$$4\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right) \text { is equal to }$$

$$ \text { If } \sin A=\frac{4}{5} \text { and } \cos B=\frac{-12}{13} \text { where } A \text { and } B \text { lie in first and third quadrant respectively. Then } \cos (A+B)= $$

If $$\cos A=m \cos B$$ and $$\cot \left(\frac{A+B}{2}\right)=\lambda \tan \left(\frac{B-A}{2}\right)$$, then $$\lambda$$ is equal to

The expression $$\frac{2 \tan A}{1-\cot A}+\frac{2 \cot A}{1-\tan A}$$ can be written as