# Answer to Question #4280 in Trigonometry for D Sudha

Question #4280

Prove: 1-(sec^6a+cos^6a)=3sin^2acos^2a

Expert's answer

To prove the equality means to show that it holds for every value of a. In this case it's not true

For example. we can substitute a=pi/4:

left part& 1-((sec pi/4)^6+(cos pi/4)^6)=1-((sqrt(2))^6+1/(sqrt(2))^6)=1-(8+1/8)=-7-1/8=-57/8

right part 3*(sin pi/4)^2 * (cos pi/4)^2=3*1/2*1/2=3/4

-57/8 = 3/4 that's false so given equality doesn't hold for a=pi/4 . It means that equality is not true for every a

For example. we can substitute a=pi/4:

left part& 1-((sec pi/4)^6+(cos pi/4)^6)=1-((sqrt(2))^6+1/(sqrt(2))^6)=1-(8+1/8)=-7-1/8=-57/8

right part 3*(sin pi/4)^2 * (cos pi/4)^2=3*1/2*1/2=3/4

-57/8 = 3/4 that's false so given equality doesn't hold for a=pi/4 . It means that equality is not true for every a

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