If ^@ sec\theta + tan\theta = y, ^@ simplify ^@ \dfrac{ y^2 -1 } { y^2 + 1 } ^@ in terms of ^@ \theta. ^@


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Answer:

^@ sin\theta ^@

Step by Step Explanation:
  1. ^@\begin{align} \dfrac{ y^2 -1 } { y^2 + 1 } & = \dfrac{ (\sec \theta + \tan \theta)^2 - 1 } { (\sec \theta + \tan \theta)^2 + 1 } \\ & = \dfrac{ \sec^2 \theta + \tan^2 \theta + 2\tan \theta \sec \theta - 1 } { \sec^2 \theta + \tan^2 \theta + 2\tan \theta \sec \theta + 1 } \\ & = \dfrac{ (\sec^2 \theta -1 ) + \tan^2 \theta + 2\tan \theta \sec \theta } { \sec^2 \theta + (1 + \tan^2 \theta) + 2\tan \theta \sec \theta } \\ & = \dfrac{ 2\tan^2 \theta + 2\tan \theta \sec \theta } { 2\sec^2 \theta + 2\tan \theta \sec \theta } \\ & = \dfrac{ 2\tan \theta ( \tan \theta + \sec \theta) } { 2\sec \theta (\tan \theta + \sec \theta) } \\ & = \sin \theta \end{align}^@

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