given that angles A and B are such that, sec^2A-tanA = 13 and sinBsec^2B=27cosBcosec^2B
First of all key elements required for this question is the knowledge or retrival of these equations: tan2 + 1 =sec2 A tan(A-B) = (tanA-tanB)/(1+tanAtanB) so for the first equation sec2 A - tanA =13 we substitute sec2 A for tan2A+1 then we get tan2A - tanA +1 = 13 which can then form tan2A - tan A-12=0 from this factorising to get (tanA-4)(tanA+3)=0, so tanA=4,-3 For the second equation sinBsec2B = 27cosBcosec2B first relaise that sec2B is just 1/cos2B and cosecB is just sinB then simplify by bring the denomenators of the equations to the opposite sides so you get sin3B=27cos3B, we know that sinB/cosB, so tan3B= 27 from this we can then cube root both sides to get tanB = 3, 3 being the cube root of 27. now we have the values of both tanA and tanB, we just need to substitue them into the euqtaion tanA-tanB/(1+tanAtanB)=tan(A-B) and then we get tan A =4 we get the value of (4-3)/(1+4x3) = 1/14 and when tanA = -3 we shoudl get a value of (-3-3)/(1+(-3)*3) = 3/4
