AB = a
⇒ AD + DB = a
⇒ AD + AD = a
⇒ 2 AD = a
⇒ AD = a2
Thus, AD = DB = a2
By Pythagoras theorem, we have
AC2=AB2+BC2
⇒b2=a2+BC2
⇒BC2=b2−a2
⇒BC=√b2−a2
Thus, in ΔBCD, we have
Base = BC = √b2−a2 and Perpendicular = BD = a2
Applying Pythagoras theorem in Δ BCD, we have
⇒BC2+BD2=CD2
⇒(√b2−a2)+(a2)2=CD2
⇒CD2=b2−a2+a24
⇒CD2=4b2−4a2+a24
⇒CD2=4b2−3a24
⇒CD=√4b2−3a22
Now,
sin2θ+cos2θ=(a√4b2−3a2)2+(2√b2−a2√4b2−3a2)2
⇒sin2θ+cos2θ=a24b2−3a2+4(b2−a2)4b2−3a2
⇒sin2θ+cos2θ=4b2−3a24b2−3a2=1.