Question

AD=BD angle B=90 find sin^2+cos^2

Answer 1

Solution: If AB=BD angle B=90.
Sin A=P/H=AB/AD...1
Cos A=B/H=BD/AD...2
Since, Sin^2 +Cos^2
So, from 1 and 2 we get,
Sin^2+Cos^2=(AB/AD)^2+(BD/AD)^2
=AB^2+BD^2/AD^2
=AD^2/AD^2 {By using
Pythagoras theorem}
=AD^2/AD^2
=1 Answer