A) Only I
B) Only II
C) Both I and II
D) None of these
Correct Answer: B
Solution :
As,\[A+B+C+D=360{}^\circ \] |
\[\therefore \] \[A+B=360{}^\circ -\,\,(C+D)\] |
\[\therefore \]\[\sin \,\,(A+B)=\sin \,\,[360{}^\circ -\,\,(C+D)]=-\sin \,\,(C+D)\] |
\[\therefore \]\[\sin \,\,(A+B)+\sin \,\,(C+D)=0\] |
Also, \[\cos \,\,(A+B)=cos\,\,[360{}^\circ -(C+D)]\] |
\[\cos \,(A+B)=\cos \,\,(C+D)\] |
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