Question

Please help me to prove (1-sin x)/(1+sin x)= tan^2 (pi/4 - x/2).

Answer 1

sad... nobody helps me.....

Answer 2

Wait, I will help you. Needs a lot of typing.

Answer 3

I almost want to close this question...

Answer 4

\[\frac{1-\sin (x)}{\sin
(x)+1}=\frac{(1-\sin (x)) (\sin
(x)+1)}{(\sin (x)+1) (\sin
(x)+1)}=\frac{1-\sin ^2(x)}{(\sin
(x)+1)^2}=

\]

Answer 5

\[\frac{1-\sin ^2(x)}{(\sin
(x)+1)^2}= \frac {\cos^2(x)}{(\sin
(x)+1)^2}
\]

Answer 6

We need to show that now
\[
\frac{\cos (x)}{\sin
(x)+1}=\tan \left(\frac{\pi
}{4}-\frac{x}{2}\right)
\]

Answer 7

\[\tan \left(\frac{\pi
}{4}-\frac{x}{2}\right)=\frac{\sin \left(\frac{\pi
}{4}-\frac{x}{2}\right)}{\cos \left(\frac{\pi
}{4}-\frac{x}{2}\right)}=\frac{\cos
\left(\frac{x}{2}\right)-\sin
\left(\frac{x}{2}\right)}{\sin
\left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)}=\\
\frac{\left(\cos
\left(\frac{x}{2}\right)-\sin
\left(\frac{x}{2}\right)\right) \left(\sin
\left(\frac{x}{2}\right)+\cos
\left(\frac{x}{2}\right)\right)}{\left(\sin
\left(\frac{x}{2}\right)+\cos
\left(\frac{x}{2}\right)\right) \left(\sin
\left(\frac{x}{2}\right)+\cos
\left(\frac{x}{2}\right)\right)}=\\
\frac{\cos^2\left(\frac{x}{2}\right)-\
\sin^2\left(\frac{x}{2}\right)}{
\left(\sin
\left(\frac{x}{2}\right)+\cos
\left(\frac{x}{2}\right)\right)^2}=
\frac{\cos (x)}{\sin
^2\left(\frac{x}{2}\right)+\
cos
^2\left(\frac{x}{2}\right)+2
\sin
\left(\frac{x}{2}\right)
\cos
\left(\frac{x}{2}\right)}=\\

\frac{\cos (x)}{1+2
\sin
\left(\frac{x}{2}\right)
\cos
\left(\frac{x}{2}\right)}=\frac{\cos (x)}{1+\sin(x)}
\]

Answer 8

We are done.

Answer 9

Wow ~ Thanks!!

Answer 10

yw

Answer 11

Another quicker way
\[
\tan \left(\frac{\pi
}{4}-\frac{x}{2}\right)=\frac{1-\tan
\left(\frac{x}{2}\right)}{\tan
\left(\frac{x}{2}\right)+1}=\frac{\cos
\left(\frac{x}{2}\right)-\sin
\left(\frac{x}{2}\right)}{\sin
\left(\frac{x}{2}\right)+\cos
\left(\frac{x}{2}\right)}=\frac{\cos
(x)}{\sin (x)+1}
\]