Trigonometric Identities Question 161
Question: $ \frac{1+\sin A-\cos A}{1+\sin A+\cos A} $ =
Options:
A) $ \sin \frac{A}{2} $
B) $ \cos \frac{A}{2} $
C) $ \tan \frac{A}{2} $
D) $ \cot \frac{A}{2} $
Show Answer
Answer:
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},( \sin \frac{A}{2}+\cos \frac{A}{2} )}{2\cos \frac{A}{2},( \cos \frac{A}{2}+\sin \frac{A}{2} )} $ = $ \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.
BETA
