Question

If $\tan \left( A-B \right)=1,\sec \left( A+B \right)=\dfrac{2}{\sqrt{3}}$, 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: Use trigonometric identities relating tangent and secant function to convert from one form to another. Use inverse trigonometric functions to find the value of angles which satisfy the given equations. Also use formulas for tangent of sum and difference of two angles to find the value of angle B.

Complete step-by-step answer:
We have the trigonometric equations $\tan \left( A-B \right)=1,\sec \left( A+B \right)=\dfrac{2}{\sqrt{3}}$. We have to calculate the smallest positive value of angle B which satisfies the above equations.
We know the trigonometric identity ${{\tan }^{2}}x+1={{\sec }^{2}}x$.
Substituting $x=A+B$ in the above equation, we have ${{\tan }^{2}}\left( A+B \right)+1={{\sec }^{2}}\left( A+B \right)$.
We know that $\sec \left( A+B \right)=\dfrac{2}{\sqrt{3}}$.
Thus, we have ${{\tan }^{2}}\left( A+B \right)+1={{\sec }^{2}}\left( A+B \right)=\dfrac{4}{3}$.
Rearranging the terms of the above equation, we have ${{\tan }^{2}}\left( A+B \right)=\dfrac{4}{3}-1$.
Thus, we have ${{\tan }^{2}}\left( A+B \right)=\dfrac{1}{3}$.
So, we have $\tan \left( A+B \right)=\pm \dfrac{1}{\sqrt{3}}$.
We know that $\tan \left( A-B \right)=1$.
So, the smallest possible value of $A-B$ is $A-B={{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$.
If $\tan \left( A+B \right)=\dfrac{1}{\sqrt{3}}$, the smallest possible value of $A+B$ is $A+B={{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)=\dfrac{\pi }{6}$.
We know that $A-B=\dfrac{\pi }{4}$ and $A+B=\dfrac{\pi }{6}$. Subtracting the first equation from the second one, we have $\left( A+B \right)-\left( A-B \right)=\dfrac{\pi }{6}-\dfrac{\pi }{4}$.
Thus, we have $2B=\dfrac{-\pi }{12}\Rightarrow B=\dfrac{-\pi }{24}$.
If $\tan \left( A+B \right)=\dfrac{-1}{\sqrt{3}}$, the smallest possible value of $A+B$ is $A+B={{\tan }^{-1}}\left( \dfrac{-1}{\sqrt{3}} \right)=\dfrac{11\pi }{6}$.
We know that $A-B=\dfrac{\pi }{4}$ and $A+B=\dfrac{11\pi }{6}$. Subtracting the first equation from the second one, we have $\left( A+B \right)-\left( A-B \right)=\dfrac{11\pi }{6}-\dfrac{\pi }{4}$.
Thus, we have $2B=\dfrac{19\pi }{12}\Rightarrow B=\dfrac{19\pi }{24}$.
Hence, the smallest positive value of angle B is $B=\dfrac{19\pi }{24}$, which is option (b).

Note: We can also solve this question by expanding the equations using the trigonometric functions $\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$ and $\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}$ and then simplifying the equations to get smallest positive value of angle B. It’s also necessary to keep in mind that we must consider only positive values of angle B. We can check if the calculated values of angles are correct or not by substituting the value of angles in the equations and check if they satisfy the given equations.