A) I and II
B) I, II and III
C) I, II and IV
D) All of these
Correct Answer: C
Solution :
As, \[A+B+C=180{}^\circ \] |
I. \[\sin \,\,(A+B)=\sin \,\,(180{}^\circ -C)=\sin C\] |
II. \[\sin \left( \frac{A+B}{2} \right)=\sin \left( \frac{180{}^\circ -C}{2} \right)\] |
\[\Rightarrow \]\[\sin \left( 90{}^\circ -\frac{C}{2} \right)=\cos \frac{C}{2}\] |
III. \[=\tan \left( \frac{A+B+C}{2} \right)=\tan \frac{(180{}^\circ -C)-C}{2}\] |
\[=\tan \frac{(180{}^\circ -2\,C)}{2}\] |
IV. \[\tan \frac{(A-B-C)}{2}=\tan \frac{A-(B+C)}{2}\] |
\[=\tan \frac{A-(180{}^\circ -A)}{2}\] |
\[=\tan [-(90{}^\circ -A)]\] |
\[=-\tan \,\,(90{}^\circ -A)=-\cot A\] |
You need to login to perform this action.
You will be redirected in
3 sec