Question

prove that (tan^2 a/sec a +1)+1= sec a​

Answer 1

Answer:

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Step-by-step explanation:

$let \: lhs = (\tan \: square \: a \div sec \: a )+ 1$

$rhs = sec \: a$

$so \: considering \: lhs \: first \\ we \: get \\ (tan \: square \: a \div sec \: a+1) + 1 \\ \\ so \: we \: know \: that \: \\ tan \: square \: a = sec \: square - 1$

$= (sec \: square - 1 \div sec \: a - 1) + 1$

$= ((sec \: a + 1)(sec \: a - 1) \div (sec \: a + 1)) + 1$

$= sec \: a - 1 + 1$

$= sec \: a$

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