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sec^6x-tan^6x-3sec^2x*tan^2x=1 how to prove

At 1st I will used the way I know clearly.

Note that sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x).

With this, we can transform the left side of this equation to

1/cos^6(x) - sin^6(x)/cos^6(x) - [3/cos^2(x)][sin^2(x)/cos^2(x)].

To put all terms over cos^6(x), we just need to multiply the last term by cos^(x)/cos^2(x).

That gives 1/cos^6(x) - sin^6(x)/cos^6(x) - 3*sin^2(x)cos^2(x)/cos^6(x).

Combining the fractions gives [1 - sin^6(x) - 3*sin^2(x)cos^2(x)]/cos^6(x).

Now the cos^2(x) in the denominator can be changed to 1 - sin^2(x). This gives

[1 - sin^6(x) - 3*sin^2(x){1 - sin^2(x)}]/cos^6(x).

Multiplying out that 3rd term gives

[1 - sin^6(x) - 3*sin^2(x) + 3sin^4(x)]/cos^6(x).

Reorganizing the numerator gives

[1 - 3*sin^2(x) + 3sin^4(x) - sin^6(x)]/cos^6(x).

The top looks like 1 - 3b + 3b^2 - b^3 where b = sin^2(x).

The expression just given factors into (1-b)^3.

Since b = sin^2(x), this becomes [1 - sin^2(x)]^3.

It is known that sin^2(x) + cos^2(x) = 1, so subtracting sin^2(x) from both sides gives

cos^2(x) = 1 - sin^2(x). Noting that this is the same as what's inside the brackets,

[1 - sin^2(x)]^3 become [cos^2(x)]^3. That is the same as cos^6(x).

Now we get cos^6(x) in the numerator, and that is the same thing as is in the denominator,

so they both cancel and the result is 1.

Now for the 2nd, this could have been done using 1 + tan^2(x) = sec^2(x).

To do it, just put in 1 + tan^2(x) for sec^2(x).

This gives (1 + tan^2)^3 - tan^6(x) - 3(1 + tan^2(x))tan^2(x).

Since (1+a)^3 = 1 + 3a + 3a^2 + a^3, this can be used on the 1st term where a is tan^2(x).

That is, 1 + 3tan^2(x) + 3tan^4(x) + tan^6(x) - tan^6(x) - 3tan^2(x) - 3tan^4(x).

Looking at the whole expression, the 2nd term is the negative of the 7th term, the 3rd term is the negative of the 6th term, and the 4th term is the negative of the 5th term. Cancelling term 2, 7, 3, 7, 4, and 5 leaves us with term 1, and the value of term 1 is 1.

As can be seen, this is easier, but it requires more to be known about the trig functions.

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