prove that (sinA+cosA)2+(sinA-cosA)2=2
Anonymous:
___k off
Answers
Answered by
4
On expansion we get 2(sin^2 A+COS^2A)
=2(1)=2
can you explain me it in easy way??please
use formula (a+b) ^2 +(a-b) ^2 =2(a^2 +b^2)
ok
Answered by
5
Solution:
(sinA+cosA)^2+(sinA-cosA)^2=2
LHS = sin^2A + 2sinAcosA + cos^2A + sin^2A - 2sinAcosA + cos^2A
(+ 2sinAcosA) and ( - 2sinAcosA) are cancelled.
= sin^2A + cos^2A + sin^2A + cos^2A
As we know that,
sin^2A + cos^2A = 1
Putting the value of sin^2A + cos^2A , we get
= sin^2A + cos^2A + sin^2A + cos^2A
= 1 + 1
= 2 = RHS
Therefore, LHS = RHS
Hence proved.
(sinA+cosA)^2+(sinA-cosA)^2=2
LHS = sin^2A + 2sinAcosA + cos^2A + sin^2A - 2sinAcosA + cos^2A
(+ 2sinAcosA) and ( - 2sinAcosA) are cancelled.
= sin^2A + cos^2A + sin^2A + cos^2A
As we know that,
sin^2A + cos^2A = 1
Putting the value of sin^2A + cos^2A , we get
= sin^2A + cos^2A + sin^2A + cos^2A
= 1 + 1
= 2 = RHS
Therefore, LHS = RHS
Hence proved.
thank u sir
Similar questions