A) \[\sin \frac{A}{2}\]
B) \[\cos \frac{A}{2}\]
C) \[\tan \frac{A}{2}\]
D) \[\cot \frac{A}{2}\]
Correct Answer: C
Solution :
\[\frac{1+\sin A-\cos A}{1+\sin A+\cos A}\] \[=\frac{2\,{{\sin }^{2}}\frac{A}{2}+2\,\sin \frac{A}{2}\cos \frac{A}{2}}{2\,{{\cos }^{2}}\frac{A}{2}+2\,\sin \frac{A}{2}\cos \frac{A}{2}}\] \[=\frac{2\,\,\sin \frac{A}{2}\,\left( \sin \frac{A}{2}+\cos \frac{A}{2} \right)}{2\,\,\cos \frac{A}{2}\,\left( \cos \frac{A}{2}+\sin \frac{A}{2} \right)}\]=\[\tan \frac{A}{2}\]. Trick: Put \[A={{60}^{o}}.\] Then \[\frac{1+(\sqrt{3}/2)-(1/2)}{1+(\sqrt{3}/2)+(1/2)}=\frac{1+\sqrt{3}}{3+\sqrt{3}}=\frac{1}{\sqrt{3}}\] which is given by option , i.e. \[\tan \frac{{{60}^{o}}}{2}=\frac{1}{\sqrt{3}}\] Note: Students should remember at the time of assuming the values of A, B, q, ..... etc. that, for the assumed values, the options must have different values.
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