find a positive value of \[B\] at first we get negative value while solving the equations in case (I) so we proceed to case (2) where we get a positive value. If they asked only the value of \[B\] and not the positive value we can stop the solution at case (I) only. . If we don’t get positive value even in case (2). We move to case (3) by changing the angle again.