Math, asked by sandip19, 1 year ago

sec6A-tan6A= 1+3tan2A+3tan4A


siddhartharao77: Is it tan 2a (or)tan^2a?
sandip19: sec^6A-tan^6A= 1+3tan^2A+3tan^4A

Answers

Answered by siddhartharao77
16
Given LHS = sec^6 A - tan^6 A.

W.K.T a^3 - b^3 = (a - b)(a^2 + ab + b^2)

                          = (sec^2 A - tan^2 A)((sec^2 A)^2 + sec^2 A tan^2 A + tan^ 4A)


W.K.T sec^2 theta = 1 + tan^2 theta (or) sec^2 theta - tan^2 theta = 1.

 
                           = (1)((sec ^2A)^2 + (1 + tan^2 A)(tan^2 A) + (tan^4 A)

                           = (sec^2 A)^2 + (1 + tan^2 A)(tan^2 A) + (tan^4 A)

                           = (1 + tan^2 A)^2 + (1 + tan^2 A)(tan^2 A) + (tan^4 A)

                           = (1 + tan^2 A)[(1 + tan^2 A) + tan^2 A] + tan^4 A

                           = (1 + tan^2 A)(1 + 2 tan^2 A) + tan^4 A

                           = 1 + 2tan^2 A + tan^2 A + 2 tan^4 A + tan^4 A

                           = 1 + 3 tan^2 A + 3 tan^4 A


LHS = RHS.


Hope this helps!

Inferno11: Could you re check you solution again
siddhartharao77: What happened dude?
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