Expression : $$sec^6 A - tan^6 A - 3 sec^2 A tan^2A$$
= $$[(sec^2A)^3-(tan^2A)^3]-3sec^2Atan^2A$$
Using, $$x^3-y^3=(x-y)(x^2+y^2+xy)$$
= $$[(sec^2A-tan^2A)(sec^4A+tan^4A+sec^2Atan^2A)]-3sec^2Atan^2A$$
Also, $$(sec^2A-tan^2A=1)$$
= $$sec^4A+tan^4A-2sec^2Atan^2A$$
Using, $$x^2+y^2-2xy=(x-y)^2$$
= $$(sec^2A-tan^2A)^2$$
=Â $$(1)^2=1$$
=> Ans - (C)
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