Respuesta :
Answer:
C and D.
Step-by-step explanation:
Check each one using x = 60 degrees:
A. tan(x/2) = tan 30 = 0.5774
1 - cos 60 / sin 60 = 0.5774 So A is an identity
B. sin 60 / (1 + cos 60) = 0.5774 :- B is an identity.
C. cos 60 / ( 1 - sin 60) = 3.732 so This is NOT an identity.
D +/-sqrt( (1 - cos60)/(1 + cos 60)) = +/- 0.5774. Because oif the +/- I don't think this is an identity. Sorry I can't be sure.
Answer:
Options C and D.
Step-by-step explanation:
A. [tex]\frac{(1-cox)}{sinx}=\frac{1-(2cos^{2}\frac{x}{2}-1)}{2sin\frac{x}{2}cos\frac{x}{2}}[/tex]
= [tex]\frac{(2-2cos^{2}\frac{x}{2})}{2sin\frac{x}{2}cos\frac{x}{2}}[/tex]
= [tex]\frac{2(1-cos^{2}\frac{x}{2})}{2sin\frac{x}{2}cos\frac{x}{2}}[/tex]
= [tex]\frac{2sin^{2}\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}[/tex]
= [tex]\frac{sin\frac{x}{2} }{cos\frac{x}{2}}[/tex]
= [tex]tan\frac{x}{2}[/tex]
Therefore, it's an identity for [tex]tan\frac{x}{2}[/tex]
B. [tex]\frac{sinx}{1+cosx}=\frac{2sin\frac{x}{2}cos\frac{x}{2}}{1+2cos^{2}\frac{x}{2}-1}[/tex]
[tex]=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}[/tex]
[tex]=tan\frac{x}{2}[/tex]
Therefore, it's an identity [tex]tan\frac{x}{2}[/tex]
C. [tex]\frac{cosx}{1-sinx}=\frac{cos^{2}\frac{x}{2}-sin^{2}\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}[/tex]
[tex]=\frac{cos\frac{x}{2}}{2sin\frac{x}{2}}-\frac{sin\frac{x}{2}}{2cos\frac{x}{2}}[/tex]
[tex]=\frac{1}{2}[cot\frac{x}{2}-tan\frac{x}{2}][/tex]
Therefore, it's not an identity for [tex]tan\frac{x}{2}[/tex]
D. [tex]\pm \sqrt{\frac{1-cosx}{1+cosx}}=\pm {\sqrt{\frac{1-(1-2sin^{2}\frac{x}{2})}{1+(2cos^{2}\frac{x}{2}-1)}}}[/tex]
[tex]=\pm \sqrt{\frac{sin^{2}\frac{x}{2}}{cos^{2}\frac{x}{2}}}[/tex]
[tex]=\pm \sqrt{tan^{2}\frac{x}{2}}[/tex]
[tex]=\pm tan\frac{x}{2}[/tex]
Therefore, it's not an identity for [tex]tan\frac{x}{2}[/tex]
Options C and D are not the identities.
