Question

The value of the expression $\frac{\log _8(1+2 \cos 2 \theta)^{3 / 4}+\log _2(\sin \theta)^{1 / 4}}{\log _{16}(\operatorname{cosec} 3 \theta)}$ (whenever defined) is equal to

• A. $\log _7 6 \cdot \log _6 7$
• B. $(\sqrt{2}+1) \cot \frac{5 \pi}{8}$
• C. $\log _7 6 \cdot \log _6\left(\frac{1}{7}\right)$
• D. $(\sqrt{2}-1) \tan \frac{3 \pi}{8}$

Answer 1

Answer: (B), (C)

$E =\frac{\frac{1}{4} \log _2(1+2 \cos 2 \theta)+\frac{1}{4} \log _2 \sin \theta}{-\frac{1}{4} \log _2 \sin 3 \theta} $
$=\frac{\log _2(\sin \theta+2 \cos 2 \theta \sin \theta)}{-\log _2 \sin 3 \theta}=\frac{\log _2(\sin \theta+\sin 3 \theta-\sin \theta)}{-\log _2 \sin 3 \theta}=-1$
Now,
(A) $\log _7 6 \cdot \log _6 7=1$
(B) $(\sqrt{2}+1) \cos \frac{5 \pi}{8}=(\sqrt{2}+1)(-(\sqrt{2}-1))=-1$
(D) $(\sqrt{2}-1) \tan \frac{3 \pi}{8}=(\sqrt{2}-1)(\sqrt{2}+1)=1$
Hence, options (B) and (C) are correct