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.

Answer 1

`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.