Prove the identity

[Math Tutor]

Prove the identity

[tex]\frac{1+\sin x-\cos x}{1+\sin x+\cos x}=\tan\frac{x}{2}[/tex]

for every [tex]x[/tex] for which both sides are defined.

Then state exactly which values of [tex]x[/tex] must be excluded.

[nycmath]

Prove the identity

[tex]\frac{1+\sin x-\cos x}{1+\sin x+\cos x}=\tan\frac{x}{2}[/tex]

for every [tex]x[/tex] for which both sides are defined.

Then state exactly which values of [tex]x[/tex] must be excluded.

I will work on this on my days off to provide a detailed answer.

[nycmath]

Math Tutor

This one is a bit tough. The expression tan (pi/2) DNE because the tangent function has vertical asymptotes at odd multiples of pi/2.

Stuck here. ....

[Math Tutor]

Put everything in terms of [tex]\frac{x}{2}[/tex], but use a different form of the double angle for cosine top and bottom: [tex]\cos x=1-2\sin^2\frac{x}{2}[/tex] in the numerator, [tex]\cos x=2\cos^2\frac{x}{2}-1[/tex] in the denominator. Both 1's cancel and what's left factors.

[Guest]

tg[tex]\frac{x}{2}[/tex]=tg[tex]\frac{x}{2}[/tex]

[tex]\frac{sin \frac{x}{2} }{cos \frac{x}{2} }[/tex]=tg[tex]\frac{x}{2}[/tex]

[tex]\frac{sin \frac{x}{2}.2(sin \frac{x}{2}+cos \frac{x}{2}) }{cos \frac{x}{2}.2(cos \frac{x}{2} +sin \frac{x}{2}) }[/tex]=tg[tex]\frac{x}{2}[/tex]

[tex]\frac{2 sin^{2 } \frac{x}{2} + 2sin \frac{x}{2} .cos\frac{x}{2} }{2 cos^{2 } \frac{x}{2} +2sin \frac{x}{2}.cos \frac{x}{2} }[/tex] =tg[tex]\frac{x}{2}[/tex]

[tex]\frac{1-cosx+sinx}{1+cosx+sinx}=tg \frac{x}{2}[/tex]

[tex]\frac{1+sinx-cosx}{1+sinx+cosx}=tg \frac{x}{2}[/tex]

decision from Bulgaria

[nycmath]

tg[tex]\frac{x}{2}[/tex]=tg[tex]\frac{x}{2}[/tex]

[tex]\frac{sin \frac{x}{2} }{cos \frac{x}{2} }[/tex]=tg[tex]\frac{x}{2}[/tex]

[tex]\frac{sin \frac{x}{2}.2(sin \frac{x}{2}+cos \frac{x}{2}) }{cos \frac{x}{2}.2(cos \frac{x}{2} +sin \frac{x}{2}) }[/tex]=tg[tex]\frac{x}{2}[/tex]

[tex]\frac{2 sin^{2 } \frac{x}{2} + 2sin \frac{x}{2} .cos\frac{x}{2} }{2 cos^{2 } \frac{x}{2} +2sin \frac{x}{2}.cos \frac{x}{2} }[/tex] =tg[tex]\frac{x}{2}[/tex]

[tex]\frac{1-cosx+sinx}{1+cosx+sinx}=tg \frac{x}{2}[/tex]

[tex]\frac{1+sinx-cosx}{1+sinx+cosx}=tg \frac{x}{2}[/tex]

decision from Bulgaria

Thank you. Nicely-done!

[nycmath]

Put everything in terms of [tex]\frac{x}{2}[/tex], but use a different form of the double angle for cosine top and bottom: [tex]\cos x=1-2\sin^2\frac{x}{2}[/tex] in the numerator, [tex]\cos x=2\cos^2\frac{x}{2}-1[/tex] in the denominator. Both 1's cancel and what's left factors.

Thanks. Will try again on days off.