Question

sec theta + tan theta is equal to p so prove that sin theta = p^2-1/p^2+1​

Answer 1

Step-by-step explanation:

$sec \: \theta + tan \: \theta = p \\ \\ \therefore \: \frac{1}{cos \: \theta} + \frac{sin\: \theta}{cos \: \theta} = p \\ \\ \therefore \: \frac{1 + sin\: \theta}{cos \: \theta} = p \\ \\ \therefore \: (1 + sin\: \theta) = p \: cos \: \theta \\ \\ \therefore \: (1 + sin \: \theta) ^{2}= p \: cos ^{2} \: \theta \\ \\ \therefore \: (1 + sin \: \theta) ^{2}= p^{2} \: (1 - sin ^{2} \: \theta) \\ \\ \therefore \: (1 + sin \: \theta) ^{2}= p^{2} \: (1 - sin \: \theta) (1 + sin \: \theta) \\ \\ \therefore \: (1 + sin \: \theta) = p^{2} \: (1 - sin \: \theta) \\ \\ \therefore \: 1 + sin \: \theta =p^{2} - p^{2} sin \: \theta \\ \\ \therefore \: (1 + p^{2}) sin \: \theta = p^{2} - 1 \\ \\ sin \: \theta = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1}$