Question

$(1+tan\alpha tan\beta {)}^2+(tan\alpha -tan\beta {)}^2$ is equal to

• A. $ta{n}^2\alpha +ta{n}^2\beta$
• B. $co{s}^2\alpha co{s}^2\beta$
• C. $se{c}^2\alpha se{c}^2\beta$
• D. $ta{n}^2\alpha ta{n}^2\beta$

Answer 1

Answer: (C)

$se{c}^2\alpha se{c}^2\beta$

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

$=1+ta{n}^2\alpha ta{n}^2\beta +ta{n}^2\alpha +ta{n}^2\beta$

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

$=(1+ta{n}^2\alpha )(1+ta{n}^2\beta )$
$=se{c}^2\alpha se{c}^2\beta$