sec six theta=tan6theta+3tan square theta sec square theta +1
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Consider, tan6A + 3tan2A.sec2A + 1
= tan6A + 3tan2A(1+ tan2A ) + 1 [since sec2A = 1+ tan2A]
= tan6A + 3tan4A + 3 tan2A + 1
= (tan2A)3 + 3(tan2A)2+ 3 (tan2A) + 1
It is in the form of a3 + 3a2b + 3ab2 +b3 = (a + b)3
∴ (tan2A)3 + 3(tan2A)2+ 3 (tan2A) + 1 = (tan2A + 1)3
= (sec2A)3
= sec6A
= tan6A + 3tan2A(1+ tan2A ) + 1 [since sec2A = 1+ tan2A]
= tan6A + 3tan4A + 3 tan2A + 1
= (tan2A)3 + 3(tan2A)2+ 3 (tan2A) + 1
It is in the form of a3 + 3a2b + 3ab2 +b3 = (a + b)3
∴ (tan2A)3 + 3(tan2A)2+ 3 (tan2A) + 1 = (tan2A + 1)3
= (sec2A)3
= sec6A
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