Question

Prove that cos theta divided by one plus sin theta minus one minus sin theta dived by cos theta equals 0

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \\ (cos \: theta \div 1 + sin \: theta) - (1 + sin \: theta \div cos \: theta) = 0$

$let \: lhs = (cos \: theta \div 1 + sin \: theta) - (1 - sin \: theta \div cos) \\ rhs = 0$

$considering \: lhs \: first \\ = cos \: theta \div 1 + sin \: theta - 1 - sin \: theta \div cos \: theta \\ = cos \: theta.cos \: theta - (1 - sin \: theta)(1 + sin \: theta) \div (1 + sin \: theta).cos \: theta$

$= cos \: square \: theta - (1 - sin \: square \: theta) \div (1 + sin \: theta).cos \: theta \\ = cos \: square - cos \: square \: theta \div (1 + sin \: theta).cos \: theta$

$= 0 \div (1 + sin \: theta).cos \: theta \\ = 0$

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