to find minimum value of $\sec A + \sec B$

Question

to find minimum value of $\sec A + \sec B$ given $A+B=\frac{\pi}{3}$ , A>0 , $B < \frac{\pi}{3}$.

i put $B = \frac{\pi}{3} - A $ but how to proceed? i don't have to use differentiation

thanks

Answer 1

We make use of the following identities: $$(1): 2 \cos A \cos B = \cos (A+B) + \cos (A-B) $$ $$(2): \cos A \cos B + \sin A \sin B = \cos (A-B) $$

Now, $$\sec A + \sec B = \frac {1}{\cos A} + \frac{1}{\cos B}$$ $$=\frac {1}{\cos A} + \frac {1}{\cos ( 60^\circ - A)} $$ $$= \frac {\cos 60^\circ \cos A + \sin 60^\circ \sin A + \cos A}{\cos A \cos (60^\circ -A)}$$ $$=\frac {\sqrt {3}[\frac {1}{2}\sin A + \frac {\sqrt {3}}{2} \cos A] }{\cos A \cos (60^\circ -A)}$$ $$=\frac {2\sqrt {3}\cos (30^\circ -A)}{2\cos A \cos (60^\circ -A)} $$ $$=\frac {2\sqrt {3}\cos (30^\circ -A)}{\cos 60^\circ + \cos (60^\circ -2A)}\tag {a} $$

Now observe we have that if $(30^\circ -A) =k $, then $(60^\circ - 2A) =2k $. Thus, we have $$(a) = \frac {2\sqrt {3} \cos k}{0.5 + \cos 2k} $$ Hope you can take it from here.