Question 821540
{{{2/(1+cosx) - tan^2(x/2) = 1}}}
There are three variations of tan((1/2)x):<ul><li>{{{tan((1/2)x)}}} = <u>+</u>{{{sqrt((1-cos(x))/(1+cos(x)))}}}</li><li>{{{tan((1/2)x) = sin(x)/(1+cos(x))}}}</li><li>{{{tan((1/2)x) = (1-cos(x))/sin(x)}}}</li></ul>Since our tan is squared, I'm going to use the first one (and, since we're squaring it, the <u>+</u> is not needed):
{{{2/(1+cosx) - (sqrt((1-cos(x))/(1+cos(x))))^2 = 1}}}
Simplifying:
{{{2/(1+cosx) - (1-cos(x))/(1+cos(x)) = 1}}}
Changing the subtraction between the fractions into an equivalent addition (since this situation is a source of much confusion and many errors):
{{{2/(1+cosx) +  (- (1-cos(x)))/(1+cos(x)) = 1}}}
which simplifies to:
{{{2/(1+cosx) +  (-1+cos(x))/(1+cos(x)) = 1}}}
The denominators are equal so we can add the fractions:
{{{(1+cos(x))/(1+cos(x)) = 1}}}
And since the numerator and denominator are the same:
{{{1=1}}}