Find the value of p2+1p2, if p+1p=7.
49
47
36
20
a2+b2=(a+b)2–2ab
p2+1p2=(p+1p)2–2p×1p=(7)2–2=49–2=47
If p=3−2√2, then p2+1p2= ______.
If tan A, tan B are the roots of x2−Px+Q=0 the value of sin2 (A+B)=(where P, Q ϵ R)