Question

sec⁶x-tan⁶x=1+3sec²x* tan²x Prove it?

Answer 1

Answer:

$sec^6x - tan^6x = 1 + 3sec^2x tan^2x$

Step-by-step explanation:

We have to prove:

$sec^6x - tan^6x = 1 + 3sec^2x tan^2x$

LHS:

$sec^6x - tan^6x$

$(sec^2x)^3 - (tan^2x )^3$

By the formula of:

$a^3 -b^3 = (a- b)(a^2 +ab  + b^2)$

So,

$(sec^2x - tan^2x)[(sec^2x)^2 + sec^2xtan^2x  + (tan^2x)^2]$

$(sec^2x - tan^2x) =1$ is the identity

$1 \times (sec^4x + sec^2xtan^2x  + tan^4x)$

Now add and subtract $2tan^2x sec^2x$

$(sec^4x + tan^4x + sec^2xtan^2x +  2sec^2xtan^2x  - 2sec^2xtan^2x )\\(sec^4x + tan^4x - sec^2xtan^2x +  3sec^2xtan^2x)$

By the formula of $(a-b)^2 = a^2 + b^2 -2ab$

$(sec^2x - tan^2x )^2 + 3sec^2xtan^2x)\\1 + 3sec^2xtan^2x$

=R.H.S.

Hence Proved

Answer 2

Answer:

Step-by-step explanation :

= sec⁶x - tan⁶x

= (sec²x)³ - (tan²x)³

USE THE FORMULA OF a³-b³= (a-b)³+3ab(a-b)

= (sec²x - tan²x)³+3×sec²×tan²x (sec²x -tan²x)

USE IDENTITY (sec²x - tan²x) = 1

= (1) + 3×sec²x×tan²x

= RHS

LHS=RHS

Hence proved.