Question

Prove that 2(sin^6A + cos^6A) - 3(sin^4A+ cos^4A) +1 = 0

Answer 1

here the answer................

Answer 2

Answer:

Step-by-step explanation:

The given equation is:

$2(sin^{6}A+cos^{6}A)-3(sin^{4}A+cos^{4}A)+1$

Using $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$

⇒$2(sin^{2}A+cos^{2}A)(sin^{4}A-sin^{2}Acos^{2}B+cos^{2}A)-3(sin^{4}A+cos^{4}A)+1$

⇒$2(1)(sin^{4}A-sin^{2}Acos^{2}B+cos^{4}A)-3(sin^{4}A+cos^{4}A)+1$

⇒$2sin^{4}A-2sin^{2}Acos^{2}B+2cos^{4}A-3sin^{4}A-3cos^{4}A+1$

⇒$-sin^{4}A-2sin^{2}Acos^{2}B-cos^{4}A+1$

⇒$-(sin^{4}A+2sin^{2}Acos^{2}B+cos^{4}A)+1$

⇒$-(sin^{2}A+cos^{2}B)^{2}+1$

⇒$-1+1=0$= RHS

Hence proved.