prove that sec^6A-tan^6A=1+3tan^2A sec ^2A
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sec⁶A - tan⁶A = 1-sin⁶A /cos⁶A = 1³-(sin²A)³ /cos⁶A
= (1-sin²A)(1 + sin⁴A + sin²A) / cos⁶A
applied [ a³ - b³ = (a-b) (a² + b² + ab) ]
= cos²A (1+sin⁴A + sin²A) / cos⁶A = 1 + sin⁴A + sin²A / cos⁴A
= sec⁴A + tan⁴A + tan²Asec²A ...........(1)
Now (sec²A - tan²A)² = sec⁴A + tan⁴A - 2tan²Asec²A
= 1² + 2tan²Asec²A = sec⁴A + tan⁴A
putting this value in (1)
1 + 2tan²Asec²A + tan²Asec²A = 1 + 3tan²Asec²A
Hence proved.
= (1-sin²A)(1 + sin⁴A + sin²A) / cos⁶A
applied [ a³ - b³ = (a-b) (a² + b² + ab) ]
= cos²A (1+sin⁴A + sin²A) / cos⁶A = 1 + sin⁴A + sin²A / cos⁴A
= sec⁴A + tan⁴A + tan²Asec²A ...........(1)
Now (sec²A - tan²A)² = sec⁴A + tan⁴A - 2tan²Asec²A
= 1² + 2tan²Asec²A = sec⁴A + tan⁴A
putting this value in (1)
1 + 2tan²Asec²A + tan²Asec²A = 1 + 3tan²Asec²A
Hence proved.
anna19:
Ohh ok thank you Soo much
Its alright .
also in the step where u marked (1)how did u get tan^4?
I think u took 1+sin^4A as sec^4A
I divided all terms by cos^4x.
and sin^2A÷cos^2A as tan^2A and the 1÷cos^2A u took as sec^2A
Ohh ok ok
I'm so sorry for troubling so much but I was tyring to fully understand
thank you so much for the immense help
Its OK . You should get the basics, that's important.
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