Question

Which of the following is not an identity for tan(x/2)

Answer 1

Answer:

Options C and D.

Step-by-step explanation:

A. $\frac{(1-cox)}{sinx}=\frac{1-(2cos^{2}\frac{x}{2}-1)}{2sin\frac{x}{2}cos\frac{x}{2}}$

= $\frac{(2-2cos^{2}\frac{x}{2})}{2sin\frac{x}{2}cos\frac{x}{2}}$

= $\frac{2(1-cos^{2}\frac{x}{2})}{2sin\frac{x}{2}cos\frac{x}{2}}$

= $\frac{2sin^{2}\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}$

= $\frac{sin\frac{x}{2} }{cos\frac{x}{2}}$

= $tan\frac{x}{2}$

Therefore, it's an identity for $tan\frac{x}{2}$

B. $\frac{sinx}{1+cosx}=\frac{2sin\frac{x}{2}cos\frac{x}{2}}{1+2cos^{2}\frac{x}{2}-1}$

$=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}$

$=tan\frac{x}{2}$

Therefore, it's an identity $tan\frac{x}{2}$

C. $\frac{cosx}{1-sinx}=\frac{cos^{2}\frac{x}{2}-sin^{2}\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}$

$=\frac{cos\frac{x}{2}}{2sin\frac{x}{2}}-\frac{sin\frac{x}{2}}{2cos\frac{x}{2}}$

$=\frac{1}{2}[cot\frac{x}{2}-tan\frac{x}{2}]$

Therefore, it's not an identity for $tan\frac{x}{2}$

D. $\pm \sqrt{\frac{1-cosx}{1+cosx}}=\pm {\sqrt{\frac{1-(1-2sin^{2}\frac{x}{2})}{1+(2cos^{2}\frac{x}{2}-1)}}}$

$=\pm \sqrt{\frac{sin^{2}\frac{x}{2}}{cos^{2}\frac{x}{2}}}$

$=\pm \sqrt{tan^{2}\frac{x}{2}}$

$=\pm tan\frac{x}{2}$

Therefore, it's not an identity for $tan\frac{x}{2}$

Options C and D are not the identities.

Answer 2

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.