A) I and II
B) I and III
C) None of these
D) All of these
Correct Answer: A
Solution :
I. \[{{\sin }^{2}}\,\,(n+1)A-{{\sin }^{2}}nA\] |
\[=\sin \,\,[(n+1)A+n\,A]\sin \,\,[(n+1)A-nA]\] |
\[[\because {{\sin }^{2}}A-{{\sin }^{2}}B=\sin \,\,(A+B)\sin \,\,(A-B)]\] |
\[=\sin \,\,(2n+1)A\sin A\] |
II. \[{{\cos }^{2}}\,\,(45{}^\circ +x)-{{\sin }^{2}}(45{}^\circ -x)\] |
\[=\cos \,\,(45{}^\circ +x+45{}^\circ -x)\cos (45{}^\circ +x-45{}^\circ +x)\] |
\[[\because {{\cos }^{2}}A-{{\sin }^{2}}B=\cos \,\,(A+B)\cos \,\,(A-B)]\] |
\[=\cos 90{}^\circ \cdot \cos 2x=0\] |
Which is independent of x. |
You need to login to perform this action.
You will be redirected in
3 sec