Prove identity: Sec^6x-tan^6x= 1+3tan^2xsec^2x
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a³-b³=(a-b)³+3ab(a-b)
1+tan²x=sec²x
(sec²x)³-(tan²x)³=(sec²x-tan²x)³+3sec²xtan²x(sec²x-tan²x)
=1³+3sec²xtan2x(1)
=1+3tan²xsex²x
hence proved
1+tan²x=sec²x
(sec²x)³-(tan²x)³=(sec²x-tan²x)³+3sec²xtan²x(sec²x-tan²x)
=1³+3sec²xtan2x(1)
=1+3tan²xsex²x
hence proved
mukundkale:
thx
a^3-b^3=(a-b)^3-3ab(a-b)
how 1+3sec^2x,tan^2x
Plz how 1+3sec^2x.tan^2x
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