Question is too big need to type it in chat related to trignometry

[molten temple]

EEE
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[modern warren]

yes

$\frac{\tan^2A}{\tan A-1}+\frac{1}{\tan A(1-\tan A)}=\frac{\tan^2A}{\tan A-1}-\frac{1}{\tan A(\tan A-1)}$

[molten temple]

wolfqz

[modern warren]

is this what you mean or am i misinterpreting

oh

yeah

$\frac{1}{\tan A(1-\tan A)}=\frac{1}{-\tan A(\tan A-1)}$

[molten temple]

wolfqz

[modern warren]

because the entire term changed from + to -

you do realize that $\frac{a}{-b}=\frac{-a}{b}=-\frac{a}{b}$

[molten temple]

wolfqz

[modern warren]

so that's why they're equal

dont worry, it happesn

nah nah add it here

[molten temple]

EEE

EEE

[modern warren]

Hi

sec^2A=1-tan^2A