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3 years ago

Sec^6x(secxtanx)-sec^4(secxtanx)=sec^5xtan^3x Verify the trigonometric identity

Mathematics

1 answer:

mario62 [17]3 years ago

7 0

I guess you mean

$\sec^6x(\sec x\tan x)-\sec^4x(\sec x\tan x)=\sec^5x\tan^3x$

On the left side, we have a common factor of $\sec^4x(\sec x\tan x)=\sec^5x\tan x$ , so that

$\sec^6x(\sec x\tan x)-\sec^4x(\sec x\tan x)=\sec^5x\tan x(\sec^2x-1)$

Recall that

$\sec^2x=1+\tan^2x$

from which it follows that

$\sec^5x\tan x(\sec^2x-1)=\sec^5x\tan x\tan^2x=\sec^5x\tan^3x$

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