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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. - CBSE Class 10 - Mathematics

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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.

  Is there an error in this question or solution?
Solution Prove that (sinθ−cosθ+1)/(sinθ+cosθ−1)=1/(secθ−tanθ) using the identity sec^2 θ = 1 + tan^2 θ. Concept: Trigonometric Identities.
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