Question

tanx/2=tanx/secx+1
Need help.

Answer 1

\[
\cos(2x) = 2\cos^2(x) - 1 \\
\cos^2(x) = \frac 12(1 + \cos(2x)) \\
\text{ } \\
\cos(2x) = 1 - 2\sin^2(x) \\
\sin^2(x) = \frac 12(1 - \cos(2x)) \\
\tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)} = \frac{1-\cos(2x)}{1+\cos(2x)} \\
\tan^2\left(\frac x2\right) = \frac{1-\cos(x)}{1+\cos(x)} =
\frac{1-\cos(x)}{1+\cos(x)}*\frac{1+\cos(x)}{1+\cos(x)} =
\frac{1-\cos^2(x)}{(1+\cos(x))^2} \\
\text{ } \\
\tan^2\left(\frac x2\right) = \frac{\sin^2(x)}{(1+\cos(x))^2} \\
\tan\left(\frac x2\right) = \frac{\sin(x)}{1+\cos(x)} =
\large \frac{\frac{\sin(x)}{\cos(x)}}{\frac{1+\cos(x)}{\cos(x)}} = \normalsize
\frac{\tan(x)}{\sec(x)+1}
\]