Question

sin⁶A + cos⁶A = 1 – 3sin²A . cos²A हे सिद्ध करा.

Answer 1

डावी बाजू = sin⁶A + cos⁶A

= (sin²A)³ + (cos²A)³   

= (1 – cos²A)³ + (cos²A)³    ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2A")]`

= 1 – 3cos²A + 3(cos²A)² – (cos²A)³ + cos⁶A   ......[∵ (a – b)³ = a³ – 3a²b + 3ab² – b³] 

= 1 – 3 cos²A(1 – cos²A) – cos⁶A + cos⁶A

= 1 – 3 cos²A sin²A

= उजवी बाजू

∴ sin⁶A + cos⁶A = 1 – 3sin²A . cos²A