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what is the solution for 10th qstn

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Secondary SchoolMath 5+3 pts

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

Report by Bijukumarmnj6312 26.11.2018

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jashanjatana12274

Jashanjatana12274 Ambitious

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anil6610

hello

Boffeemadrid Ambitious

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.