Question

If $tan{\theta }_{1}tan{\theta }_2=k,$ then $\frac{cos({\theta }_1-{\theta }_2)}{cos({\theta }_1+{\theta }_2)}$

• A. $\frac{1+k}{1-k}$
  
• B. $\frac{1-k}{1+k}$
  
• C. $\frac{k+1}{k-1}$
  
• D. $\frac{k-1}{k+1}$

Answer 1

The correct option is A $\frac{1+k}{1-k}$


$\frac{cos({\theta }_1-{\theta }_2)}{cos({\theta }_1+{\theta }_2)}$
= $\frac{cos{\theta }_{1}cos\theta 2+sin{\theta }_{1}sin{\theta }_2}{cos{\theta }_{1}cos{\theta }_2-sin{\theta }_{1}sin{\theta }_2}$
Dividing numerator and denominator by $cos{\theta }_{1}cos{\theta }_2$, we get:
$\frac{1+tan{\theta }_{1}tan{\theta }_2}{1-tan{\theta }_{1}tan{\theta }_2}$
$=\frac{1+k}{1-k}$