A) 0
B) 1
C) -1
D) 2
Correct Answer: C
Solution :
Let\[\frac{\beta +\gamma -\alpha }{4}=A,\frac{\gamma +\alpha -\beta }{4}=B\And \frac{\alpha +\beta -\gamma }{4}=C\] \[\therefore \]\[A+B+C=\frac{\alpha +\beta +\gamma }{4}=\frac{\pi }{4}\Rightarrow \pi \tan A=1\] \[\Rightarrow \]\[\frac{\sin A.\sin B}{\cos A.\cos B}=\frac{1}{\tan C}\] \[\Rightarrow \]\[-\frac{\cos (A+B)}{\cos (A-B)}=\tan \left( \frac{\pi }{4}-C \right)\] (on applying compodendo & dividendo) \[\Rightarrow \]\[2\sin \left( \frac{\pi }{4}-C \right)\cos (A-B)+2cos\left( \frac{\pi }{4}-C \right)\cos (A+B)=0\] \[\Rightarrow \]\[\sin \left( \frac{\pi }{4}-C+A-B \right)+\sin \left( \frac{\pi }{4}-C-A+B \right)+\] \[\sin \left( \frac{\pi }{4}-C-A-B \right)+\sin \left( \frac{\pi }{4}-C+A+B \right)=0\] As\[A-B-C=\frac{\pi }{4}-\alpha ,B-C-A=\frac{\pi }{4}-\beta ,\] \[C-A-B=\frac{\pi }{4}-\gamma ,A+B+C=\frac{\pi }{4}\] \[\Rightarrow \]\[\sin \left( \frac{\pi }{4}-\alpha \right)+\sin \left( \frac{\pi }{2}-\beta \right)+\sin \left( \frac{\pi }{2}-\gamma \right)+\sin \frac{\pi }{2}=0\] \[\therefore \]\[\Sigma \cos \alpha =-1\]
You need to login to perform this action.
You will be redirected in
3 sec
