If cos(A-B)=3/5 and tanAtanB=2, then
cosAcosB=1/5
sinAsinB=-2/5
cosAcosB=-1/5
sinAsinB=-1/5
Explanation for the correct option:
Step 1. Given that cos(A-B)=3/5
⇒ cos(A–B)=3/5…(i)
⇒ tanAtanB=2
⇒(sinA/cosA).(sinB/cosB)=2/1
Step 2. Using componendo and dividendo rule,
sinAsinB+cosAcosBsinAsinB–cosAcosB=2+12-1
⇒ cos(A–B)-cos(A+B)=3
⇒ -3/5cos(A+B)=3[From(i)]
⇒ cos(A+B)=-1/5….(ii)
Step 3. Adding (i) and (ii),
cos(A–B)+cos(A+B)=(3/5)–(1/5)
2cosAcosB=2/5
⇒cosAcosB=15
The correct answer is option A
If 5a+2b5a−2b=53, then a : b =