Question

solve this question
first change into sin and cos
and then solving with releasizing ​

Answer 1

Answer:

answer for the given problem is given

Answer 2

Answer:

$= \frac{\sec \: 0 - 1 }{ \sec \: 0 + 1} \: \: since \: \sec \: 0 = \frac{1}{ \cos \: 0 } \\ = \frac{ \frac{1}{ \cos \: 0 } - 1}{ \frac{1}{ \cos \: 0 } + 1 } \\ = \frac{ \frac{1 - \cos \: 0 }{ \cos \: 0 } }{ \frac{1 + \cos \: 0}{ \cos \: 0 } } \\ = \frac{1 - \cos \: 0}{1 + \cos \: 0} \: \: now \: multiply \: by \: (1 + \cos \: 0)on \: both \: sides \\ = \frac{(1 - \cos \: 0)(1 + \cos \: 0)}{(1 + \cos \: 0)(1 + \cos \: 0)} \\ = \frac{1 - \cos ^{2} \: 0 }{(cos \: 0 {)}^{2} } \\ = \frac {\sin{}^{2} \: 0}{(1 + \cos \: 0)}$

Hence proved