Question

prove that sin into 90 degree minus theta divided by 1 + sin theta + cos theta divided by 1 minus Cos into 90 minus theta is equal to 2 sec theta ​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\cos \: theta \div \sec \: theta + \sin \: theta \div \csc \: theta \: \\ \cos \: theta \div 1 \div \csc \: theta + \sin \: theta \: \div 1 \div \sin \: theta \: \\ \cos \: theta \times \cos \: theta \div 1 + \sin \: theta \: \times \sin \: theta \div 1 \\ = 1 \\ \cos {}^{2} theta + sin {}^{2} theta \: \\ = 1$

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