Prove that sec6a=tan6a+3tan2asec2a+1
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sec⁶A
=(sec²A)³
=(tan²A+1)³ [, sec²A-tan²A=1]
=(tan²A)³+3(tan²A)².1+3tan²A.1+1³
=tan⁶A+3tan⁴A+3tan²A+1
=tan⁶A+3tan²A(tan²A+1)+1
=tan⁶A+3tan²Asec²A+1 (Proved)
=(sec²A)³
=(tan²A+1)³ [, sec²A-tan²A=1]
=(tan²A)³+3(tan²A)².1+3tan²A.1+1³
=tan⁶A+3tan⁴A+3tan²A+1
=tan⁶A+3tan²A(tan²A+1)+1
=tan⁶A+3tan²Asec²A+1 (Proved)
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