# Prove that (sinθ−cosθ+1)/(sinθ+cosθ−1)=1/(secθ−tanθ) using the identity sec^2 θ = 1 + tan^2 θ. - Mathematics

#### Question

Prove that  \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta } using the identity sec2 θ = 1 + tan2 θ.

#### Solution

LHS=\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta-1}=\frac{\tan \theta -1+\sec \theta }{\tan \theta +1-\sec \theta }

=\frac{(\tan \theta +\sec \theta )-1}{(\tan \theta -\sec \theta )+1}

=\frac{\{(\tan \theta +\sec \theta )-1\}(\tan \theta -\sec \theta)}{\{(\tan \theta -\sec \theta )+1\}(\tan \theta -\sec \theta )}

=\frac{(\tan ^{2}\theta -\sec ^{2}\theta )-(\tan \theta -\sec\theta )}{\{\tan \theta -\sec \theta +1\}(\tan \theta -\sec \theta )}

=\frac{-1-\tan \theta +\sec \theta }{(\tan \theta -\sec \theta+1)(\tan \theta -\sec \theta )}

=\frac{-1}{\tan \theta -\sec \theta }=\frac{1}{\sec \theta -\tan \theta }

which is the RHS of the identity, we are required to prove.

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Prove that (sinθ−cosθ+1)/(sinθ+cosθ−1)=1/(secθ−tanθ) using the identity sec^2 θ = 1 + tan^2 θ. Concept: Trigonometric Identities.