prove that sec^6 A = tan ^6 A + 3 tan ^2 A sec ^2 A + 1
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Given, RHS = tan^6 A + 3tan^2 A sec^2 A + 1
= tan^6 A + 3tan^2 A(1+ tan^2 A ) + 1
= tan^6 A + 3tan^4 A + 3 tan^2 A + 1
= (tan^2 A)^3 + 3(tan^2 A)^2+ 3 (tan^2 A) + 1
We know that a^3 + 3a^2b + 3ab^2 +b^3 = (a + b)^3
(tan^2 A)^3 + 3(tan^2 A)^2+ 3 (tan^2 A) + 1 = (tan^2 A + 1)^3
= (sec^2 A)^3
= sec^6 A.
Hope this helps!
= tan^6 A + 3tan^2 A(1+ tan^2 A ) + 1
= tan^6 A + 3tan^4 A + 3 tan^2 A + 1
= (tan^2 A)^3 + 3(tan^2 A)^2+ 3 (tan^2 A) + 1
We know that a^3 + 3a^2b + 3ab^2 +b^3 = (a + b)^3
(tan^2 A)^3 + 3(tan^2 A)^2+ 3 (tan^2 A) + 1 = (tan^2 A + 1)^3
= (sec^2 A)^3
= sec^6 A.
Hope this helps!
kiran78:
thanks
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