Question

prove that sec a -1/sec A+1=1-cos a/1+cos a​

Answer 1

$\mathtt\green{answer \: is \: in \: given \: attachment}$

Answer 2

Answer:

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Step-by-step explanation:

$so \: here \: given \: trigonometric \: equation \: is \: \\ sec \: a - 1 \div sec \: a + 1 = 1 - cos \: a \div 1 + cos \: a$

$let \: lhs = sec \: a - 1 \div sec \: a + 1 \\ rhs = 1 - cos \: a \div 1 + cos \: a$

$so \: considering \: lhs \\ sec \: a - 1 \div sec \: a + 1 \\ = (1 \div cos - 1) \div (1 \div cos \: a + 1) \\ since \:sec \: a = 1 \div cos \: a \\ \\ = 1 - cos \: a \div cos \: a \div 1 + cos \: a \div cos \: a \\ = 1 - cos \: a \div 1 \ + cos \: a$

$hence \: lhs = rhs \\ thus \: proved$