If cosec(a-b) = 2/3 and sec(a+b) = 2, find the least positive values of a and b - Maths - Trigonometric Functions

Question

If cosec(a-b) = 2/3 and sec(a+b) = 2, find the least positive
values of a and b

Santosh Kumar · Accepted Answer

$co sec (a - b) = \frac{2}{\sqrt{3}} \Rightarrow \sin (a - b) = \frac{\sqrt{3}}{2}$ â‡’ sin (a â€“ b) = sin 60Â° = sin 120Â° â‡’ a â€“ b = 120Â° (i) $\sec (a + b) = \sqrt{2} \Rightarrow \cos (a + b) = \frac{1}{\sqrt{2}} \Rightarrow \cos (a + b) = \cos 45 ° = \cos 315^{\circ} \Rightarrow a + b = 315 ° (ii)$ Adding (i) and (ii), a â€“ b + a + b = 120Â° + 315Â° â‡’ 2a = 435Â° â‡’ a = 217 5Â° â‡’ b = 315Â° â€“ 217 5Â° [ from equation (ii) ] â‡’ b = 97 5Â° âˆ´ a = 217 5Â° and b = 97 5Â°

If cosec(a-b) = 2/3 and sec(a+b) = 2, find the least positive values of a and b.

If cosec(a-b) = 2/ $scriptstyle sqrt{,,}$ 3 and sec(a+b) = $scriptstyle sqrt{,,}$ 2, find the least positive values of a and b.