| \[\sin \,a+7\,\sin \,b=4\,(\sin \,c+2\,\sin \,d)\] |
| \[\cos \,a+7\,\cos \,b=4\,(\cos \,c+2\,\cos \,d)\] |
| and the numerical value of \[\frac{7\,\cos \,(b-c)}{\cos \,(a-d)}\] is m. the correct statement is/are |
A) m is a prime number
B) n is a composite number
C) m, n both are perfect square
D) m is divisible by n.
Correct Answer: A
Solution :
\[(1+\tan \,{{1}^{o}})\,(1+\tan \,{{2}^{o}})\,(1+\tan \,{{2}^{o}})\] \[....(1+\tan \,{{45}^{o}})={{2}^{n}}\] \[[(1+\tan \,{{1}^{o}})\,(1+\tan \,{{44}^{o}})]\,\,[(1+\tan \,{{2}^{o}})\] \[(1+\tan \,{{43}^{o}})]...[(1+\tan \,{{22}^{o}})\] \[(1+\tan \,{{23}^{o}})].\,(1+\tan \,{{45}^{o}})\] \[\left[ \begin{align} & \because \,\,(1+\tan \,{{1}^{o}})\,(1+\tan \,{{44}^{o}}) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,1+\,\tan \,{{1}^{o}}\,+\tan \,{{44}^{o}}+\tan \,{{1}^{o}}\,\tan \,{{44}^{o}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=1+\tan \,{{1}^{o}}\,\tan \,{{44}^{o}}+(\tan \,{{1}^{o}}+\tan \,{{44}^{o}}) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,1+\tan \,{{45}^{o}}=2 \\ \end{align} \right]\] \[=\,\underbrace{2.\,2.\,2....2.2.(2)={{2}^{23}}}_{22\,\text{times}}\] \[\Rightarrow \] \[n=23\] \[\sin \,a+7\,\sin \,b=4\,\sin \,c+8\,\sin \,d\] \[\sin \,a-8\,\sin \,d=4\,\sin \,c-7\,\sin \,b\] ?(i) and second equaiton is \[\cos \,a-8\,\cos \,d=4\,\cos \,c-7\,\sin \,b\] ?(ii) On squaring and adding Eqs. (i) and (ii), we get \[1+{{8}^{2}}-16\,(\sin \,a\,\sin \,d\,+\cos \,a\,\cos \,d)\] \[=16+49-56\,(\sin \,b\,\,\sin \,c+\,\cos \,b\,\cos \,c)\] \[\Rightarrow \] \[-16\,\cos \,(a-d)=-\,56\,\cos \,(b-c)\] \[\Rightarrow \] \[\frac{\cos \,(b-c)}{\cos \,(a-d)}=\frac{2}{7}\] \[\Rightarrow \] \[\frac{7\,\cos \,(b-c)}{\cos \,(a-d)}=2\]
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