Question

Cos0 (1+tan^2 0) simplifies to

Answer 1

Explanation:

`\cos \theta(1 + { \tan}^(2) \theta) \\ = \cos \theta * { \sec}^(2) \theta \\ = \cos \theta * \frac{1}{{ \cos}^(2) \theta} \\ = \frac{1}{{ \cos} \theta} \\ = \sec \theta \\ \\ \red{ \boxed{ \bold{\therefore \: \cos \theta(1 + { \tan}^(2) \theta)= \sec \theta }}} \\`