Question

prove that : (1 + sin a - cos a)/1+sin a + cos a = tan 1/2 a

Answer 1

\[\frac{(1 + \sin a - \cos a)}{1+\sin a + \cos a} = \tan \frac{1}{2} a\]
\[LHS=\frac{(1 + \sin a - \cos a)}{1+\sin a + \cos a} \]
\[=\frac{(1 + 2\sin \frac{a}{2}\cos \frac{a}{2} - (1-2\sin^2 \frac{a}{2})}{1+2\sin \frac{a}{2}\cos \frac{a}{2} +2\cos^2 \frac{a}{2}-1} \]
\[=\frac{(1 + 2\sin \frac{a}{2}\cos \frac{a}{2} - 1+2\sin^2 \frac{a}{2})}{1+2\sin \frac{a}{2}\cos \frac{a}{2} +2\cos^2 \frac{a}{2}-1} \]
\[=\frac{( 2\sin \frac{a}{2}\cos \frac{a}{2} +2\sin^2 \frac{a}{2})}{2\sin \frac{a}{2}\cos \frac{a}{2} +2\cos^2 \frac{a}{2}} \]
\[=\frac{2\sin \frac{a}{2}( \cos \frac{a}{2} +\sin \frac{a}{2})}{2\cos \frac{a}{2} (\sin \frac{a}{2}+\cos \frac{a}{2})} \]
\[=\frac{\sin \frac{a}{2}}{\cos \frac{a}{2} } = \tan \frac{a}{2} = RHS\]
Thus proved

@ninik

Answer 2

or \[= \tan \frac{1}{2}a\]