Question

${(1+\tan \alpha \tan \beta )}^2+{(\tan \alpha -\tan \beta )}^2=$

• A. ${\sec }^2\alpha {\sec }^2\beta$
  
• B. ${\sin }^2\alpha {\sin }^2\beta$
  
• C. ${\cos }^2\alpha {\cos }^2\beta$
  
• D. $1$

Answer 1

Answer: (A)

${\sec }^2\alpha {\sec }^2\beta$

L.H.S$={(1+\tan \alpha \tan \beta )}^2+{(\tan \alpha -\tan \beta )}^2$

$=1+{\tan }^2\alpha {\tan }^2\beta +2\tan \alpha \tan \beta +{\tan }^2\alpha +{\tan }^2\beta -2\tan \alpha \tan \beta$

$=1+{\tan }^2\alpha {\tan }^2\beta +{\tan }^2\alpha +{\tan }^2\beta$

$=(1+{\tan }^2\alpha )+{\tan }^2\beta ({\tan }^2\alpha +1)$

$={\sec }^2\alpha +{\tan }^2\beta {\sec }^2\alpha$

$={\sec }^2\alpha (1+{\tan }^2\beta )$

$={\sec }^2\alpha {\sec }^2\beta$