Question #fbd74

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Answer 1 · 2017-03-19T03:09:52

$(cos^2x + tan^2x - 1)/tan^2x =sin^2x$

Work on left side:

$(cos^2x + tan^2x - (cos^2x+sin^2x))/tan^2x$

$cos^2x$ cancels, leaving:

$(tan^2x-sin^2x)/tan^2x$

Change tan to $sin/cos$ :

$((sin^2x/cos^2x)-(Sin^2x/1))/(sin^2x/cos^2x)$

Multiply $(Sin^2x/1)$ by $(cos^2x/cos^2x)$ to get a common denominator:

$((sin^2x/cos^2x)-((Sin^2xcos^2x)/cos^2x))/(sin^2x/cos^2x)$

$((Sin^2x-sin^2xcos^2x)/cos^2x)/(sin^2x/cos^2x)$

$cos^2x$ cancels:

$(sin^2x-sin^2xcos^2x)/sin^2x$

Factor the numerator:

$(sin^2x(1-cos^2x))/sin^2x$

$sin^2x$ cancels:

$1-cos^2x$ which is also $sin^2x$ = right side

Answer 2 · 2017-03-19T04:45:35

$LHS=(cos^2x + tan^2x - 1)/tan^2x$

$=( tan^2x - (1-cos^2x))/tan^2x$

$=( tan^2x - sin^2x)/tan^2x$

$= tan^2x/tan^2x - sin^2x/tan^2x$

$=1 - sin^2x/(sin^2x/cos^2x)$

$=1 - cos^2x=sin^2x=RHS$