Question

Prove:
${sec}^{6} x - {tan}^{6} x = 1 + {sec}^{2} x. {tan}^{2} x$
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Answer 1

${\huge {\mathfrak {Answer :-}}}$

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L. H. S = sec^6 x - tan^6 x

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

= ( sec^2 x - tan^2 x )( sec^4 x + sec^2 x. tan^2 x + tan^4 x)

= sec^4 x + tan^4 x + sec^2 x. tan^2 x

= (sec^2 x - tan^2 x)^2 + 2 sec^2 x. tan^2 x + sec^2 x. tan^2 x

= 1 + 3. sec^2 x. tan^2 x

I guess there's something wrong in the question . Please check :)

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Thanks..

Answer 2

Answer:

hey mate please refer to the attachment