Question

prove that sin®A+cos®A=1....plz ans

Answer 1

consider a triangle PQR right angled at

Q

so,

BY PYTHAGORUS THEUROM

PR^2=PQ^2+QR^2

sin R=PQ/PR

sq we get

sin^2 R=PQ^2/PR^2

similarly cos^2=QR^2/PR^2

ADDING THEM

SIN^2+COS^2= PQ^2/PR^2+QR^2/PR^2

=PQ^2+QR^2/PR^2

=PR^2/PR^2

=1