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Question

Prove that

sin6x .- tan6 x - 3 sec2x •

tan2x = 1

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Solution

It should be sec instead of sin

As,
a³-b³​​a³-b³=(a-b)³+3ab(a-b)
And, 1+tan²x=sec²x

(sec²x)³-(tan²x)³
=(sec²x-tan²x)³+3sec²xtan²x(sec²x-tan²x)
=1+3sec²xtan2x(1)
=1+3tan²xsex²x-3tan²xsex²x
=1
hence proved

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