Question

if tan(A-B)=1 and sec(A+B)=2/underroot3 then the smallst positive value of B is_____________??

plz explain

Answer 1

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Suppose $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$. What is the smallest positive value of $B$?

Solution. The identity

$\tan(A-B)=1$

implies that

$A-B=\frac{\pi}{4}+\pi k,\qquad k\in\mathbb{Z}.$

Also, from

$\sec(A+B)=2/\sqrt{3}$

we get

$\frac{1}{\cos(A+B)}=\frac{2}{\sqrt{3}}$

$\cos(A+B)=\frac{\sqrt{3}}{2}$

$A+B=\pm\frac{\pi}{6}+2\pi m,\qquad m\in\mathbb{Z}.$

Hence

$A+B-(A-B)=\pm\frac{\pi}{6}+2\pi m-\frac{\pi}{4}-\pi k,\qquad k,m\in\mathbb{Z}.$

$2B=\pm\frac{\pi}{6}+2\pi m-\frac{\pi}{4}-\pi k,\qquad k,m\in\mathbb{Z}.$

$B=\pm\frac{\pi}{12}+\pi m-\frac{\pi}{8}-\frac{\pi}{2}k,\qquad k,m\in\mathbb{Z}.$

$B=t\,\frac{\pi}{12}-\frac{\pi}{8}+\frac{\pi}{2}(2m-k),\qquad k,m\in\mathbb{Z},t=\pm 1.$

When $(k,m)$ runs over all pairs of integers, the number $2m-k$ runs over all integers, therefore

$B=t\,\frac{\pi}{12}-\frac{\pi}{8}+\frac{\pi}{2}n,\qquad n\in\mathbb{Z},t=\pm 1.$

Thus $B$ depends of two parameters $(n,t)$ and we can write it as a function

$B(k,m,t)=t\,\frac{\pi}{12}-\frac{\pi}{8}+\frac{\pi}{2}n,\qquad n\in\mathbb{Z},t=\pm 1.$

We should find values of $(n,t)$ that give the smallest positive value for $B$.

$B(0,1)=\frac{\pi}{12}-\frac{\pi}{8}=\frac{2-3}{24}=-\frac{1}{24}<0,$

$B(0,-1)=-\frac{\pi}{12}-\frac{\pi}{8}=\frac{-2-3}{24}=-\frac{5}{24}<0,$

Thus negative values of $n$ will decrease $B$, and therefore the smallest positive value of $B$ is achieved for positive $n$:

$B(1,1)=\frac{\pi}{12}\frac{\pi}{8}+\frac{\pi}{2}=-\frac{1}{24}+\frac{\pi}{2}=\frac{-1+12}{24}=\frac{11}{24},$

$B(1,-1)=-\frac{\pi}{12}-\frac{\pi}{8}+\frac{\pi}{2}=-\frac{5}{24}+\frac{\pi}{2}=\frac{-5+12}{24}=\frac{7}{24}.$

Since for fixed $t$ the number $B(n,t)$ increases with $n$, the smallest positive value of $B$ is achieved for positive $n=1$, $t=-1$, and is equal to

$\frac{7}{24}.$