Question

prove that bracket sin theta + 1 + cos theta bracket close bracket sin theta minus 1 + cos theta bracket close into sec theta cosec theta is equal to 2​

Answer 1

Answer:

$\huge\mathfrak{question}$

$(sin\theta + cos \theta + 1)(sin \theta \: - 1 + cos \theta)sec \theta \: cosec \theta \\ = 2 \\ \\ taking \: left \: hand \: side \\ (sin \theta + cos \theta + 1)(sin \theta + cos \theta - 1)sec \theta \: cosec \theta \\ = \: \bigg({(sin \theta + cos \theta)}^{2} - {1}^{2} \bigg)sec \theta \: cosec \: theta \: \\ \\ = (1 + 2sin \theta \: cos \theta - 1)sec \theta \: cosec \theta \\ \\ = 2sin \theta \: cos \theta \: \frac{1}{cos \theta} \frac{1}{sin \theta} \\ \\ = 2 \\ \\ hence \: proved$

Answer 2

Answer:

hence 0roved by getting 2