Question

prove LHS equals to RHS

Answer 1

LHS = sin⁶ A + cos⁶ A
= (sin² A)³ + (cos² A)³
=  (sin² A + cos² A)³ − 3 sin² A cos² A (sin² A + cos² A) 
 [Since, (a+b)³ = a³ + b³ +3ab(a+b)]
= (1)³ - 3 sin² A cos² A (sin² A + cos² A)......( since sin² A + cos² A = 1 )
= 1 - 3 sin² A cos² A (1)
= 1 - 3 sin²A cos²A
= RHS

∴LHS = RHS
Hence, proved.

Hope this helps !!
#Dhruvsh

Answer 2

Arrange in this type to prove :-

••Sin^6A+Cos^6A+3Sin²ACos²A= 1

From L•H•S :-

(Sin²A)³+(Cos²A)³+3 Sin²A Cos²A

a³+b³=(a + b )(a²+b²-ab)

=(Sin²A+Cos²A) [(Sin²A)²+(Cos²A)²-(Sin²A)

Cos²A] + 3 Sin²A Cos²A

Then

Sin²A + Cos²A = 1,

=1 {(Sin²A)²+(Cos²A)²+2 Sin²ACos²A }

It is in the form of a²+ b²+2ab=(a + b)²

So :)

(Sin²A)²+(Cos²A)²+2 Sin²A Cos²A=(Sin²A + Cos²A)²

We have :-

Sin²A + Cos²A = 1

=>( 1 )²= 1

L•H•S=R•H•S

So :)

Sin^6A + Cos^6A + 3 Sin²A Cos²A = 1

=Sin^6A + Cos^6A = 1 - 3 Sin²A Cos²A..... Proved