Question

Only for Brainly moderators , Brainly stars and brainly best users !▬▬ஜ۩۞۩ஜ▬▬ 1 + cot A + tan a into Sin A minus Cos A is equal to sec a upon cos square A minus Cos A upon sec square A..​ Solve on paper​

Answer 1

Answer:

hey here is your proof in above pics

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Step-by-step explanation:

$to \: prove = \\ (1 + cot \: A + tan \: A)(sin \: A - cos \: A) = \frac{sec \:A }{cosec {}^{2} \:A } - \frac{cosec \: A}{sec \: {}^{2} A} \\ \\ so \: let \:then \\ LHS = (1 + tan \: A + cot \: A)(sin \: A - cos \: A) \\ \\ RHS = \frac{sec \:A}{cosec {}^{2} \:A } - \frac{cosec \:A}{sec {}^{2} \:A } \\ \\ considering \: LHS \: first \\ = (1 + tan \: A + cot \: A)(sin \: A - cos \: A) \\ \\ = (1 + \frac{sin \:A }{cos \: A} \: + \: \frac{cos \:A }{sin \:A } \: )(sin \: A - cos \: A \: ) \\ \\ = (1 + \frac{sin {}^{2} \: A + cos {}^{2} \:A }{sin \:A.cos \: A } \: )(sin \: A - cos \: A) \\ \\ =( \: 1 + \frac{1}{sin \: A.cos \:A } \: )(sin \: A - cos \: A) \\ \\ =( \: \frac{1 + sin \:A.cos \: A}{sin \: A.cos \:A } \: )(sin \: A - cos \: A)$

$now \: considering \: RHS \\ \\ = \frac{sec \: A}{cosec \: {}^{2} \: A } \: - \: \frac{cosec \:A }{sec {}^{2} \: A } \\ \\ = \frac{sec {}^{3} \:A - cosec {}^{3} \: A}{cosec {}^{2} \: A.sec {}^{2} \: A} \\ \\ = \frac{ \frac{1}{cos {}^{3} \: A } - \frac{1}{sin {}^{3} \: A} }{ \frac{1}{sin {}^{2} \:A } \times \frac{1}{cos {}^{2} \: A } } \\ \\ = \frac{ \frac{ \frac{sin {}^{3} \: A \: - \: cos {}^{3} \: A }{sin {}^{3} \: A.cos {}^{3} \: A } }{1} }{sin {}^{2} \: A.cos {}^{2} \:A } \\ \\ = \frac{sin {}^{3} \: A - cos {}^{3} \:A }{sin \:A.cos \: A}$

$so \: here \: \\ sin {}^{3} \: A - cos {}^{3} \: A \: \: is \: in \: form \: a {}^{3} \: - \: b {}^{3} \\ \\ so \: we \: know \: that \\ a {}^{3} \: - \: b {}^{3} = (a - b)(a {}^{2} + b {}^{2} + ab) \\ \\ hence \: accordingly \\ \\ = \frac{(sin \: A - cos \: A)(sin {}^{2} \:A + cos {}^{2} \: A + sin \: A.cos \:A) }{sin \: A.cos \:A } \\ \\ = \frac{(sin \: A - cos \: A)(1 + sin \:A.cos \: A) }{sin \: A.cos \:A } \\ \\ hence \: then \\ \: LHS=RHS \\ \\ thus \: proved$