Question

# Prove that: $$(\sec \theta +\tan \theta-1) (\sec \theta -\tan \theta+1)=2\tan \theta$$

Solution

## $$\begin{array} { *{ 20 }{ l } }{ L.H.S=\left( { \sec \theta +\tan \theta -1 } \right) \left( { \sec \theta -\tan \theta +1 } \right) } \\ { =\left( { \dfrac { 1 }{ { \cos \theta } } +\frac { { \sin \theta } }{ { \cos \theta } } -1 } \right) \left( { \dfrac { 1 }{ { \cos \theta } } -\frac { { \sin \theta } }{ { \cos \theta } } +1 } \right) } \\ { =\dfrac { { \left( { 1+\sin \theta -cos\theta } \right) \left( { 1-\sin \theta +\cos \theta } \right) } }{ { { { \cos }^{ 2 } }\theta } } } \\ { =\dfrac { { 1-\sin \theta +\cos \theta +\sin \theta -si{ n^{ 2 } }\theta +\sin \theta \cos \theta -\cos \theta +\sin \theta \cos \theta -{ { \cos }^{ 2 } }\theta } }{ { { { \cos }^{ 2 } }\theta } } } \\ { =\dfrac { { 1-\left( { { { \sin }^{ 2 } }\theta +{ { \cos }^{ 2 } }\theta } \right) +2{ { \sin }^{ 2 } }\theta \cos \theta } }{ { { { \cos }^{ 2 } }\theta } } } \\ { =\dfrac { { 1-1+2\sin \theta cos\theta } }{ { { { \cos }^{ 2 } }\theta } } } \\ { =2\tan \theta } \\ { =R.H.S } \end{array}$$ Mathematics

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