Problem: If b>1,sinx>0,cosx>0b>1, \sin x>0, \cos x>0b>1,sinx>0,cosx>0 and bsinx=a\log _{b} \sin x=alogbsinx=a, then bcosx\log _{b} \cos xlogbcosx equals
Answer Choices:
A. 2b(1−ba/2)2 \log _{b}\left(1-b^{a / 2}\right)2logb(1−ba/2)
B. 1−a2\sqrt{1-a^{2}}1−a2
C. ba2b^{a^{2}}ba2
D. 12b(1−b2a)\dfrac{1}{2} \log _{b}\left(1-b^{2 a}\right)21logb(1−b2a)
E. none of these
Solution:
bsinx=a;sinx=ba\log _{b} \sin x=a ; \sin x=b^{a}logbsinx=a;sinx=ba;
2x=b2a;cosx=(1−b2a)1/2;\sin ^{2} x=b^{2 a} ; \cos x=\left(1-b^{2 a}\right)^{1 / 2} ;sin2x=b2a;cosx=(1−b2a)1/2;
bcosx=12b(1−b2a)\log _{b} \cos x=\dfrac{1}{2} \log _{b}\left(1-b^{2 a}\right)logbcosx=21logb(1−b2a).