Question

Let L denotes the value of $\cos^2(\alpha - \beta)$, if $\sin 2\alpha + \sin 2\beta = \frac{1}{2}$ and $\cos 2\alpha + \cos 2\beta = \frac{\sqrt{3}}{2}$. If M denotes the value of expression $\frac{1}{\log_{xy}xyz} + \frac{1}{\log_{yz}xyz} + \frac{1}{\log_{zx}xyz}$, (Here $x, y, z \in R^+ - \{1\}$) then the value of $(16L^2 + M^2)$, is

• A. 5
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (A)

5

Answer 2

1. Calculate L: Given $\sin 2\alpha + \sin 2\beta = \frac{1}{2}$ and $\cos 2\alpha + \cos 2\beta = \frac{\sqrt{3}}{2}$. Squaring and adding these equations: $(\sin 2\alpha + \sin 2\beta)^2 + (\cos 2\alpha + \cos 2\beta)^2 = (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$ $2 + 2(\cos 2\alpha \cos 2\beta + \sin 2\alpha \sin 2\beta) = \frac{1}{4} + \frac{3}{4} = 1$ $2 + 2\cos(2\alpha - 2\beta) = 1 \implies \cos(2(\alpha - \beta)) = -\frac{1}{2}$. Using $\cos(2\theta) = 2\cos^2\theta - 1$: $-\frac{1}{2} = 2\cos^2(\alpha - \beta) - 1 \implies 2\cos^2(\alpha - \beta) = \frac{1}{2} \implies L = \cos^2(\alpha - \beta) = \frac{1}{4}$.

2. Calculate M: $M = \frac{1}{\log_{xy}xyz} + \frac{1}{\log_{yz}xyz} + \frac{1}{\log_{zx}xyz}$ Using $\frac{1}{\log_a b} = \log_b a$: $M = \log_{xyz} xy + \log_{xyz} yz + \log_{xyz} zx$ Using $\log_b A + \log_b B + \log_b C = \log_b (ABC)$: $M = \log_{xyz} (xy \cdot yz \cdot zx) = \log_{xyz} (x^2 y^2 z^2) = \log_{xyz} (xyz)^2$ Using $\log_b A^k = k \log_b A$: $M = 2 \log_{xyz} (xyz) = 2 \cdot 1 = 2$.

3. Calculate $16L^2 + M^2$: $16L^2 + M^2 = 16\left(\frac{1}{4}\right)^2 + (2)^2 = 16\left(\frac{1}{16}\right) + 4 = 1 + 4 = 5$.