Question

If $3{\tan }^{2}A{\tan }^{2}B=1$, then $(2-\cos 2A)(2-\cos 2B)=$

• A. $5$
• B. $4$
• C. $3$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $3$

$(2-\cos 2A)(2-\cos 2B)=\left(\frac{1+3{\tan }^{2}A}{1+{\tan }^{2}A}\right)\left(\frac{1+3{\tan }^{2}B}{1+{\tan }^{2}B}\right)$

$=\frac{1+3({\tan }^{2}A+{\tan }^{2}B)+9{\tan }^{2}A{\tan }^{2}B}{1+{\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}A{\tan }^{2}B}$

$=\frac{4+3({\tan }^{2}A+{\tan }^{2}B)}{\frac{4}{3}+{\tan }^{2}A+{\tan }^{2}B}=3$